Rd Sharma 2020 2021 Solutions for Class 9 Maths Chapter 13 Quadrilaterals are provided here with simple step-by-step explanations. These solutions for Quadrilaterals are extremely popular among Class 9 students for Maths Quadrilaterals Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2020 2021 Book of Class 9 Maths Chapter 13 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2020 2021 Solutions. All Rd Sharma 2020 2021 Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

#### Page No 13.19:

#### Question 1:

Two opposite angles of a parallelogram are (3*x* − 2)° and (50 − *x*)°. Find the measure of each angle of the parallelogram.

#### Answer:

It is given that the two opposite angles of a parallelogram are and .

We know that the opposite angles of a parallelogram are equal.

Therefore,

…… (i)

Thus, the given angles become

Also, .

Therefore the sum of consecutive interior angles must be supplementary.

That is;

Since opposite angles of a parallelogram are equal.

Therefore,

And

Hence the four angles of the parallelogram are , , and .

#### Page No 13.20:

#### Question 2:

If an angle of a parallelogram is two-third of its adjacent angle, find the angles of the parallelogram.

#### Answer:

Let one of the angle of the parallelogram as

Then the adjacent angle becomes

We know that the sum of adjacent angles of the parallelogram is supplementary.

Therefore,

Thus, the angle adjacent to

Since, opposite angles of a parallelogram are equal.

Therefore, the four angles in sequence are ,,and.

#### Page No 13.20:

#### Question 3:

Find the measure of all the angles of a parallelogram, if one angle is 24° less than twice the smallest angle.

#### Answer:

Let the smallest angle of the parallelogram be

Therefore, according to the given statement other angle becomes .

Also, the opposite angles of a parallelogram are equal.

Therefore, the four angles become ,,and.

According to the angle sum property of a quadrilateral:

Thus, the other angle becomes

Hence, the four angles of the parallelogram are , , and .

#### Page No 13.20:

#### Question 4:

The perimeter of a parallelogram is 22 cm. If the longer side measures 6.5 cm what is the measure of the shorter side?

#### Answer:

Let the shorter side of the parallelogram be *cm*.

The longer side is given as*cm**.*

Perimeter of the parallelogram is given as 22 *cm*

Therefore,

Hence, the measure of the shorter side is * cm.*

#### Page No 13.20:

#### Question 5:

In a parallelogram *ABCD*, ∠*D* = 135°, determine the measures of ∠*A* and ∠*B*.

#### Answer:

It is given that *ABCD* is a parallelogram with

We know that the opposite angles of the parallelogram are equal.

Therefore,

Also, and are adjacent angles, which must be supplementary.

Therefore,

Hence , and .

#### Page No 13.20:

#### Question 6:

*ABCD* is a parallelogram in which ∠*A* = 70°. Compute ∠*B*, ∠*C* and ∠*D*.

#### Answer:

It is given that *ABCD* is a parallelogram with

We know that the opposite angles of the parallelogram are equal.

Therefore,

Also, and are adjacent angles, which must be supplementary.

Therefore,

Also, and are opposite angles of a parallelogram.

Therefore,

Hence, the angles of a parallelogram are , , and .

#### Page No 13.20:

#### Question 7:

In the given figure, *ABCD* is a parallelogram in which ∠*DAB* = 75° and ∠*DBC* = 60°. Compute ∠*CDB* and ∠*ADB*.

#### Answer:

The figure is given as follows:

It is given that *ABCD* is a parallelogram.

Thus

And are alternate interior opposite angles.

Therefore,

…… (i)

We know that the opposite angles of a parallelogram are equal. Therefore,

Also, we have

Therefore,

…… (ii)

In

By angle sum property of a triangle.

From (i) and (ii),we get:

Hence, the required value for is

And is .

#### Page No 13.20:

#### Question 8:

Which of the following statements are true (T) and which are false (F)?

(i) In a parallelogram, the diagonals are equal.

(ii) In a parallelogram, the diagonals bisect each other.

(iii) In a parallelogram, the diagonals intersect each other at right angles.

(iv) In any quadrilateral, if a pair of opposite sides is equal, it is a parallelogram.

(v) If all the angles of a quadrilateral are equal, it is a parallelogram.

(vi) If three sides of a quadrilateral are equal, it is a parallelogram.

(vii) If three angles of a quadrilateral are equal, it is a parallelogram.

(viii) If all the sides of a quadrilateral are equal it is a parallelogram.

#### Answer:

(i) Statement: In a parallelogram, the diagonals are equal.

False

(ii) Statement: In a parallelogram, the diagonals bisect each other.

True

(iii) Statement: In a parallelogram, the diagonals intersect each other at right angles.

False

(iv) Statement: In any quadrilateral, if a pair of opposite sides is equal, it is a parallelogram.

False

(v) Statement: If all the angles of a quadrilateral are equal, then it is a parallelogram.

True

(vi) Statement: If three sides of a quadrilateral are equal, then it is not necessarily a parallelogram.

False

(vii) Statement: If three angles of a quadrilateral are equal, then it is no necessarily a parallelogram.

False

(viii) Statement: If all sides of a quadrilateral are equal, then it is a parallelogram.

True

#### Page No 13.20:

#### Question 9:

In the given figure, *ABCD* is a parallelogram in which ∠*A* = 60°. If the bisectors of ∠*A** *and ∠*B* meet at P, prove that *AD* = *DP*, *PC* = *BC* and *DC* = 2*AD*.

#### Answer:

The figure is given as follows:

It is given that *ABCD* is a parallelogram.

Thus,

Opposite angles of a parallelogram are equal.

Therefore,

Also, we have *AP* as the bisector of

Therefore,

…… (i)

Similarly,

…… (ii)

We have ,

From (i)

Thus, sides opposite to equal angles are equal.

Similarly,

From (ii)

Thus, sides opposite to equal angles are equal.

Also,

#### Page No 13.20:

#### Question 10:

In the given figure, *ABCD* is a parallelogram and *E* is the mid-point of side *BC*. If *DE* and *AB* when produced meet at *F*, prove that *AF* = 2*AB*.

#### Answer:

Figure is given as follows:

It is given that *ABCD* is a parallelogram.

*DE* and *AB* when produced meet at *F*.

We need to prove that

It is given that

Thus, the alternate interior opposite angles must be equal.

In and , we have

(Proved above)

(Given)

(Vertically opposite angles)

Therefore,

(By ASA Congruency )

By corresponding parts of congruent triangles property, we get

DC = BF …… (i)

It is given that *ABCD* is a parallelogram. Thus, the opposite sides should be equal. Therefore,

…… (ii)

But,

From (i), we get:

From (ii), we get:

Hence proved.

#### Page No 13.4:

#### Question 1:

Three angles of a quadrilateral are respectively equal to 110°, 50° and 40°. Find its fourth angles.

#### Answer:

Let the measure of the fourth angles be *x*°. We know that the sum of the angles of a quadrilateral is 360°.

Therefore,

Hence the measure of the fourth angle is .

#### Page No 13.4:

#### Question 2:

In a quadrilateral *ABCD*, the angles *A*, *B*, *C* and *D* are in the ratio 1 : 2 : 4 : 5. Find the measure of each angles of the quadrilateral.

#### Answer:

We have ,.

So, let ,

,

And

By angle sum property of a quadrilateral, we get:

Also,

And

Similarly,

Hence, the four angles are ,,and .

#### Page No 13.4:

#### Question 3:

The angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.

#### Answer:

We have, .

So, let ,

,

and

By angle sum property of a quadrilateral, we get:

Also,

And

Similarly,

Hence, the four angles are , , and .

#### Page No 13.4:

#### Question 4:

In a quadrilateal *ABCD*, *CO* and *DO* are the bisectors of ∠*C* and ∠*D* respectively. Prove that ∠*COD* =$\frac{1}{2}$ (∠*A* + ∠*B*).

#### Answer:

The quadrilateral can be drawn as follows:

We have CO and DO as the bisectors of angles and respectively.

We need to prove that.

In ,We have,

…… (I)

By angle sum property of a quadrilateral, we have:

Putting in equation (I):

Hence proved.

#### Page No 13.42:

#### Question 1:

In a parallelogram *ABCD*, determine the sum of angles ∠*C* and ∠*D*.

#### Answer:

The parallelogram can be drawn as:

We have ,thus and are consecutive interior angles.

These must be supplementary.

Therefore,

#### Page No 13.42:

#### Question 2:

In a parallelogram *ABCD*, if ∠*B* = 135°, determine the measure of its other angles.

#### Answer:

Since *ABCD* is a parallelogram with .

Opposite angles of a parallelogram are equal.

Therefore,

Also, let

Similarly,

We know that the sum of the angles of a quadrilateral is .

Hence the measure of other angles are , and .

#### Page No 13.42:

#### Question 3:

*ABCD* is a square. *AC* and *BD* intersect at *O*. State the measure of ∠*AOB*.

#### Answer:

The figure can be drawn as follows:

In and,

(Sides of a square are equal)

(Diagonals of a parallelogram bisect each other)

(Common)

So, by SSS Congruence rule, we have

Also,

(Corresponding parts of congruent triangles are equal)

But, (Linear pairs)

We have,

Hence, the required measure of is .

#### Page No 13.42:

#### Question 4:

*ABCD* is a rectangle with ∠*ABD* = 40°. Determine ∠*DBC*.

#### Answer:

The rectangle is given as follows with

We have to find .

An angle of a rectangle is equal to .

Therefore,

Hence, the measure for is .

#### Page No 13.42:

#### Question 5:

The sides *AB* and CD of a parallelogram *ABCD* are bisected at *E* and *F*. Prove that *EBFD* is a parallelogram.

#### Answer:

Figure is given as follows:

It is given that *ABCD* is a parallelogram.

*E *is the mid point of *AB*

Thus,

,

...... (i)

Similarly,

……(ii)

From (i) and (ii)

Also,

Thus,

Therefore, *EBFD* is a parallelogram.

#### Page No 13.42:

#### Question 6:

*P* and *Q* are the points of trisection of the diagonal *BD* of a parallelogram *ABCD*. Prove that *CQ* is parallel to *AP*. Prove also that *AC* bisects *PQ*.

#### Answer:

Figure can be drawn as follows:

We have *P* and *Q* as the points of trisection of the diagonal *BD* of parallelogram ABCD.

We need to prove that *AC* bisects *PQ.* That is, .

Since diagonals of a parallelogram bisect each other.

Therefore, we get:

and

*P* and *Q* as the points of trisection of the diagonal *BD.*

Therefore*,*

and

Now, and

Thus,

AC bisects PQ.

Hence proved.

#### Page No 13.43:

#### Question 7:

*ABCD* is a square *E*, *F*, *G* and *H* are points on *AB*, *BC*, *CD* and *DA* respectively. such that *AE* = *BF* = *CG* = *DH*. Prove that *EFGH* is a square.

#### Answer:

Square *ABCD* is given:

*E, F, G* and *H* are the points on *AB, BC, CD* and *DA* respectively, such that :

We need to prove that *EFGH* is a square.

Say,

As sides of a square are equal. Then, we can also say that:

In and ,we have:

(Given)

(Each equal to 90°)

(Each equal to *y* )

By SAS Congruence criteria, we have:

Therefore, EH = EF

Similarly, EF= FG, FG= HG and HG= HE

Thus, HE=EF=FG=HG

Also,

and

But,

and

Therefore,

i.e.,

Similarly,

Thus,* EFGH* is a square.

Hence proved.

#### Page No 13.43:

#### Question 8:

*ABCD* is a rhombus, *EABF* is a straight line such that *EA* = *AB* = *BF*. Prove that *ED* and *FC* when produced meet at right angles.

#### Answer:

Rhombus *ABCD* is given:

We have

We need to prove that

We know that the diagonals of a rhombus bisect each other at right angle.

Therefore,

,,

In *A* and *O *are the mid-points of *BE* and *BD* respectively.

By using mid-point theorem, we get:

Therefore,

In *A* and *O *are the mid-points of *BE* and *BD* respectively.

By using mid-point theorem, we get:

Therefore,

Thus, in quadrilateral *DOCG*,we have:

and

Therefore,* DOCG *is a parallelogram.

Thus, opposite angles of a parallelogram should be equal.

Also, it is given that

Therefore,

Or,

Hence proved.

#### Page No 13.43:

#### Question 9:

*ABCD* is a parallelogram, *AD* is produced to *E* so that *DE* = *DC* and *EC* produced meets *AB* produced in F. Prove that *BF* = *BC*

#### Answer:

*ABCD *is a parallelogram, *AD* produced to *E *such that .

Also , *AB* produced to *F.*

We need to prove that

In , *D* and *O* are the mid-points of *AE* and *AC* respectively.

By using Mid-point Theorem, we get:

Since, *BD* is a straight line and *O *lies on *AC*.

And, *C* lies on *EF*

Therefore,

…… (i)

Also, is a parallelogram with .

Thus,

In and ,we have:

So, by ASA Congruence criterion, we have:

By corresponding parts of congruence triangles property, we get:

From (i) equation, we get:

Hence proved.

#### Page No 13.62:

#### Question 1:

In a Δ*ABC*, *D*, *E* and *F* are, respectively, the mid-points of *BC*, *CA* and *AB*. If the lengths of side *AB*, *BC* and *CA* are 7 cm, 8 cm and 9 cm, respectively, find the perimeter of Δ*DEF*.

#### Answer:

is given with *D,E* and *F *as the mid-points of *BC , CA* and *AB* respectively as shown below:

Also, , and .

We need to find the perimeter of

In , *E *and *F *are the mid-points of *CA* and *AB* respectively.

Theorem states, the line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.

Therefore, we get:

Similarly, we get

And

Perimeter of

Hence, the perimeter of is .

#### Page No 13.62:

#### Question 2:

In a triangle ∠*ABC*, ∠*A* = 50°, ∠*B* = 60° and *C* =∠70°. Find the measures of the angles of the triangle formed by joining the mid-points of the sides of this triangle.

#### Answer:

It is given that *D, E* and *F *be the mid-points of *BC , CA* and *AB* respectively.

Then,

, and .

Now, and transversal *CB* and *CA* intersect them at *D* and *E* respectively.

Therefore,

[(Given)]

and

[(Given)]

Similarly,

Therefore,

[(Given)]

and

[(Given)]

Similarly,

Therefore,

[(Given)] and

[(Given)]

Now *BC* is a straight line.

Similarly,

and

Hence the measure of angles are , and.

#### Page No 13.63:

#### Question 3:

In a triangle, *P*, *Q* and *R* are the mid-points of sides *BC*, *CA* and *AB* respectively. If *AC* = 21 cm, *BC* = 29 cm and *AB* = 30 cm, find the perimeter of the quadrilateral *ARPQ*.

#### Answer:

It is given that *P, Q* and *R *are the mid-points of *BC, CA* and *AB* respectively.

Also, we have , and

We need to find the perimeter of quadrilateral *ARPQ*

In , *P *and *R *are the mid-points of *CB* and *AB* respectively.

Theorem states, the line segment joining the mid-points of any two sides of a traingle is parallel to the third side and equal to half of it.

Therefore, we get:

Similarly, we get

We have *Q* and *R* as the mid points of *AC* and *AB* respectively.

Therefore,

And

Perimeter of

Hence, the perimeter of quadrilateral *ARPQ *is.

#### Page No 13.63:

#### Question 4:

In a Δ*ABC* median *AD* is produced to *X *such that *AD* = *DX*. Prove that *ABXC* is a parallelogram.

#### Answer:

is given with *AD* as the median extended to point *X* such that .

Join *BX* and *CX*.

We get a quadrilateral ABXC, we need to prove that it’s a parallelogram.

We know that *AD* is the median.

By definition of median we get:

Also, it is given that

Thus, the diagonals of the quadrilateral *ABCX* bisect each other.

Therefore, quadrilateral *ABXC *is a parallelogram.

Hence proved.

#### Page No 13.63:

#### Question 5:

In a *ABC*, *E* and *F* are the mid-points of *AC* and *AB* respectively. The altitude *AP* to *BC* intersects *FE* at *Q*. Prove that *AQ* = *QP*.

#### Answer:

is given with *E* and *F* as the mid points of sides *AB* and *AC.*

Also, intersecting *EF* at *Q*.

We need to prove that

In , *E* and *F *are the mid-points of *AB* and *AC* respectively.

Theorem states, the line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.

Therefore, we get:

Since, *Q* lies on *EF*.

Therefore,

This means,

*Q* is the mid-point of *AP*.

Thus, (Because, F is the mid point of AC and)

Hence proved.

#### Page No 13.63:

#### Question 6:

In a Δ*ABC*, *BM* and *CN* are perpendiculars from *B* and *C* respectively on any line passing through *A*. If* L* is the mid-point of *BC*, prove that *ML* =* NL*.

#### Answer:

In , *BM* and *CN* are perpendiculars on any line passing through *A*.

Also.

We need to prove that

From point *L* let us draw

It is given that , and

Therefore,

Since, L is the mid points of BC,

Therefore intercepts made by these parallel lines on MN will also be equal

Thus,

Now in ,

And . Thus, perpendicular bisects the opposite sides.

Therefore, is isosceles.

Hence

Hence proved.

#### Page No 13.63:

#### Question 7:

In the given figure, triangle *ABC* is right-angled at *B*. Given that *AB* = 9 cm, *AC* = 15 cm and *D*, *E *are the mid-points of the sides *AB *and *AC *respectively, calculate

(i) The length of *BC*

(ii) The area of Δ*ADE*

#### Answer:

We have right angled at B.

It is given that and

*D* and *E* are the mid-points of sides *AB* and *AC* respectively.

(i) We need to calculate length of *BC.*

In right angled at B:

By Pythagoras theorem,

Hence the length of *BC* is .

(ii) We need to calculate area of *.*

In right angled at *B*, *D* and *E *are the mid-points of *AB* and *AC *respectively.

Theorem states, the line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.

Therefore, .

Thus, (Corresponding angles of parallel lines are equal)

And

area of

*D* is the mid-point of side *AB* .

Therefore, area of

Hence the area of * *is .

#### Page No 13.63:

#### Question 8:

In the given figure, *M*, *N* and *P* are the mid-points of *AB*, *AC* and *BC* respectively. If *MN* = 3 cm, *NP* = 3.5 cm and *MP* = 2.5 cm, calculate *BC*, *AB* and *AC*.

#### Answer:

We have as follows:

*M, N* and *P* are the mid-points of sides *AB ,AC *and* BC* respectively.

Also, , and

We need to calculate *BC, AB* and *AC.*

In ,* M* and *N *are the mid-points of *AB* and *AC *respectively.

Therefore,

Similarly,

And

Hence, the measure for *BC, AB* and *AC* is , and respectively*.*

#### Page No 13.63:

#### Question 9:

In the given figure, *AB* = *AC* and *CP* || *BA* and *AP* is the bisector of exterior ∠*CAD* of Δ*ABC*. Prove that (i) ∠*PAC* = ∠*BCA* (ii) *ABCP* is a parallelogram.

#### Answer:

We have the following given figure:

We have and and *AP* is the bisector of exterior angle of .

(i) We need to prove that

In ,

We have (Given)

Thus, (Angles opposite to equal sides are equal)

By angle sum property of a triangle, we get:

…… (i)

Now,

(*AP* is the bisector of exterior angle )

(Linear Pair)

…… (ii)

From equation (i) and (ii),we get:

(ii) We need to prove that is a parallelogram.

We have proved that

This means,

Also it is given that

We know that a quadrilateral with opposite sides parallel is a parallelogram.

Therefore, is a parallelogram.

#### Page No 13.63:

#### Question 10:

*ABCD* is a kite having *AB *= *AD* and *BC* = *CD*. Prove that the figure formed by joining the mid-points of the sides, in order, is a rectangle.

#### Answer:

ABCD is a kite such that and

Quadrilateral *PQRS* is formed by joining the mid-points *P,Q,R* and *S *of sides *AB,BC,CD* and *AD* respectively.

We need to prove that Quadrilateral *PQRS *is a rectangle*.*

In , *P* and *Q* are the mid-points of *AB* and *BC* respectively.

Therefore,

and

Similarly, we have

and

Thus,

and

Therefore, *PQRS *is a parallelogram.

Now,

But, *P* and *S* are the mid-points of *AB* and *AD*

…… (I)

In :* P *and* S *are the mid-point* *of side *AB *and* AD*

By mid-point Theorem, we get:

Or,

In ,* P *is the mid-point* *of side *AB* and

By Using the converse of mid-point theorem, we get:

*M* is the mid-point of *AO*

Thus*,*

…… (II)

In and , we have:

(Common)

[From (I)]

[From (II)]

By SSS Congruence theorem, we get:

By corresponding parts of congruent triangles property, we get:

But,

and

Therefore,

(, Corresponding angles should be equal)

Or,

We have proved that

Similarly, .

Then we can say that and

Therefore, is a parallelogram with

Or, we can say that is a rectangle.

We get:

Also,* PQRS *is a parallelogram.

Therefore,* PQRS *is a rectangle.

Hence proved.

#### Page No 13.64:

#### Question 11:

Let *ABC* be an isosceles triangle in which *AB* = *AC.* If *D*, *E*, *F* be the mid-points of the sides BC, *CA* and *AB* respectively, show that the segment *AD* and *EF* bisect each other at right angles.

#### Answer:

, an isosceles triangle is given with *D,E* and *F *as the mid-points of *BC, CA* and *AB* respectively as shown below:

We need to prove that the segment *AD* and *EF* bisect each other at right angle.

Let’s join *DF* and *DE*.

In , *D* and *E *are the mid-points of *BC* and *AC* respectively.

Therefore, we get: Or

Similarly, we can get

Therefore, *AEDF* is a parallelogram

We know that opposite sides of a parallelogram are equal.

and

Also, from the theorem above we get

Thus,

Similarly,

It is given that , an isosceles triangle

Thus,

Therefore,

Also,

Then, *AEDF* is a rhombus.

We know that the diagonals of a rhombus bisect each other at right angle.

Therefore, *M* is the mid-point of *EF *and* *

Hence proved.

#### Page No 13.64:

#### Question 12:

Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

#### Answer:

Figure can be drawn as:

Let *ABCD* be a quadrilateral such that *P,Q ,R* and *S* are the mid-points of side *AB,BC,CD* and *DA* respectively.

In , *P* and *Q* are the mid-points of *AB* and *BC* respectively.

Therefore,

and

Similarly, we have

and

Thus,

and

Therefore, *PQRS *is a parallelogram.

Since, diagonals of a parallelogram bisect each other.

Therefore, *PR* and *QS* bisect each other.

Hence proved.

#### Page No 13.64:

#### Question 13:

Fill in the blanks to make the following statements correct:

(i) The triangle formed by joining the mid-points of the sides of an isosceles triangle is........

(ii) The triangle formed by joining the mid-points of the sides of a right triangle is ........

(iii) The figure formed by joining the mid-points of consecutive sides of a quadrilateral is ..........

#### Answer:

(i) The triangle formed by joining the mid-points of the sides of an isosceles triangle is isosceles.

Explanation:

Figure can be drawn as: A

, an isosceles triangle is given.

*F* and *E* are the mid-points of *AB* and *AC* respectively.

Therefore,

…… (I)

Similarly,

…… (II)

And

…… (III)

Now, is an isosceles triangle.

From equation (II) and (III), we get:

Therefore, in two sides are equal.

Therefore, it is an isosceles triangle.

(ii) The triangle formed by joining the mid-points of the sides of a right triangle is right triangle.

Explanation:

Figure can be drawn as: A

right angle at *B* is given.

*F* and *E* are the mid-points of *AB* and *AC* respectively.

Therefore,

…… (I)

Similarly,

…… (II)

And

…… (III)

Now, *DE || AB* and transversal *CB* and *CA* intersect them at *D* and *E* respectively.

Therefore,

and

Similarly,

Therefore,

and

Similarly,

Therefore,

Now *AC* is a straight line.

Now, by angle sum property of ,we get:

Therefore,

But,

Then we have:

(iii) The figure formed by joining the mid-points of the consecutive sides of a quadrilateral is **parallelogram**.

Explanation:

Figure can be drawn as:

Let *ABCD* be a quadrilateral such that *P, Q, R* and *S* are the mid-points of side *AB, BC, CD* and *DA* respectively.

In , *P* and *Q* are the mid-points of *AB* and *BC* respectively.

Therefore,

and

Similarly, we have

and

Thus,

and

Therefore, *PQRS *is a parallelogram.

#### Page No 13.64:

#### Question 14:

ABC is a triangle and through A, B, C lines are drawn parallel to BC, CA and AB respectively intersecting at P, Q and R. Prove that the perimeter of ΔPQR is double the perimeter of ΔABC.

#### Answer:

We have as follows:

Through *A*,*B* and *C* lines are drawn parallel to *BC,CA* and *AB* respectively intersecting at *P,Q* and *R* respectively.

We need to prove that perimeter of is double the perimeter of .

and

Therefore, is a parallelogram.

Thus,

Similarly,

is a parallelogram.

Thus,

Therefore,

Then, we can say that *A* is the mid-point of *QR*.

Similarly, we can say that *B* and* C *are the mid-point of *PR *and* PQ *respectively.

In ,

Theorem states, the line drawn through the mid-point of any one side of a triangle is parallel to the another side, intersects the third side at its mid-point.

Therefore,

Similarly,

Perimeter of is double the perimeter of

Hence proved.

#### Page No 13.64:

#### Question 15:

In the given figure, *BE* ⊥ *AC*. *AD* is any line from *A* to *BC* intersecting BE in *H*.* P,* *Q* and* R* are respectively the mid-points of *AH*,* AB* and *BC*. Prove that ∠*PQR* = 90°

#### Answer:

is given with

*AD* is any line from *A* to *BC* intersecting *BE* in *H*.

*P,Q* and *R *respectively are the mid-points of *AH,AB* and *BC*.

We need to prove that

Let us extend *QP* to meet *AC* at *M*.

In , *R *and* Q *are the mid-points of *BC* and *AB* respectively.

Therefore, we get:

…… (i)

Similarly, in,

…… (ii)

From (i) and (ii),we get:

and

We get, is a parallelogram.

Also,

Therefore, is a rectangle.

Thus,

Or,

Hence proved.

#### Page No 13.64:

#### Question 16:

*ABC* is a triangle, *D* is a point on *AB* such that *AD* = $\frac{1}{4}$ *AB* and *E* is a point on *AC* such that *AE *= $\frac{1}{4}$ *AC*. Prove that *DE* = $\frac{1}{4}$ *BC*.

#### Answer:

is given with *D* a point on *AB* such that .

Also, *E* is point on *AC* such that.

We need to prove that

Let *P* and *Q* be the mid points of *AB* and *AC* respectively.

It is given that

and

But, we have taken *P* and *Q* as the mid points of *AB* and *AC* respectively.

Therefore,* D *and *E* are the mid-points of *AP* and *AQ* respectively.

In , *P* and *Q* are the mid-points of *AB* and *AC* respectively.

Therefore, we get and …… (i)

In , *D *and *E* are the mid-points of *AP* and *AQ* respectively.

Therefore, we get and …… (ii)

From (i) and (ii),we get:

Hence proved.

#### Page No 13.64:

#### Question 17:

In the given figure, *ABCD* is a parallelogram in which *P* is the mid-point of *DC* and *Q* is a point on *AC* such that* CQ* = $\frac{1}{4}$ *AC*. If *PQ* produced meets *BC* at *R*, prove that *R* is a mid-point of *BC*.

#### Answer:

Figure is given as follows:

*ABCD* is a parallelogram, where *P* is the mid-point of *DC* and *Q* is a point on *AC* such that

.

*PQ* produced meets *BC* at *R*.

We need to prove that *R* is a mid-point of *BC.*

Let us join *BD* to meet *AC* at *O.*

It is given that *ABCD* is a parallelogram.

Therefore, (Because diagonals of a parallelogram bisect each other)

Also,

Therefore,

In , *P* and *Q *are the mid-points of *CD* and *OC* respectively.

Therefore, we get:

Also, in , Q is the mid-point of *OC *and* *

Therefore, *R* is a mid-point of *BC.*

Hence proved.

#### Page No 13.64:

#### Question 18:

In the given figure, *ABCD* and *PQRC* are rectangles and *Q* is the mid-point of *AC*. Prove that

(i) *DP* = *PC*

(ii) *PR* = $\frac{1}{2}$ *AC*

#### Answer:

Rectangles *ABCD* and *PQRC *are* *given* *as follows:

*Q* is the mid-point of *AC*.

In , *Q* is the mid-point of *AC* such that

Using the converse of mid-point theorem, we get:

*P *is the mid-point of *DC*

That is;

Similarly, *R* is the mid-point of *BC*.

Now, in , *P *and* R *are the mid-points of *DC *and* BC *respectively.

Then, by mid-point theorem, we get:

Now, diagonals of a rectangle are equal.

Therefore putting ,we get:

Hence Proved.

#### Page No 13.65:

#### Question 19:

*ABCD* is a parallelogram,* E* and *F* are the mid-points of *AB* and *CD* respectively. *GH* is any line intersecting *AD*, *EF* and *BC* at *G,* *P* and *H* respectively. Prove that *GP* = *PH*.

#### Answer:

*ABCD* is a parallelogram with *E *and *F* as the mid-points of *AB* and *CD* respectively.

We need to prove that

Since *E* and *F* are the mid-points of *AB* and *CD* respectively.

Therefore,

,

And

,

Also,* ABCD* is a parallelogram. Therefore, the opposite sides should be equal.

Thus,

Also, (Because )

Therefore, *BEFC *is a parallelogram

Then, and …… (i)

Now,

Thus, (Because as *ABCD* is a parallelogram)

We get,

*AEFD *is a parallelogram

Then, we get:

…… (ii)

But, *E* is the mid-point of *AB.*

Therefore,

Using (i) and (ii), we get:

Hence proved.

#### Page No 13.65:

#### Question 20:

*RM* and *CN* are perpendiculars to a line passing through the vertex *A* of a triangle *ABC*. IF *L* is the mid-point of *BC*, prove that* LM* = *LN*.

#### Answer:

In *BM* and *CN* are perpendiculars on any line passing through *A*.

Also.

We need to prove that

From point *L* let us draw

It is given that , and

Therefore,

Since, L is the mid points of BC,

Therefore intercepts made by these parallel lines on MN will also be equal

Thus,

Now in ,

And . Thus, perpendicular bisects the opposite sides.

Therefore, is isosceles.

Hence

Hence proved.

#### Page No 13.68:

#### Question 1:

*PQRS* is a quadrilateral, *PR* and *QS* intersect each other at *O*. In which of the following cases, *PQRS* is a parallelogram?

(a) ∠*P* = 100°, ∠*Q* = 80°, ∠*R* = 95°

(b) ∠*P* =85°, ∠*Q* = 85°, ∠*R* = 95°

(c) *PQ* = 7 cm, *QR* = 7 cm, *RS* = 8 cm, *SP* = 8 cm

(d) *OP* = 6.5 cm, *OQ* = 6.5 cm, *OR* = 5.2 cm, *OS* = 5.2 cm

#### Answer:

Let us analyze each case one by one.

We have a quadrilateral named *PQRS*, with diagonals PR and QS intersecting at *O*.

(a) ,,

By angle sum property of a quadrilateral, we get:

Clearly,

And

Thus we have *PQRS *a quadrilateral with opposite angles are equal.

Therefore,

*PQRS *is a parallelogram*.*

(b) ,,

By angle sum property of a quadrilateral, we get:

Clearly,

And

Thus we have *PQRS *a quadrilateral with opposite angles are not equal.

Therefore,

*PQRS *is not a parallelogram*.*

(c) ,,,

Clearly,

And

Thus we have *PQRS *a quadrilateral with opposite sides are not equal

Therefore,

*PQRS *is not a parallelogram*.*

(c) ,,,

We know that the diagonals of a parallelogram bisect each other.

But, here we have

And

Therefore,

*PQRS *is not a parallelogram*. *

Hence, the correct choice is (a).

#### Page No 13.68:

#### Question 2:

Mark the correct alternative in each of the following:

The opposite sides of a quadrilateral have

(a) no common point

(b) one common point

(c) two common points

(d) infinitely many common points

#### Answer:

We can look at a quadrilateral as:

The opposite sides of the above quadrilateral *AB* and *CD* have no point in common.

Hence the correct choice is (a).

#### Page No 13.68:

#### Question 3:

The consecutive sides of a quadrilateral have

(a) no common point

(b) one common point

(c) two common points

(d) infinitely many common points

#### Answer:

We can look at a quadrilateral as:

The consecutive sides of the above quadrilateral *AB* and *BC* have one point in common.

Hence the correct choice is (b).

#### Page No 13.69:

#### Question 4:

Which of the following quadrilateral is not a rhombus?

(a) All four sides are equal

(b) Diagonals bisect each other

(c) Diagonals bisect opposite angles

(d) One angle between the diagonals is 60°

#### Answer:

Let us consider the rhombus *ABCD* as:

We have the following properties of a rhombus:

All four sides are equal.

Diagonals bisect each other at right angles.

Hence the correct choice is (d).

#### Page No 13.69:

#### Question 5:

Diagonals necessarily bisect opposite angles in a

(a) rectangle

(b) parallelogram

(c) isosceles trapezium

(d) square

#### Answer:

From the given choices, only in a square the diagonals bisect the opposite angles.

Let us prove it.

Take the following square *ABCD* with diagonal *AD.*

In and:

(Opposite sides of a square are equal.)

(Common)

(Opposite sides of a square are equal.)

Thus,

(By SSS Congruence Rule)

By Corresponding parts of congruent triangles property we have:

Therefore, in a square the diagonals bisect the opposite angles.

Hence the correct choice is (d).

#### Page No 13.69:

#### Question 6:

The two diagonals are equal in a

(a) parallelogram

(b) rhombus

(c) rectangle

(d) trapezium

#### Answer:

Two diagonals are equal only in a rectangle.

This can be proved as follows:

The rectangle is given as *ABCD*, with the two diagonals as *AD* and *BC*.

In and:

(Opposite sides of a rectangle are equal.)

(Common)

(Each angle in a rectangle is a right angle)

Thus,

(By SAS Congruence Rule)

By Corresponding parts of congruent triangles property we have:

Therefore, in a rectangle the two diagonals are equal.

Hence the correct choice is (c).

#### Page No 13.69:

#### Question 7:

We get a rhombus by joining the mid-points of the sides of a

(a) parallelogram

(b) rhombus

(c) rectangle

(d) triangle

#### Answer:

We get a rhombus by joining the mid-points of the sides of a rectangle.

It is given a rectangle *ABCD* in which *P,Q,R* and *S* are the mid-points *AB*,*BC*,*CD* and *DA* respectively.

*PQ*,*QR,RS* and *SP* are joined.

In ,* P *and* Q* are the mid-points *AB* and *BC* respectively.

Therefore,

and ……(i)

Similarly, In ,* R *and* S* are the mid-points *CD* and *AD* respectively.

Therefore,

and ……(ii)

From (i) and (ii), we get

and

Therefore, is a parallelogram. …… (iii)

Now is a rectangle.

Therefore,

…… (iv)

In and , we have:

(*P* is the mid point of *AB*)

(Each is a right angle)

(From equation (iv))

So, by SAS congruence criteria, we get:

By Corresponding parts of congruent triangles property we have:

…… (v)

From (iii) and (v) we obtain that is a parallelogram such that and

Thus, the two adjacent sides are equal.

Thus, is a rhombus.

Hence the correct choice is (c).

#### Page No 13.69:

#### Question 8:

The bisectors of any two adjacent angles of a parallelogram intersect at

(a) 30°

(b) 45°

(c) 60°

(d) 90°

#### Answer:

Let the figure be as follows:

*ABCD *is a parallelogram.

We have to find

*AN* and *AD* is the bisectors of and .

Therefore,

…… (i)

And

…… (ii)

We know that .

Therefore, the sum of consecutive interior angles must be supplementary.

From (i) and (ii), we get

…… (iii)

By angle sum property of a triangle:

Hence the correct choice is (d).

#### Page No 13.69:

#### Question 9:

The bisectors of the angle of a parallelogram enclose a

(a) parallelogram

(b) rhombus

(c) rectangle

(d) square

#### Answer:

We have *ABCD*, a parallelogram given below:

Therefore, we have

Now, and transversal *AB* intersects them at *A* and *B* respectively. Therefore,

Sum of consecutive interior angle is supplementary. That is;

We have AR and BR as bisectors of and respectively.

…… (i)

Now, in , by angle sum property of a triangle, we get:

From equation (i), we get:

Similarly, we can prove that.

Now, and transversal *AD*intersects them at *A* and *D* respectively. Therefore,

Sum of consecutive interior angle is supplementary. That is;

We have AR and DP as bisectors of and respectively.

…… (ii)

Now, in , by angle sum property of a triangle, we get:

From equation (i), we get:

We know that and are vertically opposite angles, thus,

Similarly, we can prove that.

Therefore, *PQRS* is a rectangle.

Hence, the correct choice is (c).

#### Page No 13.69:

#### Question 10:

The figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a

(a) parallelogram

(b) rectangle

(c) square

(d) rhombus

#### Answer:

Figure can be drawn as:

Let *ABCD* be a quadrilateral such that *P,Q,R* and *S* are the mid-points of side *AB,BC,CD* and *DA* respectively.

In,*P* and *Q* are the mid-points of *AB* and *BC* respectively.

Therefore,

and

Similarly, we have

and

Thus,

and

Therefore, *PQRS*is a parallelogram.

The figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a

parallelogram.

Hence the correct choice is (a).

#### Page No 13.69:

#### Question 11:

The figure formed by joining the mid-points of the adjacent sides of a rectangle is a

(a) square

(b) rhombus

(c) trapezium

(d) none of these

#### Answer:

We get a rhombus by joining the mid-points of the sides of a rectangle.

It is given that rectangle *ABCD* in which *P,Q,R* and *S* are the mid-points *AB*,*BC*,*CD* and *DA* respectively.

*PQ*,*QR,RS* and *SP* are joined.

In ,* P *and* Q* are the mid-points *AB* and *BC* respectively.

Therefore,

and ……(i)

Similarly, In ,* R *and* S* are the mid-points *CD* and *AD* respectively.

Therefore,

and ……(ii)

From (i) and (ii), we get

and

Therefore, is a parallelogram. …… (iii)

Now is a rectangle.

Therefore,

…… (iv)

In and , we have:

(*P* is the mid point of *AB*)

(Each is a right angle)

(From equation (iv))

So, by SAS congruence criteria, we get:

By Corresponding parts of congruent triangles property we have:

…… (v)

From (iii) and (v) we obtain that is a parallelogram such that .

Thus, the two adjacent sides are equal.

Thus, is a rhombus .

Hence the correct choice is (b).

#### Page No 13.69:

#### Question 12:

The figure formed by joining the mid-points of the adjacent sides of a rhombus is a

(a) square

(b) rectangle

(c) trapezium

(d) none of these

#### Answer:

Figure is given as :

A rhombus *ABCD *is given in which *P, Q, R* and *S* are the mid-points of sides *AB, BC, CD* and *DA* respectively.

In ,* P *and* Q* are the mid-points *AB* and *BC *respectively.

Therefore,

and ……(i)

Similarly, In ,* R *and* S* are the mid-points *CD* and *AD* respectively.

Therefore,

and ……(ii)

From (i) and (ii), we get

and

Therefore, is a parallelogram. …… (iii)

Now, we shall find one of the angles of a parallelogram.

Since *ABCD *is a rhombus

Therefore,

(Sides of rhombus are equal)

(*P *and* Q* are the mid-points *AB* and *BC* respectively)

In , we have

(Angle opposite to equal sides are equal)

Therefore*, ABCD *is a rhombus

…… (iii)

Also,

…… (iv)

Now, in and , we have

[From (iii)]

[From (iv)]

And ( is a parallelogram)

So by SSS criteria of congruence, we have

By Corresponding parts of congruent triangles property we have:

…… (v)

Now,

And

Therefore,

From (ii), we get

From (v), we get

Therefore, …… (vi)

Now, transversal *PQ* cuts parallel lines *SP* and *RQ* at *P* and *Q* respectively.

[Using (vi)]

Thus, is a parallelogram such that .

Therefore, is a rectangle.

Hence the correct choice is (b).

#### Page No 13.69:

#### Question 13:

The figure formed by joining the mid-points of the adjacent sides of a square is a

(a) rhombus

(b) square

(c) rectangle

(d) parallelogram

#### Answer:

We get a square by joining the mid-points of the sides of a square.

It is given a square *ABCD* in which *P, Q, R* and *S* are the mid-points *AB*, *BC*, *CD* and *DA* respectively.

*PQ*, *QR, RS* and *SP* are joined.

Join *AC* and *BD*.

In ,* P *and* Q* are the mid-points *AB* and *BC* respectively.

Therefore,

and ……(i)

Similarly, In Δ*ADC*,* R *and* S* are the mid-points *CD* and *AD* respectively.

Therefore,

and ……(ii)

From (i) and (ii), we get

and

Therefore, is a parallelogram. …… (iii)

Now is a square.

Therefore,

…… (iv)

Similarly,

…… (v)

In and , we have:

(From equation (iv))

(Each is a right angle)

(From equation (v))

So, by SAS congruence criteria, we get:

By Corresponding parts of congruent triangles property we have:

…… (vi)

From (iii) and (vi) we obtain that is a parallelogram such that .

But, is a parallelogram.

So,

and …… (vii)

Now, (From equation (i))

Therefore,

…… (viii)

Since* P *and* S* are the mid-points *AB* and *AD* respectively

Therefore,

…… (ix)

Thus, in quadrilateral *PMON*, we have:

and (From equation (viii) and (ix))

Therefore, quadrilateral *PMON *is a parallelogram.

Also,

(Because )

(Because diagonals of a square are perpendicular)

Therefore, is a quadrilateral such that ,

and .Also, .

Hence, is a square.

Hence the correct choice is (b).

#### Page No 13.69:

#### Question 14:

The figure formed by joining the mid-points of the adjacent sides of a parallelogram is a

(a) rectangle

(b) parallelogram

(c) rhombus

(d) square

#### Answer:

It is given a parallelogram *ABCD* in which *P,Q,R* and *S* are the mid-points *AB*, *BC*, *CD* and *DA* respectively.

*PQ*, *QR, RS* and *SP* are joined.

In ,* P *and* Q* are the mid-points *AB* and *BC* respectively.

Therefore,

and ……(i)

Similarly, In ,* R *and* S* are the mid-points *CD* and *AD* respectively.

Therefore,

and ……(ii)

From (i) and (ii), we get

and

Therefore, is a parallelogram.

Hence the correct choice is (b).

#### Page No 13.69:

#### Question 15:

If one angle of a parallelogram is 24° less than twice the smallest angle, then the measure of the largest angle of the parallelogram is

(a) 176°

(b) 68°

(c) 112°

(d) 102°

#### Answer:

Let the smallest angle of the parallelogram be

Therefore, according to the given statement other angle becomes.

Also, the opposite angles of a parallelogram are equal.

Therefore, the four angles become ,,and.

According to the angle sum property of a quadrilateral:

Thus, the other angle becomes

Thus, the largest angle of the parallelogram are is .

Hence the correct choice is (c).

#### Page No 13.69:

#### Question 16:

In a parallelogram ABCD, if ∠DAB = 75° and ∠DBC = 60°, then ∠BDC =

(a) 75°

(b) 60°

(c) 45°

(d) 55°

#### Answer:

Parallelogram can be drawn as :

We know that the opposite angles of a parallelogram are equal.

Therefore,

By angle sum property of a triangle:

Thus, measures.

Hence the correct choice is (c).

#### Page No 13.69:

#### Question 17:

*ABCD* is a parallelogram and *E* and *F* are the centroids of triangles *ABD* and *BCD* respectively, then *EF* =

(a) *AE*

(b) *BE*

(c) *CE*

(d) *DE*

#### Answer:

Parallelogram *ABCD* is given with *E *and *F* are the centroids of and.

We have to find EF.

We know that the diagonals of a parallelogram of bisect each other.

Thus, *AC* and *BD* bisect each other at point *O*.

Also, median is the line segment joining the vertex to the mid-point of the opposite side of the triangle. Therefore, the centroids *E* and *F *lie on *AC.*

Now, the centroid divides each median into two segments whose lengths are in the ratio 2:1, with the longest one nearest the vertex.

Then, in Δ*ABD*, we get:

Or,

and …… (I)

Similarly, in ,we get:

and …… (II)

Also,

From (I) and (II), we get:

And …… (III)

Also, from (II) and (III), we get :

…… (IV)

Now, from (I),

From (IV), we get:

From(III):

Hence the correct choice is (a).

#### Page No 13.69:

#### Question 18:

ABCD is a parallelogram, M is the mid-point of BD and BM bisects ∠B. Then ∠AMB =

(a) 45°

(b) 60°

(c) 90°

(d) 75°

#### Answer:

Figure is given as follows:

*ABCD* is a parallelogram.

It is given that :

BM bisects

Therefore,

But,

(Alternate interior opposite angles as)

Therefore,

In,

Sides opposite to equal angles are equal.

Thus,

Also,

(Opposite sides of a parallelogram are equal)

Thus,

*ABCD* is a parallelogram with

Therefore,

*ABCD* is a rhombus.

And we know that diagonals of the rhombus bisect each other at right angle.

Thus,

Hence the correct choice is (c).

#### Page No 13.70:

#### Question 19:

If an angle of a parallelogram is two-third of its adjacent angle, the smallest angle of the parallelogram is

(a) 108°

(b) 54°

(c) 72°

(d) 81°

#### Answer:

Let one of the angle of the parallelogram as

Then the adjacent angle becomes

We know that the sum of adjacent angles of the parallelogram is supplementary.

Therefore,

Thus, the angle adjacent to

Therefore, the smallest angle of the parallelogram as

Hence, the correct choice is (c).

#### Page No 13.70:

#### Question 20:

If the degree measures of the angles of quadrilateral are 4x, 7x, 9x and 10x, what is the sum of the measures of the smallest angle and largest angle?

(a) 140°

(b) 150°

(c) 168°

(d) 180°

#### Answer:

We have,

,

,

and

By angle sum property of a quadrilateral, we get:

Smallest angle:

Also, Largest angle:

Thus, the sum of the smallest and the largest becomes:

Hence, the correct choice is (c).

#### Page No 13.70:

#### Question 21:

If the diagonals of a rhombus are 18 cm and 24 cm respectively, then its side is equal to

(a) 16 cm

(b) 15 cm

(c) 20 cm

(d) 17 cm

#### Answer:

Let *ABCD* be rhombus with diagonals *AC* and *BD* 18cm and 24cm respectively.

We know that diagonals of the rhombus bisect each other at right angles.

Therefore,

Similarly,

Also, is a right angled triangle.

By Pythagoras theorem, we get:

Hence the correct choice is (b).

#### Page No 13.70:

#### Question 22:

*ABCD* is a parallelogram in which diagonal *AC* bisects ∠*BAD*. If ∠*BAC* = 35°, then ∠*ABC* =

(a) 70°

(b) 110°

(c) 90°

(d) 120°

#### Answer:

*ABCD* is a parallelogram in which *AC* bisects.

It is given that

Therefore,

Since,

Therefore,

Hence, the correct choice is (b).

#### Page No 13.70:

#### Question 23:

In a rhombus ABCD, if ∠ACB = 40°, then ∠ADB =

(a) 70°

(b) 45°

(c) 50°

(d) 60°

#### Answer:

Rhombus ABCD is given as follows:

It is given that.

Therefore, (Because *O* lies on *AC*)

We know that the diagonals of a rhombus intersect at right angle.

Therefore,

By angle sum property of a triangle, we get:

Since, *O* lies on *BD*

Also ,

Therefore,

Hence, the correct choice is (c).

#### Page No 13.70:

#### Question 24:

In Δ*ABC*, ∠*A* = 30°, ∠*B* = 40° and ∠*C* = 110°. The angles of the triangle formed by joining the mid-points of the sides of this triangle are

(a) 70°, 70°, 40°

(b) 60°, 40°, 80°

(c) 30°, 40°, 110°

(d) 60°, 70°, 50°

#### Answer:

It is given that *D, E* and *F *be the mid-points of *BC , CA* and *AB* respectively.

Then,

, and .

Now, and transversal *CB* and *CA* intersect them at *D* and *E* respectively.

Therefore,

[(Given)]

and

[(Given)]

Similarly,

Therefore,

[(Given)]

and

[(Given)]

Similarly,

Therefore,

[ (Given)]

and

[ (Given)]

Now *BC* is a straight line.

Similarly,

and

Hence the correct choice is (c).

#### Page No 13.70:

#### Question 25:

The diagonals of a parallelogram *ABCD* intersect at *O*. If ∠*BOC* = 90° and ∠*BDC* = 50°, then ∠*OAB* =

(a) 40°

(b) 50°

(c) 10°

(d) 90°

#### Answer:

*ABCD* is a parallelogram with diagonals *AC* and *BD* intersect at *O*.

It is given that and.

We need to find

Now,

(Linear pair)

Since, *O* lies on *BD*.

Therefore,

By angle sum property of a triangle, we get:

Since, *O* lies on *AC*.

Therefore,

Also,

Therefore,

Hence the correct choice is (a).

#### Page No 13.70:

#### Question 26:

*ABCD* is a trapezium in which *AB* || *DC*. *M* and *N *are the mid-points of *AD* and the respectively. If *AB* = 12 cm, *MN* = 14 cm, then *CD* =

(a) 10 cm

(b) 12 cm

(c) 14 cm

(d) 16 cm

#### Answer:

The given trapezium *ABCD* can be drawn as follows:

Here, .

*M* and *N* are the mid-points of *AD* and *BC* respectively.

We have,.

We need to find *CD.*

Join* MN *to intersect* AC *at* G.*

We have and a line *MN* formed by joining the mid-points of sides *AD* and *BC.*

Thus, we can say that* *

In * M *is the mid-point of *AD *and* *

Therefore, *G* is the mid point of *AC*

By using the converse of mid-point theorem, we get:

…… (i)

In ,* N* is the mid point of *BC* and* *

By using the converse of mid-point theorem, we get:

…… (ii)

Adding (i) and (ii),we get:

…… (iii)

On substituting and in (iii),we get:

Hence the correct choice is (d).

#### Page No 13.70:

#### Question 27:

Diagonals of a quadrilateral *ABCD* bisect each other. If ∠*A*= 45°, then ∠*B* =

(i) 115°

(ii) 120°

(iii) 125°

(iv) 135°

#### Answer:

We know that the diagonals of a parallelogram bisect each other.

Thus, the given quadrilateral *ABCD* is a parallelogram.

and are consecutive interior angles, which must be supplementary.

Therefore, we have

Hence the correct choice is (d).

#### Page No 13.70:

#### Question 28:

*P* is the mid-point of side *BC* of a parallelogram *ABCD* such that ∠*BAP* = ∠*DAP*. If *AD* = 10 cm, then *CD* =

(a) 5 cm

(b) 6 cm

(c) 8 cm

(d) 10 cm

#### Answer:

Parallelogram *ABCD* is given such that

We haveand

We need to find the measure of side CD.

Since and *AP* as transversal

But it is given that

Therefore, we get:

Also, sides opposite to equal angles are equal.

Then,

…… (I)

Also,

Substituting in (I), We get:

It is given that , this means opposite side ,as* ABCD* is a parallelogram. Therefore,

Or,

Hence the correct choice is (a).

#### Page No 13.70:

#### Question 29:

In Δ*ABC*, *E* is the mid-point of median *AD* such that *BE* produced meets *AC* at *F*. *IF* *AC* = 10.5 cm, then *AF* =

(a) 3 cm

(b) 3.5 cm

(c) 2.5 cm

(d) 5 cm

#### Answer:

is given with *E* as the mid-point of median *AD*.

Also, *BE* produced meets *AC* at *F*.

We have , then we need to find *AF.*

Through *D* draw* *.

In , *E* is the mid-point of *AD* and .

Using the converse of mid-point theorem, we get:

…… (i)

In ,* D* is the mid-point of *BC* and .

…… (ii)

From (i) and (ii),we have:

, …… (iii)

Now,

From (iii) equation, we get:

…… (iv)

We have .Putting this in equation (iv), we get:

Hence the correct choice is (b).

#### Page No 13.70:

#### Question 30:

ABCD is a parallelogram and E is the mid-point of BC. DE and AB when produced meet at F. Then, AF =

(a) $\frac{3}{2}AB$

(b) 2 AB

(c) 3 AB

(d) $\frac{5}{4}AB$

#### Answer:

Parallelogram *ABCD* is given with *E* as the mid-point of *BC*.

*DE* and *AB* when produced meet at *F*

We need to find *AF*.

Since *ABCD *is a parallelogram, then

Therefore,

Then, the alternate interior angles should be equal.

Thus, …… (I)

In and :

(From(I))

(E is the mid-point of BC)

(Vertically opposite angles)

(by ASA Congruence property)

We know that the corresponding angles of congruent triangles should be equal.

Therefore,

But,

(Opposite sides of a parallelogram are equal)

Therefore,

…… (II)

Now,

From (II),we get:

Hence the correct choice is (b).

#### Page No 13.70:

#### Question 31:

In a quadrilateral *ABCD*, ∠*A* + ∠*C* is 2 times ∠*B* + ∠*D*. If ∠*A* = 140° and ∠*D* = 60°, then ∠*B* =

(a) 60°

(b) 80°

(c) 120°

(d) None of these

#### Answer:

*ABCD* is a quadrilateral, with.

By angle sum property of a quadrilateral we get:

But,we have

Then,

The two equations so formed cannot give us the value for with a given value of .

Hence the correct choice is (d).

#### Page No 13.71:

#### Question 32:

The diagonals *AC* and *BD* of a rectangle *ABCD* intersect each other at P. If ∠*ABD* = 50°, then ∠*DPC* =

(a) 70°

(b) 90°

(c) 80°

(d) 100°

#### Answer:

Figure can be drawn as follows:

The diagonals *AC* and *BD* of rectangle *ABCD* intersect at *P.*

Also, it is given that* *

We need to find

It is given that

Therefore, (Because P lies on BD)

Also, diagonals of rectangle are equal and they bisect each other.

Therefore,

Thus, (Angles opposite to equal sides are equal)

[(Given)]

By angle sum property of a triangle:

But and are vertically opposite angles.

Therefore, we get:

[(Already proved)]

Hence the correct choice is (c).

#### Page No 13.71:

#### Question 1:

Three angles of a quadrilateral are ${75}^{\circ},{90}^{\circ}$ and ${75}^{\circ}$. The measure of the fourth angle is _____________ .

#### Answer:

Let the measure of the fourth angle be *x*.

We have,

Sum of all the angles of a quadrilateral = 360°

⇒ 75° + 90° + 75° + *x* = 360°

⇒ 240° + *x* = 360°

⇒ *x* = 360° − 240°

⇒ *x* = 120°

Hence, the measure of the fourth angle is __120°__.

#### Page No 13.71:

#### Question 2:

Diagonals of a parallelogram *ABCD *intersect at *O*. If ∠*AOB *= 90° and ∠*BDC *= 40°, then ∠*OAB *is __________.

#### Answer:

Let ∠*OAB* be *x*.

Since, *AB* || *DC*

Therefore, ∠*DBA = *∠*BDC = *40°

In *∆AOB,*

∠*OAB* + ∠*ABD* + ∠*AOB* = 180°

⇒ * x* + 40° + 90° = 180°

⇒ *x* + 130° = 180°

⇒ *x* = 180° − 130°

⇒ *x* = 50°

Hence, ∠*OAB *is __50°__.

#### Page No 13.71:

#### Question 3:

The angle *A, B, C* and *D* of a quadrilateral are in the ratio 1 : 2 : 4 : 5. If bisectors of ∠*C *and ∠*D* meet of *O*, the ∠*COD* = __________.

#### Answer:

Let the angle *A, B, C* and *D* of a quadrilateral be *x*, 2*x*, 4*x* and 5*x*, respectively.

We have,

Sum of all the angles of a quadrilateral = 360°

⇒ *x* + 2*x* + 4*x* + 5*x* = 360°

⇒ 12*x* = 360°

⇒ *x* = 30°

Thus, ∠*A* = 30°

∠*B* = 60°

∠*C* = 120°

∠*D* = 150°

Now, the bisectors of ∠*C *and ∠*D* meet of *O*

Thus, in *∆COD,
$\frac{1}{2}$*∠

*DCO*+ ∠

*COD*+

*$\frac{1}{2}$*∠

*ODC*= 180°

⇒

*$\frac{1}{2}$ ×*120° + ∠

*COD*+

*$\frac{1}{2}$ ×*150° = 180°

⇒ 60° + ∠

*COD*+ 75° = 180°

⇒ 135° + ∠

*COD*= 180°

⇒ ∠

*COD*= 180° − 135°

⇒ ∠

*COD*= 45°

Hence, ∠

*COD*=

__45°__.

#### Page No 13.71:

#### Question 4:

A diagonal of a rectangle is inclined to one side of the rectangle at 25°. The acute angle between the diagonals is __________.

#### Answer:

In *∆AOB,*

∠*OAB* + ∠*ABO* + ∠*AOB* = 180°

⇒ 25° + 25° + ∠*AOB* = 180°

⇒ ∠*AOB* + 50° = 180°

⇒ ∠*AOB* = 180° − 50°

⇒ ∠*AOB* = 130° which is an obtuse angle

Now, ∠*AOB + *∠*AOD *= 180° (angles on a straight line)

⇒ 130° *+ *∠*AOD *= 180°

⇒ ∠*AOD *= 180° − 130°

⇒ ∠*AOD *= 50° which is an acute angle

Hence, the acute angle between the diagonals is __50°__.

#### Page No 13.71:

#### Question 5:

*ABCD *is a rhombus such that ∠*ACB *= 40°. Then ∠*ADB *= _________.

#### Answer:

Let ∠*ADB *be* x.*

As we know, angles formed by the diagonals of a rhombus is 90°

In *∆BOC,*

∠*OBC* + ∠*BCO* + ∠*COB* = 180°

⇒ ∠*OBC* + 40° + 90° = 180°

⇒ ∠*OBC* + 130° = 180°

⇒ ∠*OBC* = 180° − 130°

⇒ ∠*OBC* = 50°

Since, *AD* || *BC*

Thus, ∠*ADB = *∠*DBC *(alternate interior angles)

⇒ *x* = 50°

Hence, ∠*ADB *= __50°__.

#### Page No 13.71:

#### Question 6:

If angles *A, B, C* and *D *of the quadrilateral *ABCD*, taken in order, are in the ratio 3 : 7 : 6 : 4, then *ABCD *is a _________.

#### Answer:

Let the angle *A, B, C* and *D* of a quadrilateral be 3*x*, 7*x*, 6*x* and 4*x*, respectively.

We have,

Sum of all the angles of a quadrilateral = 360°

⇒ 3*x* + 7*x* + 6*x* + 4*x* = 360°

⇒ 20*x* = 360°

⇒ *x* = 18°

Thus, ∠*A* = 54°

∠*B* = 126°

∠*C* = 108°

∠*D* = 72°

Now,

∠*BCE* = 180° − 108° = 72° = ∠*ADE*

⇒ *AD* || *BC*

Hence, *ABCD *is a __trapezium__.

#### Page No 13.71:

#### Question 7:

If the diagonals of a parallelogram *ABCD *are equal then ∠*ABC *= ___________.

#### Answer:

The diagonals of a parallelogram *ABCD *are equal

⇒ *ABCD* is a rectangle

⇒ ∠*ABC *= 90°

Hence, ∠*ABC *= __90°__.

#### Page No 13.71:

#### Question 8:

Diagonals *AC *and *BD *of a parallelogram *ABCD *intersect each other at *O*. If *OA *= 3 cm and *OD *= 2 cm, then *AC *= _________ and *BD *= __________.

#### Answer:

Given:

*ABCD *is a* *parallelogram

Diagonals *AC *and *BD *intersect each other at *O.
OA *= 3 cm and

*OD*= 2 cm

As we know, diagonals of a parallelogram bisects each other.

Thus,

*AC*= 2

*OA =*2× 3 = 6 cm

and

*BD = 2OD*= 2× 2 = 4 cm

Hence,

*AC*=

__6 cm__and

*BD*=

__4 cm__.

#### Page No 13.71:

#### Question 9:

*ABCD *is a trapezium in which *AB *|| *DC *and ∠*A* = ∠*B* = 45°. Then, ∠*C* = ______ and ∠*D* = ______.

#### Answer:

Given:

*ABCD *is a trapezium

*AB *|| *DC*

∠*A* = ∠*B* = 45°

Since, *AB *|| *DC*

Thus,* AD *and* BC *are the transversals.

Therefore, sum of the interior angles must be equal to 180°.

∠*A + *∠*D* = 180°

⇒ 45° *+ *∠*D* = 180°

⇒ ∠*D* = 180° − 45°

⇒ ∠*D* = 135°

∠*B + *∠*C* = 180°

⇒ 45° *+ *∠*C* = 180°

⇒ ∠*C* = 180° − 45°

⇒ ∠*C* = 135°

Hence, ∠*C* = __135°__ and ∠*D* = __135°__.

#### Page No 13.71:

#### Question 10:

*ABCD *is a rhombus in which altitude from vertex *D *to side *AB *bisects *AB*. The measures of angles of the rhombus are ________.

#### Answer:

Given:

*ABCD *is a rhombus

Altitude from vertex *D *to side *AB *bisects *AB.*

Let *DE* biescts *AB*.

*⇒ AE = EB* ...(1)

In ∆*ADE* and ∆*BDE*

*DE = DE* (Common side)

∠*AED* = ∠*BED* (Right angle)

*AE = EB* (from (1))

By SAS property, ∆*ADE* ≅ ∆BDE

Thus, *AD = BD* (By C.P.C.T) ...(2)

Since, sides of a rhombus are equal

Thus, *AD = AB * ...(3)

From (2) and (3)

*AD = BD = AB *

Therefore,* *∆*ADB *is an equilateral triangle.

Hence, ∠*A* = 60*°*

∠*A* = ∠*C =* 60*°* (opposite angles of rhombus are equal)

Sum of adjacent angles of a rhombus is 180*°*

⇒ ∠*A* + ∠*D* = 180*°* and ∠*C + *∠*B **=* 180*°*

⇒ 60*°* + ∠*D* = 180*°* and 60*° + *∠*B **=* 180*°*

⇒ ∠*D* = 180*° − *60*°* and ∠*B **= *180*° − *60*°*

⇒ ∠*D* = 120*°* and ∠*B **= *120*°*

Hence, the measures of angles of the rhombus are __60°, 120°, 60° and 120°__.

#### Page No 13.71:

#### Question 11:

Diagonals of a quadrilateral *ABCD *bisect each other. If ∠*A* = 35°, then ∠*B *= _________.

#### Answer:

Given: Diagonals of a quadrilateral *ABCD *bisect each other

Since, the diagonals of a quadrilateral *ABCD *bisect each other

Therefore, It is a parallelogram

Sum of interior angles of a parallelogram is 180*°*

⇒ ∠*A* + ∠*B* = 180*°*

⇒ 35*°* + ∠*B* = 180*°*

⇒ ∠*B* = 180*° − *35*°*

⇒ ∠*B* = 145*°*

Hence, ∠*B *= __145°__.

#### Page No 13.71:

#### Question 12:

In ∆*ABC*, *AB *= 5 cm, *BC *= 8 cm and *CA *= 7 cm. If *D* and *E* are respectively the mid-points of *AB *and *BC*, then *DE *= __________.

#### Answer:

Given:

In ∆*ABC*, *AB *= 5 cm, *BC *= 8 cm and *CA *= 7 cm

*D* and *E* are respectively the mid-points of *AB *and *BC*

Using mid-point theorem: The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is equal to the half of it.

Therefore,

$DE=\frac{1}{2}\left(AC\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times 7\phantom{\rule{0ex}{0ex}}=3.5\mathrm{cm}$

Hence, *DE *= __3.5 cm__.

#### Page No 13.71:

#### Question 13:

Opposite angles of a quadrilateral *ABCD *are equal. If *AB *= 4 cm, then *CD *= ___________.

#### Answer:

Given:

Opposite angles of a quadrilateral *ABCD *are equal

*AB *= 4 cm

If Opposite angles of a quadrilateral *ABCD *are equal, then *ABCD *is a parallelogram.

Thus, *CD = AB *= 4 cm

Hence, *CD *= __4 cm__.

#### Page No 13.71:

#### Question 14:

The diagonals *AC *and *BD *of a parallelogram *ABCD *intersect each other at the point *O*. If ∠*DAC *= 32° and ∠*AOB *= 70°, then ∠*DBC *= _______.

#### Answer:

Given:

The diagonals *AC *and *BD *of a parallelogram *ABCD *intersect each other at the point *O*

∠*DAC *= 32°

∠*AOB *= 70°

Let ∠*DBC* be* x.*

Since, *AB* || *DC*

Therefore, ∠*DAC = *∠*ACB *= 32° (alternate angles)

Also, ∠*AOB* + ∠*BOC *= 180° (angles on a straight line)

⇒ 70° + ∠*BOC *= 180°

⇒ ∠*BOC *= 180° − 70°

⇒ ∠*BOC *= 110°

In *∆COB,*

∠*OCB* + ∠*CBO* + ∠*BOC* = 180°

⇒ * *32° + *x* + 110° = 180°

⇒ *x* + 142° = 180°

⇒ *x* = 180° − 142°

⇒ *x* = 38°

Hence, ∠*DBC *= __38°__.

#### Page No 13.71:

#### Question 15:

The quadrilateral formed by joining the mid points of the sides of a quadrilateral *PQRS*, taken in order, is a rectangle, if diagonals of *PQRS *are ________.

#### Answer:

Given:

*PQRS* is a quadrilateral

The quadrilateral formed by joining the mid points of the sides of a quadrilateral *PQRS*, is a rectangle.

Let the rectangle be *ABCD*.

*A* is the mid-point of* PQ*, *B* is the mid-point of *QR*, *C* is the mid-point of *RS* and *D* is the mid-point of *PS*.

Using mid-point theorem: The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is equal to the half of it.

In ∆*PQR*,

*AB* || *PR*

and In ∆*QRS*,

*BC* || *QS*

Since, *AB$\perp $BC*

Therefore, *PR$\perp $QS*

Hence, the quadrilateral formed by joining the mid points of the sides of a quadrilateral *PQRS*, taken in order, is a rectangle, if diagonals of *PQRS *are __perpendicular__.

#### Page No 13.72:

#### Question 16:

The quadrilateral formed joining the mid-points of the sides of quadrilaterals *PQRS*, taken in order, is a rhombus, if diagonals of *PQRS *are _____________.

#### Answer:

Given:

*PQRS* is a quadrilateral

The quadrilateral formed by joining the mid points of the sides of a quadrilateral *PQRS*, is a rhombus.

Let the rhombus be *ABCD*.

*A* is the mid-point of* PQ*, *B* is the mid-point of *QR*, *C* is the mid-point of *RS* and *D* is the mid-point of *PS*.

Using mid-point theorem: The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is equal to the half of it.

In ∆*PQR*,

*AB* = $\frac{1}{2}$*PR*

and In ∆*QRS*,

*BC* = $\frac{1}{2}$*QS*

Since, *AB = BC *(sides of a rhombus are equal)

Therefore, $\frac{1}{2}$*PR = *$\frac{1}{2}$*QS
⇒ PR = QS*

Hence, the quadrilateral formed by joining the mid points of the sides of a quadrilateral

*PQRS*, taken in order, is rhombus, if diagonals of

*PQRS*are

__equal__.

#### Page No 13.72:

#### Question 17:

The figure formed by joining the mid-points of the sides of a quadrilateral *ABCD*, taken in order, is a square only if, diagonals of *ABCD *are ___________ and _________.

#### Answer:

Given:

*ABCD* is a quadrilateral

The quadrilateral formed by joining the mid points of the sides of a quadrilateral *ABCD*, is a square.

Let the square be *PQRS*.

*P* is the mid-point of* AB*, *Q* is the mid-point of *BC*, *R* is the mid-point of *CD* and *S* is the mid-point of *AD*.

Using mid-point theorem: The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is equal to the half of it.

In ∆*ABC*,

*PQ* = $\frac{1}{2}$*AC*

and In ∆*BCD*,

*QR* = $\frac{1}{2}$*BD*

Since, *PQ = QR *(sides of a square are equal)

Therefore, $\frac{1}{2}$*AC = *$\frac{1}{2}$*BD
⇒ AC = BC*

Hence, the diagonals are equal.

In ∆

*ABC*,

*PQ*||

*AC*

and In ∆

*BCD*,

*QR*||

*BD*

Since,

*PQ$\perp $QR*

Therefore,

*AC$\perp $BD*

Hence, the diagonals are perpendicular.

Hence, the figure formed by joining the mid-points of the sides of a quadrilateral

*ABCD*, taken in order, is a square only if, diagonals of

*ABCD*are

__perpendicular__and

__equal__.

#### Page No 13.72:

#### Question 18:

If bisectors of ∠*A *and ∠*B *of a quadrilateral *ABCD *intersect each other at *P*, of ∠*B *and ∠*C *at *Q*, of ∠*C *and ∠*D *at *R *and of ∠*D *and ∠*A *at *S*, then *PQRS *is a quadrilateral whose opposite angles are __________.

#### Answer:

Given:

*ABCD* is a quadrilateral

bisectors of ∠*A *and ∠*B *intersect each other at *P*

bisectors of ∠*B *and ∠*C *intersect each other at *Q*

bisectors of ∠*C *and ∠*D *intersect each other at *R*

bisectors of ∠*D *and ∠*A *intersect each other at *S*

In ∆*DAS*,

∠*ASD + *∠*SDA + *∠*DAS *= 180° (angle sum property)

*⇒ *∠*ASD +* $\frac{1}{2}$∠*D + *$\frac{1}{2}$∠*A *= 180°

*⇒ *∠*ASD *= 180° − $\frac{1}{2}$∠*D *−* *$\frac{1}{2}$∠*A*

Also,* *∠*PSR = *∠*ASD = *180° − $\frac{1}{2}$∠*D *−* *$\frac{1}{2}$∠*A*

*⇒ *∠*PSR *= 180° − $\frac{1}{2}$∠*D *−* *$\frac{1}{2}$∠*A * ...(1)

Similarly,

In ∆*BQC*,

∠*BQC + *∠*QCB + *∠*CBQ *= 180° (angle sum property)

*⇒ *∠*BQC +* $\frac{1}{2}$∠*C + *$\frac{1}{2}$∠*B *= 180°

*⇒ *∠*BQC *= 180° − $\frac{1}{2}$∠*C *−* *$\frac{1}{2}$∠*B*

Also,* *∠*PQR = *∠*BQC = *180° − $\frac{1}{2}$∠*C *−* *$\frac{1}{2}$∠*B*

*⇒ *∠*PQR *= 180° − $\frac{1}{2}$∠*C *−* *$\frac{1}{2}$∠*B * ...(2)

Adding (1) and (2), we get

∠*PSR + *∠*PQR = *180° − $\frac{1}{2}$∠*D *−* *$\frac{1}{2}$∠*A + *180° − $\frac{1}{2}$∠*C *−* *$\frac{1}{2}$∠*B*

= 360° − $\frac{1}{2}$(∠*D + *∠*A + *∠*C + *∠*B*)

= 360° − $\frac{1}{2}$(360°)

= 360° − 180°

= 180°

Since, sum of angles of a quadrilateral is* *360°

Therefore, ∠*PSR + *∠*PQR + *∠*QRS + *∠*QPS = *360°

*⇒ *∠*QRS + *∠*QPS = *180°

Hence, the sum of the opposite angles of the quadrilateral *PQRS* is 180°.

Hence, *PQRS *is a quadrilateral whose opposite angles are __supplementary__.

#### Page No 13.72:

#### Question 19:

If *APB *and *CQD *are two parallel lines, then the bisectors of the angles *APQ*, *BPQ*, *CQP *and *PQD *form a ____________.

#### Answer:

Given:

*APB *and *CQD *are two parallel lines

Let the bisectors of the angle *APQ* and *CQP* intersects at the point *R* and the bisectors of the angle *BPQ* and *PQD* intersects at the point *S*.

Join *PR*, *RQ*, *QS* and *SP* as shown in the figure.

*APB* || *CQD*

*⇒ *∠*APQ* = ∠*PQD* (alternate angles)

*⇒ *2∠*RPQ* = 2∠*PQS
⇒ *∠

*RPQ*= ∠

*PQS*

⇒ RP || SQ

⇒ RP || SQ

*...(1)*

Similarly,

*APB*||

*CQD*

*⇒*∠

*BPQ*= ∠

*PQC*(alternate angles)

*⇒*2∠

*SPQ*= 2∠

*PQR*

⇒∠

⇒

*SPQ*= ∠

*PQR*

⇒ RQ || SP

⇒ RQ || SP

*...(2)*

From (1) and (2),

*PSQR*is a parallelogram,

Also,

*∠*

*CQP +*∠

*PQD =*180° (angles on a straight line)

*⇒*2∠

*RQP +*2∠

*PQS =*180°

*⇒*∠

*RQP +*∠

*PQS =*90°

*⇒*∠

*RQ*

*S =*90°

Therefore,

*PSQR*is a rectangle.

Hence, if

*APB*and

*CQD*are two parallel lines, then the bisectors of the angles

*APQ*,

*BPQ*,

*CQP*and

*PQD*form a

__rectangle__.

#### Page No 13.72:

#### Question 20:

*D *and* E* are the mid-points of the sides *AB *and *AC *of ∆*ABC *and *O* is any point on side *BC*. *O* is joined to *A*. If *P* and *Q* are the mid-points of *OB *and *OC *respectively, then *DEQP *is a ____________.

#### Answer:

Given:

*D *and* E* are the mid-points of the sides *AB *and *AC *of ∆*ABC.
O* is any point on side

*BC*

Pand

P

*Q*are the mid-points of

*OB*and

*OC*respectively.

Using mid-point theorem: The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is equal to the half of it.

In ∆

*ABC*,

*DE*||

*BC || PQ*...(1)

*DE*= $\frac{1}{2}$

*BC*

*⇒ DE*= $\frac{1}{2}$(

*BP + PO + OQ + QC*)

*⇒ DE*= $\frac{1}{2}$(2

*OP*+ 2

*OQ*)

*⇒ DE*=

*OP*+

*OQ*

⇒ DE=

⇒ DE

*PQ*...(2)

In ∆

*AOB*,

*DP*||

*AO*...(3)

*DP*= $\frac{1}{2}$

*AO*...(4)

In ∆

*AOC*,

*EQ*||

*AO*...(5)

*EQ*= $\frac{1}{2}$

*AO*...(6)

From (3) and (5),

*DP*||

*EQ*

From (4) and (6),

*DP = EQ*

Thus,

*DE*||

*PQ*and

*DP*||

*EQ*

DP = EQand

DP = EQ

*DE = PQ*

Hence,

*DEQP*is a

__parallelogram__.

#### Page No 13.72:

#### Question 21:

When all the sides of a quadrilateral are equal, then it is either *a *____________or *a* _____________.

#### Answer:

All the sides of a square or a rhombus are equal.

Hence, when all the sides of a quadrilateral are equal, then it is either *a *__square__ or *a* __rhombus__.

#### Page No 13.72:

#### Question 22:

The bisectors of two adjacent sides of a parallelogram *ABCD *meet at a point *P *inside the parallelogram. The angle made by these bisectors at point *P* is ____________.

#### Answer:

Given:

*ABCD* is a parallelogram

Bisectors of ∠*A *and ∠*B *intersect each other at *P*

∠*A + *∠*B *= 180° (interior angles)

*⇒ *$\frac{1}{2}$∠*A + *$\frac{1}{2}$∠*B *= $\frac{1}{2}$(180°)

*⇒ *∠*PAB + *∠*PBA *= 90° ....(1)

Now, in ∆*APB*

∠*PAB + *∠*PBA + *∠*APB *= 180° (angle sum property)

*⇒ *90°* + *∠*APB *= 180° (From (1))

*⇒ *∠*APB *= 180° − 90°

*⇒ *∠*APB *= 90°

Hence, the angle made by these bisectors at point *P* is __90°__.

#### Page No 13.72:

#### Question 23:

In the given figure, *PQRS *is a rhombus. If ∠*OPQ* = 35°, then ∠*ORS* + ∠*OQP *= ____________.

#### Answer:

Given:

*PQRS *is a rhombus

∠*OPQ* = 35°

*PQRS *is a rhombus

*PQ || RS*

∠*OPQ = *∠*ORS * (alternate angles)

*⇒ *∠*ORS* = 35° ....(1)

We know, diagonals of a rhombus intersect at right angle

∴ ∠*POQ* = 90° ....(2)

Now, in ∆*OPQ*

∠*OPQ + *∠*POQ + *∠*OQP *= 180° (angle sum property)

*⇒ *35°* + *90° + ∠*OQP *= 180°

*⇒ *125° + ∠*OQP *= 180°

*⇒ *∠*OQP** *= 180° − 125°

*⇒ *∠*OQP** *= 55° ....(3)

Adding (1) and (3), we get

∠*ORS + *∠*OQP = *35° + 55° = 90°

Hence, ∠*ORS* + ∠*OQP *= __90°__.

#### Page No 13.72:

#### Question 24:

In the given figure, *ABC *is an isosceles triangle in which *AB *= *AC*. *AEDC *is a parallelogram. If ∠*CDF *= 70° and ∠*BFE *= 100°, then ∠*FBA* = ____________.

#### Answer:

Given:

*ABC *is an isosceles triangle

*AB *= *AC*

*AEDC *is a parallelogram

∠*CDF *= 70°

∠*BFE *= 100°

*AEDC *is a parallelogram

∠*ACD + *∠*CDE = *180° (interior angles)

*⇒ *∠*ACD + *70°* = *180°

*⇒ *∠*ACD** = *180° − 70°

*⇒ *∠*ACD** = *110° ....(1)

Now, ∠*ACD + *∠*ACB = *180° (angles on a straight line)

*⇒ *110° *+ *∠*ACB = *180°

*⇒** *∠*ACB = *180° − 110°

*⇒** *∠*ACB = *70° ....(2)

Also, ∠*BFE + *∠*BFD = *180° (angles on a straight line)

*⇒ *100° *+ *∠*BFD = *180°

*⇒** *∠*BFD = *180° − 100°

*⇒** *∠*BFD = *80° ....(3)

Now, in ∆*BFD*

∠*FBD + *∠*BDF + *∠*BFD *= 180° (angle sum property)

*⇒ *∠*FBD + *70°* + *80° = 180°

*⇒ *∠*FBD +* 150°* *= 180°

*⇒ *∠*FBD** *= 180° − 150°

*⇒ *∠*FBD** *= 30° ....(4)

Since, *ABC *is an isosceles triangle with *AB *= *AC*

Thus, ∠*ABC = *∠*ACB = *70° (From (2))

∠*ABC *= ∠*ABF *+ ∠*FBD
⇒ *70° = ∠

*ABF*+ 30°

*⇒*∠

*ABF =*70° − 30°

*⇒*∠

*ABF =*40°

Hence, ∠

*FBA*=

__40°__.

#### Page No 13.73:

#### Question 1:

In a parallelogram *ABCD*, write the sum of angles *A* and *B*.

#### Answer:

In Parallelogram *ABCD*, and are adjacent angles.

Thus, .

Then, we have and as consecutive interior angles which must be supplementary.

Hence, the sum of and is .

#### Page No 13.73:

#### Question 2:

In a parallelogram *ABCD,* if ∠*D* = 115°, then write the measure of ∠*A*.

#### Answer:

In Parallelogram *ABCD* , and are Adjacent angles.

We know that in a parallelogram, adjacent angles are supplementary.

$\mathrm{Now},\angle \mathrm{A}+\angle \mathrm{D}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{A}+115\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{A}=180\xb0-115\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{A}=65\xb0\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{measure}\mathrm{of}\angle \mathrm{A}\mathrm{is}65\xb0.$

#### Page No 13.73:

#### Question 3:

*PQRS* is a square such that *PR* and *SQ* intersect at *O*. State the measure of ∠*POQ*.

#### Answer:

PQRS is a square given as:

Since the diagonals of a square intersect at right angle.

Therefore, the measure of is .

#### Page No 13.73:

#### Question 4:

If *PQRS* is a square, then write the measure of ∠*SRP**.*

#### Answer:

The square *PQRS* is given as:

Since *PQRS* is a square.

Therefore,

and

Now, in , we have

That is, (Angles opposite to equal sides are equal)

By angle sum property of a triangle.

()

Hence, the measure of is.

#### Page No 13.73:

#### Question 5:

If *ABCD* is a rhombus with ∠ABC = 56°, find the measure of ∠ACD.

#### Answer:

The figure is given as follows:

*ABCD* is a rhombus.

Therefore,

*ABCD* is a parallelogram.

Thus,

[(Given)]

[]

Now in ,we have:

Hence the measure of is .

#### Page No 13.73:

#### Question 6:

The perimeter of a parallelogram is 22 cm. If the longer side measures 6.5 cm, what is the measure of shorter side?

#### Answer:

Let the shorter side of the parallelogram be *cm*.

The longer side is given as*cm**.*

Perimeter of the parallelogram is given as 22 *cm*

Therefore,

Hence, the measure of the shorter side is * cm.*

#### Page No 13.73:

#### Question 7:

If the angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13, then find the measure of the smallest angle.

#### Answer:

We have, .

So, let ,

,

and

By angle sum property of a quadrilateral, we get:

Smallest angle is :

Hence, the smallest angle measures.

#### Page No 13.73:

#### Question 8:

In a parallelogram *ABCD*, if ∠*A* = (3*x* − 20)°, ∠*B* = (*y* + 15)°, ∠*C* = (*x* + 40)°, then find the values of *x* and* y*.

#### Answer:

In parallelogram *ABCD*, and are opposite angles.

We know that in a parallelogram, the opposite angles are equal.

Therefore,

We have and

Therefore,

Therefore,

Similarly,

Also,

Therefore,

By angle sum property of a quadrilateral, we have:

Hence the required values for *x* and *y* are and respectively.

#### Page No 13.73:

#### Question 9:

If measures opposite angles of a parallelogram are (60 − x)° and (3x − 4)°, then find the measures of angles of the parallelogram.

#### Answer:

Let ABCD be a parallelogram, with and .

We know that in a parallelogram, the opposite angles are equal.

Therefore,

Thus, the given angles become

Similarly,

Also, adjacent angles in a parallelogram form the consecutive interior angles of parallel lines, which must be supplementary.

Therefore,

Similarly,

Thus, the angles of a parallelogram are ,, and .

#### Page No 13.73:

#### Question 10:

In a parallelogram *ABCD*, the bisector of ∠*A* also bisects *BC* at *X*. Find *AB* : *AD*.

#### Answer:

Parallelogram *ABCD* is given as follows:

We have *AX* bisects bisecting *BC* at *X*.

That is,

We need to find

Since, *AX* is the bisector

That is,

…… (i)

Also, *ABCD* is a parallelogram

Therefore, and *AB* intersects them

…… (ii)

In by angle sum property of a triangle:

From (i) and (ii), we get:

…… (iii)

From (i) and (iii),we get:

Sides opposite to equal angles are equal. Therefore,

As *X* is the mid point of *BC*. Therefore,

Also, *ABCD* is a parallelogram, then,

Thus,

Hence the ratio of is .

#### Page No 13.73:

#### Question 11:

In the given figure, PQRS is an isosceles trapezium. Find x and y.

#### Answer:

Trapezium is given as follows:

We know that PQRS is a trapezium with

Therefore,

Hence, the required value for *x* is .

#### Page No 13.73:

#### Question 12:

In the given figure, *ABCD* is a trapezium. Find the values of *x *and *y*.

#### Answer:

The figure is given as follows:

We know that *ABCD* is a trapezium with

Therefore,

It is given that and .

Similarly,

Hence, the required values for *x* and *y* is and respectively.

#### Page No 13.73:

#### Question 13:

In the given figure, *ABCD* and *AEFG* are two parallelograms. If ∠*C* = 58°, find ∠*F*.

#### Answer:

*ABCD* and *AEFG* are two parallelograms as shown below:

Since *ABCD* is a parallelogram, with

We know that the opposite angles of a parallelogram are equal.

Therefore,

Similarly, *AEFG* is a parallelogram, with

We know that the opposite angles of a parallelogram are equal.

Therefore,

Hence, the required measure for is .

#### Page No 13.73:

#### Question 14:

Complete each of the following statements by means of one of those given in brackets against each:

(i) If one pair of opposite sides are equal and parallel, then the figure is ........................

(parallelogram, rectangle, trapezium)

(ii) If in a quadrilateral only one pair of opposite sides are parallel, the quadrilateral is ................ (square, rectangle, trapezium)

(iii) A line drawn from the mid-point of one side of a triangle .............. another side intersects the third side at its mid-point. (perpendicular to parallel to, to meet)

(iv) If one angle of a parallelogram is a right angle, then it is necessarily a .................

(rectangle, square, rhombus)

(v) Consecutive angles of a parallelogram are ...................

(supplementary, complementary)

(vi) If both pairs of opposite sides of a quadrilateral are equal, then it is necessarily a ...............

(rectangle, parallelogram, rhombus)

(vii) If opposite angles of a quadrilateral are equal, then it is necessarily a ....................

(parallelogram, rhombus, rectangle)

(viii) If consecutive sides of a parallelogram are equal, then it is necessarily a ..................

(kite, rhombus, square)

#### Answer:

(i) If one pair of opposite sides are equal and parallel, then the figure is parallelogram.

Reason:

In and,

(Given)

(Common)

(Because , Alternate interior angles are equal)

So, by SAS Congruence rule, we have

Also,

(Corresponding parts of congruent triangles are equal)

But, these are alternate interior angles, which are equal.

Thus, and

Hence, quadrilateral *ABCD* is parallelogram

(ii) If in a quadrilateral only one pair of opposite sides are parallel, the quadrilateral is trapezium.

(iii) A line drawn from the mid-point of one side of a triangle parallel to another side intersects the third side at its mid-point.

Reason:

This is a theorem.

(iv) If one angle of a parallelogram is a right angle, then it is necessarily a rectangle.

Reason:

Let *ABCD* be the given parallelogram.

We have,

In a parallelogram, opposite angles are equal.

Therefore,

Similarly,

Also,

Thus, a parallelogram with all the angles being right angle and opposite sides being equal is a rectangle.

(v) Consecutive angles of a parallelogram are supplementary.

Reason:

Let *ABCD* be the given parallelogram.

Thus, .

Therefore,

Consecutive angles and are supplementary.

(vi) If both pairs of opposite sides of a quadrilateral are equal, then it is necessarily a parallelogram.

Reason:

*ABCD* is a quadrilateral in which and .

We need to show that *ABCD *is a parallelogram.

In and , we have

(Common)

(Given)

(Given)

So, by SSS criterion of congruence, we have

By corresponding parts of congruent triangles property.

…… (i)

And

Now lines *AC* intersects *AB* and *DC* at *A* and *C,*such that

(From (i))

That is, alternate interior angles are equal.

Therefore, .

Similarly, .

Therefore, *ABCD* is a parallelogram.

(vii) If opposite angles of a quadrilateral are equal, then it is necessarily a **parallelogram**.

Reason:

*ABCD* is a quadrilateral in which and .

We need to show that *ABCD *is a parallelogram.

In quadrilateral *ABCD*, we have

Therefore,

…… (i)

Since sum of angles of a quadrilateral is

From equation (i), we get:

Similarly,

Now, line *AB* intersects *AD* and *BC* at *A* and *B* respectively

Such that

That is, sum of consecutive interior angles is supplementary.

Therefore, .

Similarly, we get .

Therefore, *ABCD* is a parallelogram.

(viii) If consecutive sides of a parallelogram are equal, then it is necessarily a rhombus.

We have *ABCD*, a parallelogram with .

Since *ABCD* is a parallelogram.

Thus,

And

But,

Therefore,all four sides of the parallelogram are equal, then it is a rhombus.

#### Page No 13.74:

#### Question 15:

In a quadrilateral *ABCD*, bisectors of angles *A* and *B* intersect at *O* such that ∠*AOB* = 75°, then write the value of ∠*C* + ∠*D*.

#### Answer:

The quadrilateral can be drawn as follows:

We have *AO* and *BO* as the bisectors of angles and respectively.

In ,We have,

…… (1)

By angle sum property of a quadrilateral, we have:

Putting in equation (1):

……(2)

It is given that in equation (2), we get:

Hence, the sum of and is .

#### Page No 13.74:

#### Question 16:

The diagonals of a rectangle *ABCD* meet at *O*, If ∠*BOC* = 44°, find ∠*OAD*.

#### Answer:

The rectangle *ABCD* is given as:

We have,

(Linear pair)

Since, diagonals of a rectangle are equal and they bisect each other. Therefore, in , we have

(Angles opposite to equal sides are equal.)

Therefore,

Now,in , we have

Since, each angle of a rectangle is a right angle.

Therefore,

Thus,

Hence, the measure of is.

#### Page No 13.74:

#### Question 17:

If *ABCD* is a rectangle with ∠*BAC* = 32°, find the measure of ∠*DBC*.

#### Answer:

Figure is given as :

Suppose the diagonals AC and BD intersect at O.

Since, diagonals of a rectangle are equal and they bisect each other.

Therefore, in , we have

Angles opposite to equal sides are equal.

Therefore,

But,

Now,

Hence, the measure of is .

#### Page No 13.74:

#### Question 18:

If the bisectors of two adjacent angles A and B of a quadrilateral *ABCD* intersect at a point *O *such that ∠*C* + ∠*D* = *k* ∠*AOB*, then find the value of *k*.

#### Answer:

The quadrilateral can be drawn as follows:

We have *AO* and *BO* as the bisectors of angles and respectively.

In ,We have,

…… (I)

By angle sum property of a quadrilateral, we have:

Putting in equation (I):

…… (II)

On comparing equation (II) with

We get .

Hence, the value for *k* is .

#### Page No 13.74:

#### Question 19:

In the given figure, *PQRS* is a rhombus in which the diagonal PR is produced to T. If ∠*SRT* = 152°, find *x*,* y* and *z*.

#### Answer:

Rhombus *PQRS* is given.

Diagonal *PR* is produced to *T*.

Also, .

We know that in a rhombus, the diagonals bisect each other at right angle.

Therefore,

Now,

In , by angle sum property of a triangle, we get:

Or, (Because *O* lies on *SQ*)

We have, .Thus the alternate interior opposite angles must be equal.

Therefore,

In ,we have

Since opposite sides of a rhombus are equal.

Therefore,

Also,

Angles opposite to equal sides are equal.

Thus,

But

Thus,

Hence the required values for *x,y* and *z* are , and respectively.

#### Page No 13.74:

#### Question 20:

In the given figure, *ABCD* is a rectangle in which diagonal *AC *is produced to *E*. If ∠*ECD* = 146°, find ∠*AOB*.

#### Answer:

*ABCD* is a rectangle

With diagonal *AC* produced to point *E*.

We have

(Linear pair)

We know that the diagonals of a parallelogram bisect each other.

Thus

Also, angles opposite to equal sides are equal.

Therefore,

By angle sum property of a traingle

Also, and are vertically opposite angles.

Therefore,

Hence, the required measure for is .

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