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Page No 262:
Question 1:
In a Δ ABC, if ∠A = 120° and AB = AC. Find ∠B and ∠C.
Answer:
In, it is given that
, and
We have to find, and
Since and
Then (as AB = AC)
Now
(By property of triangle)
Thus,
, as (given)
So,
Since,, so
Hence .
Page No 262:
Question 2:
In a ΔABC, if AB = AC and ∠B = 70°, find ∠A.
Answer:
In it is given that
, and
We have to find.
Since
Then (isosceles triangles)
Now
(As given)
Thus
(Property of triangle)
Hence .
Page No 262:
Question 3:
In the given figure, AB = AC and ∠ACD = 105°, find ∠BAC.
Answer:
It is given that
We have to find.
(Isosceles triangle)
Now
Since exterior angle of isosceles triangle is the sum of two internal base angles
Now
So, (By property of triangle)
Hence .
Page No 262:
Question 4:
Find the measure of each exterior angle of an equilateral triangle.
Answer:
We have to find the measure of each exterior angle of an equilateral triangle.
It is given that the triangle is equilateral
So, and
Since triangle is equilateral
So,
Now we have to find the exterior angle.
As we know that exterior angle of the triangle is sum of two interior angles
Thus
Hence each exterior angle is.
Page No 262:
Question 5:
If the base of an isosceles triangle is produced on both sides, prove that the exterior angles so formed are equal to each other.
Answer:
It is given that the base of an isosceles triangle is produced on both sides.
We have to prove that the exterior angles so formed are equal to each other.
That is we need to show that
Let the is isosceles having base and equal sides AB and AC
Then, and
(Isosceles triangles)
Now
.........(1)
And,
.......(2)
Thus
........(3)
Now from equation (2)
.........(4)
Since
Hence from equation (3) and (4)
Page No 262:
Question 6:
In a ΔPQR, if PQ = QR and L, M and N are the mid-points of the sides PQ, OR, and RP respectively. Prove that LN = MN.
Answer:
It is given that
And is the mid point of
So
And is the mid point of
So
And is the mid point of
So
We have to prove that
In we have
(Equilateral triangle)
Then
, and
, and
Similarly comparing and we have
, and
And (Since N is the mid point of )
So by congruence criterion, we have
Hence.
Page No 263:
Question 7:
Prove that the medians of an equilateral triangle are equal.
Answer:
We have to prove that the median of an equilateral triangle are equal.
Let be an equilateral triangle with as its medians.
Let
In we have
(Since similarly)
(In equilateral triangle, each angle)
And (common side)
So by congruence criterion we have
This implies that,
Similarly we have
Hence .
Page No 263:
Question 8:
In the given figure, AB = AC and DB = DC, find the ratio ∠ABD : ∠ACD.
Answer:
It is given that
We have to find the ratio.
Since
And
So we have,
So
Hence .
Page No 263:
Question 9:
Determine the measure of each of the equal angles of a right-angled isosceles triangle.
Answer:
It is given that
Is right angled triangle
And
We have to find and
Since
(Isosceles triangle)
Now
(Property of triangle)
()
So
Hence
Page No 263:
Question 10:
In the given figure, PQRS is a square and SRT is an equilateral triangle. Prove that
(i) PT = QT
(ii) ∠TQR = 15°
Answer:
It is given that
Δis a square and Δ is an equilateral triangle.
We have to prove that
(1) and (2)
(1)
Since,
(Angle of square)
(Angle of equilateral triangle)
Now, adding both
Similarly, we have
Thus in and we have
(Side of square)
And (equilateral triangle side)
So by congruence criterion we have
Hence.
(2)
Since
QR = RS ( Sides of Square)
RS = RT (Sides of Equilateral triangle)
We get
QR = RT
Thus, we get
(Angles opposite to equal sides are equal)
Now, in the triangle TQR, we have
Page No 263:
Question 11:
In the given figure, AB is a line segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B. Show that the line PQ is perpendicular bisector of AB.
Answer:
It is given that
P and Q are equidistant from A and B that is
, and
We are asked to show that line PO is perpendicular bisector of line AB.
First of all we will show that ΔAQP and ΔQBP are congruent to each other and ultimately we get the result.
Consider the triangles AQP and QBP in which
AP=BP, AQ=BQ, PQ=PQ
So by SSS property we have
Implies that
Now consider the triangles ΔAPC and ΔPCB in which
And
So by SAS criterion we find that,
So this implies that AC=BC and
But
Hence PQ is perpendicular bisector of AB.
Page No 268:
Question 1:
BD and CE are bisectors of ∠B and ∠C of an isosceles ΔABC with AB = AC. Prove that BD = CE.
Answer:
It is given that
Is bisector of and is bisector of.
And is isosceles with
We have to prove that
If will be sufficient to prove to show that
Now in these two triangles
Since, so
Now as BD and CE are bisector of the respectively, so
, and
BC=BC
So by congruence criterion we have
Hence Proved.
Page No 268:
Question 2:
Two lines AB and CD intersect at O such that BC is equal and parallel to AD. Prove that the lines AB and CD bisect at O.
Answer:
It is given that
We have to prove that the lines and bisect at.
If we prove that, then
We can prove and bisects at.
Now in and
(Given)
(Since and is transversal)
And (since and is transversal)
So by congruence criterion we have,
, so
Hence and bisect each other at.
Page No 268:
Question 3:
In the given figure, it is given that RT = TS, ∠1 = 2∠2 and ∠4 = 2∠3. Prove that ΔRBT ≅ΔSAT.
Answer:
It is given that
We have to prove that
Now
In we have
(Isosceles triangle) .......(1)
Now we have
(Vertically opposite angles)
(Since, given)
.......(2)
Subtracting equation (2) from equation (1) we have
Now in and we have
(Given)
So all the criterion for the two triangles and are satisfied to be congruent
Hence by congruence criterion we have proved.
Page No 276:
Question 1:
In two right triangles one side an acute angle of one are equal to the corresponding side and angle of the other. Prove that the triangles are congruent.
Answer:
It is given that
We are asked to show that
Let us assume
, and are right angled triangle.
Thus in and, we have
And (given)
Hence by AAs congruence criterion we haveProved.
Page No 276:
Question 2:
Prove that each angle of an equilateral triangle is 60°
Answer:
We have to prove each angle of an equilateral triangle is.
Here
(Side of equilateral triangle)
...........(1)
And
(Side of equilateral triangle)
..........(2)
From equation (1) and (2) we have
Hence
Now
That is (since)
Hence Proved.
Page No 276:
Question 3:
Angles A, B, C of a triangle ABC are equal to each other. Prove that ΔABC is equilateral.
Answer:
It is given that
We have to prove that triangle ΔABC is equilateral.
Since (Given)
So, ..........(1)
And (given)
So ........(2)
From equation (1) and (2) we have
Now from above equation if we have
Given condition satisfy the criteria of equilateral triangle.
Hence the given triangle is equilateral.
Page No 276:
Question 4:
ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.
Answer:
It is given that
We have to find and.
Since so,
Now (property of triangle)
(Since )
Here
Then
Hence
Page No 276:
Question 5:
If the bisector of the exterior vertical angle of a triangle be parallel to the base. Show that the triangle is isosceles.
Answer:
We have to prove that is isosceles.
Let Δ be such that the bisector of is parallel to
The base, we have
(Corresponding angles)
(Alternate angle)
(Since)
Hence is isosceles.
Page No 276:
Question 6:
In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle.
Answer:
In the triangle ABC it is given that the vertex angle is twice of base angle.
We have to calculate the angles of triangle.
Now, let be an isosceles triangle such that
Then
(Given)
()
Now (property of triangle)
Hence
Page No 276:
Question 7:
PQR is a triangle in which PQ = PR and S is any point on the side PQ. Through S, a line is drawn parallel to QR and intersecting PR at T. Prove that PS = PT.
Answer:
It is given that
We have to prove
In we have
(Given)
So,
Now (Given)
Since corresponding angle are equal, so
That is,
Henceproved.
Page No 276:
Question 8:
In a ΔABC, it is given that AB = AC and the bisectors of ∠B and ∠C intersect at O. If M is a point on BO produced prove that ∠MOC = ∠ABC.
Answer:
It is given that
In,
We have to prove that
Now
(Given)
Thus
........(1)
In, we have
So, {from equation (1)}
Hence Proved.
Page No 277:
Question 9:
P is a point on the bisector of an angle ∠ABC. If the line through P parallel to AB meets BC at Q, prove that triangle BPQ is isosceles.
Answer:
In the following figure it is given that sides AB and PQ are parallel and BP is bisector of
We have to prove that is an isosceles triangle.
(Since BP is the bisector of) ........(1)
(Since and are parallel) .......(2)
Now from equation (1) and (2) we have
So
Now since and is a side of.
And since two sides and are equal, so
Hence is an isosceles triangle.
Page No 277:
Question 10:
ABC is a triangle in which ∠B = 2∠C. D is a point on BC such that AD bisects ∠BAC and AB = CD. Prove that ∠BAC = 72°.
Answer:
It is given that in
And bisects
We have to prove that
Now let
(Given)
Sinceis a bisector of so let
Let be the bisector of
If we join we have
In
So
In triangle and we have
(Given)
(Proved above)
So by congruence criterion, we have
And
, and (since)
In we have
Since,
And,
So,
In we have
Here,
HenceProved.
Page No 277:
Question 11:
Bisectors of angles B and C of an isosceles triangle ABC with AB = AC intersect each other at O. Show that external angle adjacent to ∠ABC is equal to ∠BOC.
Answer:
∠ABD is the external angle adjacent to ∠ABC.
âABC is an isosceles triangle.
AB = AC (Given)
∴ ∠C = ∠ABC .....(1) (In a triangle, equal sides have equal angles opposite to them)
Also, OB and OC are the bisectors of ∠B and ∠C, respectively.
Similarly,
In âBOC,
∠OBC + ∠OCB + ∠BOC = 180º (Angle sum property of triangle)
Now,
∠ABD + ∠ABC = 180º .....(5) (Linear pair)
From (4) and (5), we have
∠ABD + ∠ABC = ∠ABC + ∠BOC
⇒ ∠ABD = ∠BOC
Thus, the external angle adjacent to ∠ABC is equal to ∠BOC.
Page No 284:
Question 1:
In the given figure, it is given that AB = CD and AD = BC. Prove that ΔADC ≅ΔCBA.
Answer:
It is given that
We have to prove that.
Now in triangles and we have
(Given)
(Given)
So (common)
Each side of is equal to .
Hence, by congruence criterion we have Proved.
Page No 284:
Question 2:
In a ΔPQR, if PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP respectively. Prove that LN = MN.
Answer:
It is given that
and L, M, N are the mid points of sides, , and respectively.
We have to prove that
Now using the mid point theorem, we have
And
Similarly we have
In triangle and we have
(Proved above)
(Proved above)
And (common)
So, by congruence criterion, we have
And
Then
HenceProved.
Page No 287:
Question 1:
Which of the following statements are true (T) and which are false (F):
(i) Sides opposite to equal angles of a triangle may be unequal.
(ii) Angles opposite to equal sides of a triangle are equal.
(iii) The measure of each angle of an equilateral triangle is 60°.
(iv) If the altitude from one vertex of a triangle bisects the opposite side, then the triangle may be isosceles.
(v) The bisectors of two equal angles of a triangle are equal.
(vi) If the bisector of the vertical angle of a triangle bisects the base, then the triangle may be isosceles.
(vii) The two altitudes corresponding to two equal sides of a triangle need not be equal.
(viii) If any two sides of a right triangle are respectively equal to two sides of other right triangle, then the two triangles are congruent.
(ix) Two right triangles are congruent if hypotenuse and a side of one triangle are respectively equal equal to the hypotenuse and a side of the other triangle.
Answer:
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
Page No 287:
Question 2:
Fill in the blanks in the following so that each of the following statements is true.
(i) Sides opposite to equal angles of a triangle are .......
(ii) Angle opposite to equal sides of a triangle are .......
(iii) In an equilateral triangle all angles are ........
(iv) In a ABC if ∠A = ∠C, then AB = ............
(v) If altitudes CE and BF of a triangle ABC are equal, then AB = ........
(vi) In an isosceles triangle ABC with AB = AC, if BD and CE are its altitudes, then BD is ...... CE.
(vii) In right triangles ABC and DEF, if hypotenuse AB = EF and side AC = DE, then ΔABC ≅ Δ .........
Answer:
(1)
(2)
(3)
(4)
(5)
(6)
(7)
Page No 287:
Question 3:
In the given figure, AD ⊥ CD and CB ⊥ CD. If AQ = BP and DP = CQ, prove that
∠DAQ = ∠CBP.
Answer:
It is given that
, and
If and
We have to prove that
In triangles and we have
(Since given)
So
And (given)
So by right hand side congruence criterion we have
So
HenceProved.
Page No 287:
Question 4:
ABCD is a quadrilateral in which AB = BC and AD = CD. Show that BD bisects both the angles ∠ABC and ∠ADC.
Answer:
In quadrilateral ABCD, AB = BC and AD = CD.
In âABD and âCBD,
AB = CB (Given)
BD = BD (Common)
AD = CD (Given)
∴ âABD ≅ âCBD (SSS congruence criterion)
So, ∠ABD = ∠CBD .....(1) (CPCT)
∠ADB = ∠CDB .....(2) (CPCT)
From (1) and (2), we conclude that
BD bisects both ∠ABC and ∠ADC.
Page No 287:
Question 5:
ABCD is a square, X and Y are points on sides AD and BC respectively such that AY= BX. Prove that BY = AX and ∠BAY = ∠ABX.
Answer:
It is given ABCD is a square and
We have to prove that and
In right angled trianglesand Δ we have
And, and
So by right hand side congruence criterion we have
So (since triangle is congruent)
HenceProved.
Page No 287:
Question 6:
ABCD is a quadrilateral such that AB = AD and CB = CD. Prove that AC is the perpendicular bisector of BD.
Answer:
In quadrilateral ABCD, AB = AD and BC = CD. Let AC and BD intersect at O.
In âABC and âADC,
AB = AD (Given)
AC = AC (Common)
BC = CD (Given)
∴ âABC ≅ âADC (SSS congruence criterion)
⇒ ∠BAC = ∠DAC (CPCT)
Now, in âABO and âADO,
AB = AD (Given)
∠BAO = ∠DAO (Proved above)
AO = AO (Common)
∴ âABO ≅ âADO (SAS congruence axiom)
⇒ OB = OD .....(1) (CPCT)
∠AOB = ∠AOD .....(2) (CPCT)
Now,
∠AOB + ∠AOD = 180º (Linear pair)
⇒ 2∠AOB = 180º [Using (2)]
⇒ ∠AOB = 90º .....(3)
From (1) and (3), we conclude that AC is the perpendicular bisector of BD.
Page No 288:
Question 7:
O is a point in the interior of a square ABCD such that OAB is an equilateral triangle. Show that âOCD is an isosceles triangle.
Answer:
O is a point in the interior of square ABCD. âOAB is an equilateral triangle.
Now,
∠DAB = ∠CBA .....(1) (Measure of each angle of a square is 90º)
∠OAB = ∠OBA .....(2) (Measure of each angle of an equilateral triangle is 60º)
Subtracting (2) from (1), we get
∠DAB − ∠OAB = ∠CBA − ∠OBA
⇒ ∠OAD = ∠OBC
In âOAD and âOBC,
OA = OB (Sides of an equilateral triangle are equal)
∠OAD = ∠OBC (Proved above)
AD = BC (Sides of a square are equal)
∴ âOAD ≅ âOBC (SAS congruence axiom)
⇒ OD = OC (CPCT)
In âOCD,
OC = OD
∴ âOCD is an isosceles triangle. (A triangle whose two sides are equal is an isosceles triangle)
Page No 288:
Question 8:
Line segment joining the mid-points M and N of parallel sides AB and DC, respectively of a trapezium ABCD is perpendicular to both the sides AB and DC. Prove that AD = BC.
Answer:
ABCD is a trapezium with AB || CD. M and N are the mid-points of sides AB and AC, respectively.
Join AN and BN.
In âAMN and âBMN,
AM = BM (M is the mid-point of AB)
∠AMN = ∠BMN (MN ⊥ AB)
MN = MN (Common)
∴ âAMN ≅ âBMN (SAS congruence axiom)
So, AN = BN .....(1) (CPCT)
∠ANM = ∠BNM (CPCT)
Now,
∠DNM = ∠CNM (90º each)
∴ ∠DNM − ∠ANM = ∠CNM − ∠BNM
⇒ ∠AND = ∠BNC .....(2)
In âAND and âBNC,
DN = CN (N is the mid-point of CD)
∠AND = ∠BNC [From (2)]
AN = BN [From (1)]
∴ âAND ≅ âBNC (SAS congruence axiom)
So, AD = BC (CPCT)
Hence proved.
Page No 298:
Question 1:
Which of the following statements are true (T) and which are false (F)?
(i) Sum of the three sides of a triangle is less than the sum of its three altitudes.
(ii) Sum of any two sides of a triangle is greater than twice the median drawn to the third side.
(iii) Sum of any two sides of a triangle is greater than the third side.
(iv) Difference of any two sides of a triangle is equal to the third side.
(v) If two angles of a triangle are unequal, then the greater angle has the larger side opposite to it.
(vi) Of all the line segments that can be drawn from a point to a line not containing it, the perpendicular line segment is the shortest one.
Answer:
(1)
(2)
(3)
(4)
(5)
(6)
Page No 298:
Question 2:
Fill in the blanks to make the following statements true.
(i) In a right triangle the hypotenuse is the ...... side.
(ii) The sum of three altitudes of a triangle is ....... than its perimeter.
(iii) The sum of any two sides of a triangle is ..... than the third side.
(iv) If two angles of a triangle are unequal, then the smaller angle has the ..... side opposite to it.
(v) Difference of any two sides of a triangle is ...... than the third side.
(vi) If two sides of a triangle are unequal, then the larger side has ...... angle opposite to it.
Answer:
(1)
(2)
(3)
(4)
(5)
(6)
Page No 299:
Question 3:
In ΔABC, if ∠A = 40° and ∠B = 60°. Determine the longest and shortest sides of the triangle.
Answer:
In the triangle ABC it is given that
We have to find the longest and shortest side.
Here
Now is the largest angle of the triangle.
So the side in front of the largest angle will be the longest side.
Hence will be the longest
Since is the shortest angle so that side in front of it will be the shortest.
And is shortest side
Hence Is longest and is shortest.
Page No 299:
Question 4:
In a ΔABC, if ∠B = ∠C = 45°, which is the longest side?
Answer:
In the triangle ABC it is given that
We have to find the longest side.
Here
(Since)
Now is the largest angle of the triangle.
So the side in front of the largest angle will be the longest side.
Hence BC will be the longest side.
Page No 299:
Question 5:
In Δ ABC, side AB is produced to D so that BD = BC. If ∠B = 60° and ∠A = 70°, prove that :
(i) AD > CD (ii) AD>AC
Answer:
It is given that
, and
We have to prove that
(1)
(2)
(1)
Now
And since BD=BC, so, and
That is,
Now
And, so
Hence (1) (Side in front of greater angle will be longer)
And (2) Proved.
Page No 299:
Question 6:
Is a possible to draw a triangle with sides of length 2 cm, 3 cm and 7 cm?
Answer:
As we know that a triangle can only be formed if
The sum of two sides is greater than the third side.
Here we have 2 cm, 3 cm and 7 cm as sides.
If we add
(Since 5 is less than 7)
Hence the sum of two sides is less than the third sides
So, the triangle will not exist.
Page No 299:
Question 7:
In Δ ABC, ∠B = 35°, ∠C = 65° and the bisector of ∠BAC meets BC in P. Arrange AP, BP and CP in descending order.
Answer:
It is given that
AP is the bisector of
We have to arrange, and in descending order.
In we have
(As AP is the bisector of)
So (Sides in front or greater angle will be greater) ........(1)
In we have
(As AP is the bisector of)
Since,
So ..........(2)
Hence
From (1) & (2) we have
Page No 299:
Question 8:
In a quadrilateral ABCD. Prove that:
(i) CD + DA + AB + BC > 2AC
(ii) CD + DA + AB > BC
Answer:
(i) We have, quadrilateral ABCD
In
Also, In
Adding (1) and (2), we get
Hence proved.
(ii)
In
Also, In
Adding (1) and (2), we get
Subtracting AC from both sides, we get
Hence proved.
Page No 299:
Question 9:
D is any point on side AC of a âABC with AB = AC. Show that CD < BD.
Answer:
It is given that, D is any point on side AC of a âABC with AB = AC.
In âABC,
AB = AC (Given)
∴ ∠ACB = ∠ABC (In a triangle, equal sides have equal angles opposite to them)
Now, ∠ABC > ∠DBC
⇒ ∠ACB > ∠DBC (∠ACB = ∠ABC)
In âBCD,
∠DCB > ∠DBC
⇒ BD > CD (In a triangle, greater angle has greater side opposite to it)
Or CD < BD
Page No 299:
Question 10:
O is any point in the interior of Δ ABC. Prove that
(i) AB + AC > OB + OC
(ii) AB + BC + CA > OA + OB + OC
(iii) OA + OB + OC > (AB + BC + CA)
Answer:
It is given that, is any point in the interior of
We have to prove that
(1) Produced to meet at.
In we have
.........(1)
And in we have
.........(2)
Adding (1) & (2) we get
HenceProved.
(2) We have to prove that
From the first result we have
..........(3)
And
.........(4)
Adding above (4) equation
HenceProved.
(3) We have to prove that
In triangles, and we have
Adding these three results
HenceProved.
Page No 299:
Question 11:
Prove that in a triangle, other than an equilateral triangle, angle opposite to the longest side is greater than of a right angle.
Answer:
Let us consider where AC is the longest side.
Since AC is the longest side
We also know
Adding (1) and (2), we get
Adding on both sides, we get
Thus, it is proved that in a triangle, other than an equilateral triangle, the angle opposite to the longest side is greater than of a right angle.
Page No 300:
Question 1:
In two congruent triangles ABC and DEF, if AB = DE and BC = EF. Name the pairs of equal angles.
Answer:
It is given that
Since, the triangles ABC and DEF are congruent, therefore,
Page No 300:
Question 2:
In two triangles ABC and DEF, it is given that ∠A = ∠D, ∠B = ∠E and ∠C =∠F. Are the two triangles necessarily congruent?
Answer:
It is given that
For necessarily triangle to be congruent, sides should also be equal.
Page No 300:
Question 3:
If ABC and DEF are two triangles such that AC = 2.5 cm, BC = 5 cm, ∠C = 75°, DE = 2.5 cm, DF = 5cm and ∠D = 75°. Are two triangles congruent?
Answer:
It is given that
Since, two sides and angle between them are equal, therefore triangle ABC and DEF are congruent.
Page No 300:
Question 4:
In two triangles ABC and ADC, if AB = AD and BC = CD. Are they congruent?
Answer:
The given information and corresponding figure is given below
From the figure, we have
And,
Hence, triangles ABC and ADC are congruent to each other.
Page No 300:
Question 5:
In triangles ABC and CDE, if AC = CE, BC = CD, ∠A = 60°, ∠C = 30° and ∠D = 90°.
Are two triangles congruent?
Answer:
For the triangles ABC and ECD, we have the following information and corresponding figure:
In triangles ABC and ECD, we have
The SSA criteria for two triangles to be congruent are being followed. So both the triangles are congruent.
Page No 300:
Question 6:
ABC is an isosceles triangle in which AB = AC. BE and CF are its two medians. Show that BE = CF.
Answer:
In the triangle ABC it is given that
, and are medians.
We have to show that
To show we will show that
In triangle ΔBFC and ΔBEC
As, so
.........(1)
BC=BC (common sides) ........(2)
Since,
As F and E are mid points of sides AB and AC respectively, so
BF = CE ..........(3)
From equation (1), (2), and (3)
HenceProved.
Page No 300:
Question 7:
Find the measure of each angle of an equilateral triangle.
Answer:
In equilateral triangle we know that each angle is equal
So
Now (by triangle property)
Hence.
Page No 300:
Question 8:
CDE is an equilateral triangle formed on a side CD of a square ABCD. Show that ΔADE ≅ ΔBCE.
Answer:
We have to prove that
Given is a square
So
Now in is equilateral triangle.
So
In and
(Side of triangle)
(Side of equilateral triangle)
And,
So
Hence from congruence Proved.
Page No 300:
Question 9:
Prove that the sum of three altitudes of a triangle is less than the sum of its sides.
Answer:
We have to prove that the sum of three altitude of the triangle is less than the sum of its sides.
In we have
, and
We have to prove
As we know perpendicular line segment is shortest in length
Since
So ........(1)
And
........(2)
Adding (1) and (2) we get
........(3)
Now, so
.......(4)
And again, this implies that
........(5)
Adding (3) & (4) and (5) we have
HenceProved.
Page No 300:
Question 10:
In the given figure, if AB = AC and ∠B = ∠C. Prove that BQ = CP.
Answer:
It is given that
, and
We have to prove that
We basically will prove to show
In and
(Given)
(Given)
And is common in both the triangles
So all the properties of congruence are satisfied
So
Hence Proved.
Page No 301:
Question 1:
If AB = QR, BC = PR and CA = PQ, then triangle _________ ≅ triangle _________.
Answer:
It is given that, AB = QR, BC = PR and CA = PQ.
So, A ↔ Q, B ↔ R and C ↔ P
∴ âABC ≅ âQRP
If AB = QR, BC = PR and CA = PQ, then triangle __ABC__ ≅ triangle __QRP__.
Page No 301:
Question 2:
In triangles ABC and DEF, AB = FD and ∠A = ∠D. The two triangles will be congruent by SAS axiom, if __________.
Answer:
SAS congruence axiom states that two triangles are congruent if two sides and the included angle of one are equal to the corresponding sides and the included angle of the other triangle.
In âABC, ∠A is included between the sides AB and AC.
In âDEF, ∠D is included between the sides DF and DE.
∴ âABC ≅ âDEF by SAS axiom if AC = DE.
In triangles ABC and DEF, AB = FD and ∠A = ∠D. The two triangles will be congruent by SAS axiom, if __AC = DE__.
Page No 301:
Question 3:
In âABC and PQR, AB = AC, ∠C = ∠P and ∠B = ∠Q. The two triangles are __________ but not __________.
Answer:
If two sides of a triangle are equal, then it is an isosceles triangle.
In âABC, AB = AC
∴ âABC is an isosceles triangle.
In âABC,
AB = AC (Given)
∴ ∠C = ∠B (In a triangle, equal sides have equal angles opposite to them)
It is given that, ∠C = ∠P and ∠B = ∠Q.
∴ ∠P = ∠Q
In âPQR,
∠P = ∠Q (Proved)
∴ QR = PR (In a triangle, equal sides have equal angles opposite to them)
So, âPQR is an isosceles triangle.
However, it cannot be proved that the corresponding sides of âABC are congruent to the corresponding sides of âPQR. Hence, the triangles are not congruent.
In âABC and âPQR, AB = AC, ∠C = ∠P and ∠B = ∠Q. The two triangles are __isosceles__ but not __congruent__.
Page No 301:
Question 4:
In âPQR, ∠P = ∠R, QR = 4 cm and PR = 5 cm. Then PQ = ________.
Answer:
In âPQR,
∠P = ∠R (Given)
∴ QR = PQ (Sides opposite to the equal angles of a triangle are equal)
⇒ PQ = 4 cm (QR = 4 cm)
In âPQR, ∠P = ∠R, QR = 4 cm and PR = 5 cm. Then PQ = ___4 cm___.
Page No 301:
Question 5:
In âABC, AB = AC and ∠B = 50°. Then, ∠C = __________.
Answer:
In âABC,
AB = AC (Given)
∴ ∠C = ∠B (Angles opposite to the equal sides of a triangle are equal)
⇒ ∠C = 50° (∠B = 50°)
In âABC, AB = AC and ∠B = 50°. Then, ∠C = __50°__.
Page No 301:
Question 6:
In âABC, AB = BC and ∠B = 80°. Then ∠A = __________.
Answer:
In âABC,
AB = BC (Given)
∴ ∠C = ∠A ..... (1) (Angles opposite to the equal sides of a triangle are equal)
Now,
∠A + ∠B + ∠C = 180° (Angle sum property of triangle)
∴ ∠A + 80° + ∠A = 180°
⇒ 2∠A = 180° − 80° = 100°
⇒ ∠A = = 50°
In âABC, AB = BC and ∠B = 80°. Then ∠A = __50°__.
Page No 301:
Question 7:
If in âPQR, ∠P = 70° and ∠R = 30°, then the longest side of âPQR is ___________.
Answer:
In âPQR,
(Angle sum property of a triangle)
So, ∠Q is the greatest angle in the âPQR.
We know that, in a triangle the greater angle has the longer side opposite to it.
∴ PR is the longest side of âPQR.
If in âPQR, ∠P = 70° and ∠R = 30°, then the longest side of âPQR is ___PR___.
Page No 301:
Question 8:
In âPQR, if ∠R > ∠Q, then __________.
Answer:
In âPQR,
∠R > ∠Q (Given)
∴ PQ > PR (In a triangle, the greater angle has the longer side opposite to it)
In âPQR, if ∠R > ∠Q, then ___PQ > PR___.
Page No 301:
Question 9:
D is a point on side BC of a âABC such that AD bisects ∠BAC. Then ________.
Answer:
In âABC, AD is the bisector of ∠A.
∴ ∠BAD = ∠CAD .....(1)
We know that exterior angle of a triangle is greater than each of interior opposite angle.
In âABD,
∠ADC > ∠BAD
⇒ ∠ADC > ∠CAD [Using (1)]
In âADC,
∠ADC > ∠CAD
∴ AC > CD (In a triangle, the greater angle has the longer side opposite to it)
Similarly, AB > BD
D is a point on side BC of a âABC such that AD bisects ∠BAC. Then ___AC > CD and AB > BD___.
Page No 301:
Question 10:
Two sides of a triangle are of lengths 5 cm and 1.5 cm. The length of the third side of the triangle lies between _________ and _________.
Answer:
In a triangle, the sum of two sides is greater than the third side.
∴ 5 cm + 1.5 cm > Third side
Or Third side < 6.5 cm
In a triangle, the difference of two sides is less than the third side.
∴ 5 cm − 1.5 cm < Third side
Or Third side > 3.5 cm
So,
3.5 cm < Third side < 6.5 cm
Thus, the length of the third side of the triangle lies between 3.5 cm and 6.5 cm.
Two sides of a triangle are of lengths 5 cm and 1.5 cm. The length of the third side of the triangle lies between __3.5 cm__ and _6.5 cm _.
Page No 301:
Question 11:
If AD is a median of âABC, then the perimeter of âABC cannot be less than or equal to ___________.
Answer:
AD is the median of the âABC.
In âABD,
AB + BD > AD .....(1) (In a triangle, the sum of any two sides is greater than the third side)
In âACD,
CD + CA > AD .....(2) (In a triangle, the sum of any two sides is greater than the third side)
Adding (1) and (2), we get
AB + BD + CD + CA > AD + AD
⇒ AB + BC + CA > 2AD (BC = BD + CD)
⇒ Perimeter of âABC > 2AD
Or the perimeter of âABC cannot be less than or equal to 2AD
If AD is a median of âABC, then the perimeter of âABC cannot be less than or equal to ___2AD___.
Page No 301:
Question 12:
If âPQR ≅ âEDF, then PR = ___________.
Answer:
If âPQR ≅ âEDF, then
P ↔ E, Q ↔ D and R ↔ F
So, PR = EF, PQ = ED and QR = DF (If two triangles are congruent, then their corresponding sides are congruent)
If âPQR ≅ âEDF, then PR = ____EF____.
Page No 301:
Question 13:
It is given that âABC ≅ âRPQ, then BC = __________.
Answer:
If âABC ≅ âRPQ, then
A ↔ R, B ↔ P and C ↔ Q
So, AB = RP, BC = PQ and CA = QR (If two triangles are congruent, then their corresponding sides are congruent)
It is given that âABC ≅ âRPQ, then BC = ___PQ___.
Page No 301:
Question 14:
In âABC and âPQR, if ∠A = ∠Q, ∠B = ∠R and PR = AC, then two triangles are congruent by __________ criterion.
Answer:
In âABC and âPQR,
∠A = ∠Q
∠B = ∠R
AC = PR
∴ âABC ≅ âQRP (AAS congruence criterion)
AAS congruence criterion states that if any two angles and a non-included side of one triangle are equal to the corresponding angles and side of another triangle, then the two triangles are congruent.
In âABC and âPQR, if ∠A = ∠Q, ∠B = ∠R and PR = AC, then two triangles are congruent by __AAS congruence__ criterion.
Page No 301:
Question 15:
In âABC and âPQR, ∠A = ∠Q, ∠B = ∠R and AB = QR, then these triangles are congruent by _______ criterion.
Answer:
In âABC and âPQR,
∠A = ∠Q
∠B = ∠R
AB = QR
∴ âABC ≅ âQRP (ASA congruence criterion)
ASA congruence criterion states that two triangles are congruent if two angles and the included side of one triangle are equal to the corresponding two angles and the included side of the other triangle.
In âABC and âPQR, ∠A = ∠Q, ∠B = ∠R and AB = QR, then these triangles are congruent by __ASA congruence criterion__.
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