R.d Sharma 2022 Mcqs Solutions for Class 9 Maths Chapter 13 - Quadrilaterals

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).

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).

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).

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).

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).

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).

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).

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).

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 ADintersects 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).

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, PQRSis 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).

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).

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).

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 ΔADC, 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).

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).

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).

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).

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 ΔABD, 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).

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).

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).

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).

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).

Answer:

ABCD is a parallelogram in which AC bisects.

It is given that

Therefore,

Since,

Therefore,

Hence, the correct choice is (b).

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).

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).

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).

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).

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).

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).

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).

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).

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).

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).

Answer:

Statement-2 (Reason): Diagonals of a parallelogram bisect each other.
Thus, Statement-2 is true.

Statement-1 (Assertion): Diagonals of a rectangle are equal.

A rectangle is a parallelogram with opposite sides equal and the measure of each angle is 90º.
Now, consider $△$ABC and $△$DCB.

BC = BC                             (Common side)
∠ABC = ∠DCB                 (Each 90º)
AB = DC                             (Opposite sides of a parallelogram are equal)

∴ â–³ABC ≅ â–³DCB             (By SAS congruence rule)

⇒ AC = DB                               (By CPCT)

Thus, Statement-1 is true.
So, Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

Answer:

Statement-2 (Reason): Diagonals of a rhombus are perpendicular to each other.
In rhombus ABCD, AC and BD intersect each other at E. 


 

Answer:

Statement-2 (Reason): Sum of the angles of a quadrilateral is 360º.

Consider a quadrilateral ABCD such that BD is the diagonal.

In $△\mathrm{ABD}$, by Angle sum property of triangle
$\angle \mathrm{ABD}+\angle \mathrm{BDA}+\angle \mathrm{DAB}=180° .....\left(1\right)$

In $△\mathrm{CBD}$, by Angle sum property of triangle
$\angle \mathrm{DBC}+\angle \mathrm{BCD}+\angle \mathrm{CDB}=180° .....\left(2\right)$

Adding (1) and (2) on both the sides
$$\begin{gathered} \angle \mathrm{ABD}+\angle \mathrm{BDA}+\angle \mathrm{DAB}+\angle \mathrm{DBC}+\angle \mathrm{BCD}+\angle \mathrm{CBD}=180°+180° .....\left(1\right) \\ \Rightarrow \left(\angle \mathrm{ABD}+\angle \mathrm{DBC}\right)+\left(\angle \mathrm{BDA}+\angle \mathrm{CDB}\right)+\angle \mathrm{DAB}+\angle \mathrm{BCD}=360° \\ \Rightarrow \angle \mathrm{ABC}+\angle \mathrm{ADC}+\angle \mathrm{DAB}+\angle \mathrm{BCD}=360° \end{gathered}$$

Thus, statement-2 is true.

Statement-1 (Assertion): All angles of a quadrilateral can be obtuse angle.

An obtuse angle is greater than $90°$ and by angle sum property, the sum of all angles of a quadrilateral is $360°$

If we take all four angles greater than $90°$  then, their sum will be obviously greater than $360°$.

Answer:

Statement-2 (Reason): Diagonals of a parallelogram bisect each other.
Thus, Statement-2 is true.

Statement-1 (Assertion): If the diagonals of a parallelogram are equal, then it is a rectangle.

Consider ABCD is parallelogram  such that AC = BD

In $△\mathrm{ACB} \mathrm{and} △\mathrm{BDA}$
AC = BD      (Given)
AB = BA      (Common)
BC = AD      (Opposite sides of a parallelogram)
$\therefore △\mathrm{ACB}\cong △\mathrm{BDA} \left(\mathrm{By} \mathrm{SSS} \mathrm{congruence} \mathrm{rule}\right)$
$$\begin{gathered} \Rightarrow \angle \mathrm{ABC}=\angle \mathrm{BAD} \left(\mathrm{By} \mathrm{CPCT}\right) ....\left(1\right) \\ \mathrm{Also}, \mathrm{AD}\left|\right|\mathrm{BC} \\ \angle \mathrm{BAD}+\angle \mathrm{ABC}=180° \left(\because \mathrm{Cointerior} \mathrm{angles} \mathrm{are} \mathrm{supplementary}\right) .....\left(2\right) \\ \mathrm{From} \left(1\right) \mathrm{and} \left(2\right) \\ 2\angle \mathrm{BAD}=180° \\ \Rightarrow \angle \mathrm{BAD} =90° \\ \therefore \angle \mathrm{ABC} =90° \left[\mathrm{From} \left(1\right)\right] \end{gathered}$$

∴ Parallelogram ABCD is a rectangle.

Thus, statement-1 is true.
So, Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

Hence, the correct answer is option (b).

 

Answer:

Statement-2 (Reason): The line through the mid-point of one side of a triangle, parallel to another side, intersects the third side at its mid-point.

According to Mid-point Theorem, the line through the mid-point of one side of a triangle, parallel to another side, intersects the third side at its mid-point.

Thus, Statement-2 is true.

Statement-1 (Assertion): In the given figure, if AD is the median of âˆ†ABC and E is the mid-point of AD. BE produced meets AC in F, then $\mathrm{AF}=\frac{1}{3}\mathrm{AC}.$

Given that, AD is the median of âˆ†ABC and E is the mid-point of AD.
Through D draw $\mathrm{DG}\parallel \mathrm{BF}$.

In âˆ†ADG
E is the midpoint of AD and $\mathrm{DG}\parallel \mathrm{BF}$
By converse of midpoint theorem, we have
F is the midpoint of AG
$\Rightarrow \mathrm{AF}=\mathrm{FG}$                          .....(1)

Similarly in âˆ†BCF
D is the midpoint of BC and $\mathrm{DG}\parallel \mathrm{BF}$
By converse of midpoint theorem, we have
G is the midpoint of CF 
$\Rightarrow \mathrm{FG}=\mathrm{GC}$                         .....(2)

From (1) and (2), we have
AF = FG = GC

Now,  AF + FG + GC = AC
⇒ AF + AF +A F = AC
$\Rightarrow \mathrm{AF}=\frac{1}{3}\mathrm{AC}$

Thus, Statement-1 is true.
So, Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is option (a).




 

Answer:

Statement-2 (Reason): In a parallelogram, the bisectors of any two consecutive angles intersect at right angle.

Consider a parallelogram ABCD.

Therefore $\mathrm{AD}\parallel \mathrm{BC}$
Now, $\mathrm{AD}\parallel \mathrm{BC}$ and transversal AB intersects them at A and B respectively. Therefore,
$$\begin{gathered} \angle \mathrm{A}+\angle \mathrm{B}=180° \left(\because \mathrm{Sum} \mathrm{of} \mathrm{consecutive} \mathrm{interier} \mathrm{angles} \mathrm{is} 180°\right) \\ \mathrm{Dividing} \mathrm{both} \mathrm{sides} \mathrm{by} 2 \\ \Rightarrow \frac{1}{2}\angle \mathrm{A}+\frac{1}{2}\angle \mathrm{B}=90° \\ \Rightarrow \angle \mathrm{BAS}+\angle \mathrm{ABS}=90° .....\left(1\right) \\ \mathrm{But} \mathrm{in} △\mathrm{ABS}, \mathrm{by} \mathrm{Angle} \mathrm{sum} \mathrm{property} \mathrm{of} \mathrm{triangle} \\ \angle \mathrm{BAS}+\angle \mathrm{ABS}+\angle \mathrm{ASB}=180° \\ \Rightarrow 90°+\angle \mathrm{ASB}=180° \\ \Rightarrow \angle \mathrm{ASB}=90° \\ \Rightarrow \angle \mathrm{RSP}=\angle \mathrm{ASB}=90° \left(\because \mathrm{Vertically} \mathrm{opposite} \mathrm{angles}\right) \\ \mathrm{Similarly}, \angle \mathrm{SRQ}=90°, \angle \mathrm{RQP}=90° \mathrm{and} \angle \mathrm{SPQ}=90° \end{gathered}$$
Therefore, the bisectors of any two consecutive angles intersect at right angle.

Thus, Statement-2 is true.

Statement-1 (Assertion): The bisectors of the angles of a parallelogram enclose a rectangle.

​Consider a parallelogram ABCD.


According to Statement-2, in a parallelogram, the bisectors of any two consecutive angles intersect at right angle.
$$\begin{gathered} \therefore \angle \mathrm{RSP}=90°,\angle \mathrm{SRQ}=90°, \angle \mathrm{RQP}=90° \mathrm{and} \angle \mathrm{SPQ}=90° \\ \mathrm{Now}, \angle \mathrm{RSP}+\angle \mathrm{SRQ}=180° \\ \Rightarrow \angle \mathrm{RSP} \mathrm{and} \angle \mathrm{SRQ} \mathrm{are} \mathrm{co}‐\mathrm{interior} \mathrm{angles} \\ \Rightarrow \mathrm{SP}\left|\right| \mathrm{RQ} \\ \mathrm{Similarly}, \angle \mathrm{RSP}+\angle \mathrm{SPQ}=180° \\ \Rightarrow \angle \mathrm{RSP} \mathrm{and} \angle \mathrm{SPQ} \mathrm{are} \mathrm{co}‐\mathrm{interior} \mathrm{angles} \\ \Rightarrow \mathrm{SR}\left|\right| \mathrm{PQ} \end{gathered}$$
⇒ PQRS is a rectangle.

Thus, Statement-1 is true.
So, Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is option (a).

Answer:

Statement-2 (Reason): Sum of the angles of a quadrilateral is 360º.

Consider a quadrilateral ABCD such that BD is the diagonal.

In $△\mathrm{ABD}$, by Angle sum property of triangle
$\angle \mathrm{ABD}+\angle \mathrm{BDA}+\angle \mathrm{DAB}=180° .....\left(1\right)$

In $△\mathrm{CBD}$, by Angle sum property of triangle
$\angle \mathrm{DBC}+\angle \mathrm{BCD}+\angle \mathrm{CDB}=180° .....\left(2\right)$

Adding (1) and (2) on both the sides
$$\begin{gathered} \angle \mathrm{ABD}+\angle \mathrm{BDA}+\angle \mathrm{DAB}+\angle \mathrm{DBC}+\angle \mathrm{BCD}+\angle \mathrm{CBD}=180°+180° .....\left(1\right) \\ \Rightarrow \left(\angle \mathrm{ABD}+\angle \mathrm{DBC}\right)+\left(\angle \mathrm{BDA}+\angle \mathrm{CDB}\right)+\angle \mathrm{DAB}+\angle \mathrm{BCD}=360° \\ \Rightarrow \angle \mathrm{ABC}+\angle \mathrm{ADC}+\angle \mathrm{DAB}+\angle \mathrm{BCD}=360° \end{gathered}$$

Thus, statement-2 is true.

Statement-1 (Assertion): If the measures of three angles of a quadrilateral are 130º, 70º and 60º, then the measure of fourth angle is 100º.
Let the measure of fourth angle be x.
According to Statement-2, the sum of the angles of quadrilateral is 360º.
⇒ 130º + 70º + 60º + x = 360º
⇒ x = 100º

Thus, Statement-1 is true.
So, Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is option (a).


 

Answer:

Statement-2 (Reason): If the diagonals of a parallelogram are equal and intersect at right angle, then it is a square.
Consider a parallelogram PQRS such that PR = SQ and ∠SOR = ∠QOR =∠QOP =∠SOP = 90º.
Since PR = SQ and diagonals of a parallelogram bisect each other.
∴ PO = OR =OQ = OS                   .....(1)


In ∆POQ and ∆ROQ     
$$\begin{gathered} \Rightarrow \mathrm{PO}=\mathrm{OR} \left[\mathrm{From}\left(1\right)\right] \\ \Rightarrow \mathrm{OQ}=\mathrm{QO} \\ \Rightarrow \angle \mathrm{POQ}=\angle \mathrm{ROQ} \left(\mathrm{Given}\right) \end{gathered}$$     

∴ ∆POQ $\cong$ ∆ROQ           (By SAS congruency rule)
⇒ PQ = RQ           (By CPCT)                     .....(2)
Since PQ = RS and RQ = PS (Opposite sides of a parallelogram are equal)                   .....(3)
From (2) and (3)
PQ = RS = RQ = PS

Now,

$$\begin{gathered} \mathrm{In}△\mathrm{OPQ} \mathrm{and} △\mathrm{OQP} \\ \mathrm{PO}=\mathrm{OQ} \\ \Rightarrow \angle \mathrm{OPQ}=\angle \mathrm{OQP} \left(\because \mathrm{Isosceles} \mathrm{property} \mathrm{of} \mathrm{triangle}\right) .....\left(4\right) \\ \mathrm{Similarly}, \mathrm{In}△\mathrm{OQR} \mathrm{and} △\mathrm{ORQ} \\ \mathrm{OR}=\mathrm{OQ} \\ \Rightarrow \angle \mathrm{OQR}=\angle \mathrm{ORQ} \left(\because \mathrm{Isosceles} \mathrm{property} \mathrm{of} \mathrm{triangle}\right) .....\left(5\right) \\ \mathrm{In} △\mathrm{PQR} \\ \Rightarrow \angle \mathrm{RPQ}+\angle \mathrm{PQR}+\angle \mathrm{QRP}=180° \\ \Rightarrow 2\left(\angle \mathrm{OQP}+\angle \mathrm{OQR}\right)=180° \\ \Rightarrow \left(\angle \mathrm{OQP}+\angle \mathrm{OQR}\right)=90° \\ \Rightarrow \angle \mathrm{PQR}=90° \end{gathered}$$

Thus, Statement-2 is true.

Statement-1 (Assertion): The diagonals of a parallelogram bisect each other.
Thus, Statement-1 is true.

So, Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

Hence, the correct answer is an option (b).

Answer:

Statement-2 (Reason): The angle bisectors of a parallelogram from a rectangle.
Consider a parallelogram ABCD.

Therefore $\mathrm{AD}\parallel \mathrm{BC}$
Now, $\mathrm{AD}\parallel \mathrm{BC}$ and transversal AB intersects them at A and B respectively. Therefore,
$$\begin{gathered} \angle \mathrm{A}+\angle \mathrm{B}=180° \left(\because \mathrm{Sum} \mathrm{of} \mathrm{consecutive} \mathrm{interier} \mathrm{angles} \mathrm{is} 180°\right) \\ \mathrm{Dividing} \mathrm{both} \mathrm{sides} \mathrm{by} 2 \\ \Rightarrow \frac{1}{2}\angle \mathrm{A}+\frac{1}{2}\angle \mathrm{B}=90° \\ \Rightarrow \angle \mathrm{BAS}+\angle \mathrm{ABS}=90° .....\left(1\right) \\ \mathrm{But} \mathrm{in} △\mathrm{ABS}, \mathrm{by} \mathrm{Angle} \mathrm{sum} \mathrm{property} \mathrm{of} \mathrm{triangle} \\ \angle \mathrm{BAS}+\angle \mathrm{ABS}+\angle \mathrm{ASB}=180° \\ \Rightarrow 90°+\angle \mathrm{ASB}=180° \\ \Rightarrow \angle \mathrm{ASB}=90° \\ \Rightarrow \angle \mathrm{RSP}=\angle \mathrm{ASB}=90° \left(\because \mathrm{Vertically} \mathrm{opposite} \mathrm{angles}\right) \\ \mathrm{Similarly}, \angle \mathrm{SRQ}=90°, \angle \mathrm{RQP}=90° \mathrm{and} \angle \mathrm{SPQ}=90° \end{gathered}$$
$$\begin{gathered} \mathrm{Now}, \angle \mathrm{RSP}+\angle \mathrm{SRQ}=180° \\ \Rightarrow \angle \mathrm{RSP} \mathrm{and} \angle \mathrm{SRQ} \mathrm{are} \mathrm{co}‐\mathrm{interior} \mathrm{angles} \\ \Rightarrow \mathrm{SP}\left|\right| \mathrm{RQ} \\ \mathrm{Similarly}, \angle \mathrm{RSP}+\angle \mathrm{SPQ}=180° \\ \Rightarrow \angle \mathrm{RSP} \mathrm{and} \angle \mathrm{SPQ} \mathrm{are} \mathrm{co}‐\mathrm{interior} \mathrm{angles} \\ \Rightarrow \mathrm{SR}\left|\right| \mathrm{PQ} \end{gathered}$$
⇒ PQRS is a rectangle.
Thus, Statement-2 is true.

Statement-1 (Assertion): Every parallelogram is a rectangle.

The given statement is not true. The correct statement is, "Every rectangle is a parallelogram, but not every parallelogram is a rectangle".
Because each corner angle of a rectangle is of 90º but each corner angle of a parallelogram need not to be 90º always.

Thus, Statement-1 is false.
So, Statement-1 is false, Statement-2 is true.

Hence, the correct answer is option (d).
 

Answer:

Statement-2 (Reason): The diagonals of a rhombus bisect each other at right angle.
$\mathrm{In} △\mathrm{AOD} \mathrm{and} △\mathrm{COD}$
⇒  OA = OC          (Diagonals of rhombus bisect each other)

⇒  2∠AOD = 180º          [From (1)]
.

∴  ∠AOD = 90º
Thus, statement-2 is true.

Statement-1 (Assertion): In a rhombus ABCD, diagonals, AC bisects the angles ∠A as well as ∠C.

In âˆ†ACB and âˆ†ACD
AB = AD (Sides of rhombus)
BC = DC (Sides of rhombus)
AC = AC (Common)
$△\mathrm{ACB}\cong △\mathrm{ACD} \left(\mathrm{By} \mathrm{SSS}\right)$

$$\begin{gathered} \Rightarrow \angle \mathrm{BAC}=\angle \mathrm{DAC} \left(\mathrm{By} \mathrm{CPCT}\right) \\ \mathrm{And} \angle \mathrm{BCA}=\angle \mathrm{DCA} \left(\mathrm{By} \mathrm{CPCT}\right) \end{gathered}$$

So, diagonal AC bisects ∠A as well as ∠C of rhombus ABCD.

Thus, statement-1 is true.
So,  Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

Hence the correct answer is option (b).