Rs Aggarwal 2017 for Class 9 Math Chapter 5 - Congruence Of Triangles And Inequalities In A Triangle

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Page No 187:

Question 1:

In a ∆ABC, if AB = AC and ∠A = 70°, find ∠B and ∠C.

Answer:

$∆ ABC is an isosceles triangle, i . e ., AB = AC . ∴ ∠ ABC = ∠ ACB In ∆ ABC, let ∠ B = ∠ C = x Now, ∠ A + ∠ B + ∠ C = 180 ° (Angle sum property) \Rightarrow 70 + x + x = 180 \Rightarrow 2 x = 110 \Rightarrow x = 55 ° ∴ ∠ B = ∠ C = 55 °$

Page No 187:

Question 2:

The vertical angle of an isosceles triangle is 100°. Find its base angles.

Answer:

Let the base angles be $x °$ .
The sum of the angles of a triangle is 180 $°$ .

$i . e ., 100 + x + x = 180 \Rightarrow 2 x = 180 - 100 \Rightarrow 2 x = 80 \Rightarrow x = 40 °$

∴ The base angles measure $40 °$ .

Page No 187:

Question 3:

In a ∆ABC, if AB = AC and ∠B = 65°, find ∠C and ∠A.

Answer:

$In ∆ ABC, AB = AC \Rightarrow ∠ ABC = ∠ ACB ∴ ∠ ACB = 65 ° Let ∠ A be x ° . Now, ∠ A + ∠ B + ∠ C = 180 ° \Rightarrow x + 65 ° + 65 ° = 180 ° \Rightarrow x = 180 ° - 130 ° \Rightarrow x = 50 ∴ ∠ C = 65 ° and ∠ A = 50 °$

Page No 187:

Question 4:

In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle.

Answer:

In an isosceles triangle, the base angles are equal to each other.
Let the base angles be $x$ $°$ .

The vertex angle is $2 (x + x) = 2 (2 x) = 4 x$ ^o
Since the sum of the angles of a triangle is 180^o, we have:
$x + x + 4 x = 180 ° \Rightarrow 6 x = 180 ° \Rightarrow x = \frac{180}{6} = 30 ° ∴ 4 x = 4 (30) = 120 °$

Thus, the base angles measure 30^o and the vertex angle measures 120^o.

Page No 187:

Question 5:

What is the measure of each of the equal angles of a right-angled isosceles triangle?

Answer:

In a right isosceles triangle, one angle is a right angle and the the other two angles are equal.
Let the equal angles measure x.

$Now, x + x + 90 ° = 180 ° (Angle sum property) \Rightarrow 2 x = 180 ° - 90 ° \Rightarrow x = 45 °$

Thus, the equal angles of the right isosceles triangle measure 45^o.

Page No 187:

Question 6:

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:

Given:
Triangle ABC is an isosceles triangle.
Or AB =AC
Or $∠ A B C = ∠ A C B (Since its an isosceles triangle)$
Now, we have:
$x + y = 180 ° . . . (1) (Linear pair) x + z = 180 ° . . . (2) (Linear pair)$
Subtracting (1) and (2), we get:
$y - z = 0 \Rightarrow y = z$
Thus, the exterior angles so formed are equal to each other.
Hence, proved.

Page No 187:

Question 7:

Find the measure of each exterior angle of an equilateral triangle.

Answer:

$In equilateral △ ABC, ∠ A = ∠ B = ∠ C = 60 ° .$
Let the exterior angle of $∆ A B C b e ∠ A C D$ .

$Now, ∠ A C D = ∠ B + ∠ A \Rightarrow ∠ A C D = 60 ° + 60 ° \Rightarrow ∠ A C D = 120 °$

Similarly, each exterior angle would be equal to $120 °$ .
Thus, all the exterior angles of an equilateral triangle are $120 °$ .

Page No 187:

Question 8:

In the given figure, O is the midpoint of each of the line segments AB and CD. Prove that AC = BD and AC || BD.

Answer:

To prove:

$A C = B D A C ∥ B D$

$In ∆ AOC and ∆ BOD, we have : OA = OB (O is the midpoint) ∠ AOC = ∠ BOD (Vertically opposite angles) OC = OD (O is the midpoint) ∴ ∆ AOC ≅ ∆ BOD (SAS criterion)$

$Also, ∠ CAO = ∠ OBD (CPCT)$
Here, $AC ∥ BD (Alternate angles formed by transversal AB)$
∴ AC = BD
Hence, proved.

Page No 187:

Question 9:

In the adjoining figure, PA ⊥ AB, QB ⊥ AB and PA = QB. If PQ intersects AB at O, show that O is the midpoint of AB as well as that of PQ.

Answer:

$In △ AOP and △ BOQ, we have : ∠ AOP = ∠ BOQ (Vertically opposite angle) ∠ OAP = ∠ OBQ (90 ° each) PA = QB (Given) △ AOP ≅ △ BOQ (AAS criterion)$
$\Rightarrow OA = OB (CPCT) \Rightarrow O P = O Q (CPCT)$
So, O is midpoint of AB and PQ.

Page No 187:

Question 10:

Let the line segments AB and CD intersect at O in such a way that OA = OD and OB = OC. Prove that AC = BD but AC may not be parallel to BD.

Answer:

$In △ OAC and △ ODB, we get : OA = OD (Given) OC = OB (Given) ∠ AOC = ∠ BOD (Vertically opposite angle) △ OAC ≅ △ ODB (SAS criterion)$
$i . e ., AC = BD (CPCT)$

However, $∠ O A C$ is not equal to $∠ O B D$ and $∠ O C A$ is not equal to $∠ O D B$ .
Thus, AC may not be parallel to BD.
Hence, proved.

Page No 188:

Question 11:

In the given figure, l || m and M is the midpoint of AB. Prove that M is also the midpoint of any line segment CD having its end points at l and m respectively.

Answer:

Consider the triangles AMC and BMD.
Since lines l and m are parallel, we have:
$∠ A C D = ∠ M D B (Alternative angles)$
$∠ A M C = ∠ B M D (Vertically opposite angles)$
It is also given that AM = MB.

$∆ A M C ≅ ∆ B M D$ (AAS criterion)
i.e., DM = MC (CPCT)
∴ M is the mid point of segment CD.
Hence, proved.

Page No 188:

Question 12:

In the given figure, AB = AC and OB = OC. Prove that ∠ABO = ∠ACO.

Answer:

Given: AB = AC
$\Rightarrow ∠ B = ∠ C$ ...(i)

$Also, O B = O C \Rightarrow ∠ O B C = ∠ O C B . . . (ii)$

Subtracting (ii) from (i), we get:
$∠ B - ∠ O B C = ∠ C - ∠ O C B$
∴ $∠ A B O = ∠ A C O$
Hence, proved.

Page No 188:

Question 13:

In the given figure, ABC is a triangle in which AB = AC and D is a point on AB. Through D, a line DE is drawn parallel to BC and meeting AC at E. Prove that AD = AE.

Answer:

In the given figure, $D E | | B C$ .
$∠ A D E = ∠ A B C ∠ A E D = ∠ A C B$ (Corresponding angles) ...(i)

But in triangle ABC, AB = AC.
$∴ ∠ A B C = ∠ A C B$ ...(ii)

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

$∠ A D E = ∠ A B C ∠ A E D = ∠ A B C$
$\Rightarrow ∠ A D E = ∠ A E D$

Therefore, triangle ADE is also isosceles.
Or AD = AE
Hence, proved.

Page No 188:

Question 14:

In the adjoining figure, X and Y are respectively two points on equal sides AB and AC of ∆ABC such that AX = AY. Prove that CX = BY.

Answer:

$In △ AXC and △ AYB, we have : AC = AB (Given) AX = AY (Given) ∠ BAC = ∠ CAB (Angle common to △ AXC and △ AYB) ∴ △ AXC ≅ △ AYB (SAS criterion) So, CX = BY (CPCT)$
Hence, proved.

Page No 188:

Question 15:

In the given figure, C is the midpoint of AB. If ∠DCA = ∠ECB and ∠DBC = ∠EAC, prove that DC = EC.

Answer:

Given: C is the midpoint of AB.
Or AC = CB

$Also, ∠ D C A = ∠ E C B (Given) \Rightarrow ∠ D C A + ∠ D C E = ∠ E C B + ∠ D C E \Rightarrow ∠ A C E = ∠ B C D . . . (i)$

$Now, in △ DBC and △ EAC, we have : ∠ DBC = ∠ EAC (Given) BC = AC (Given) ∠ BCD = ∠ ACE [From (i)] i . e ., △ DBC ≅ △ EAC (ASA criterion)$
∴ DC = EC (CPCT)
Hence, proved.

Page No 188:

Question 16:

In the given figure, BA ⊥ AC and DE ⊥ EF such that BA =DE and BF = DC. Prove that AC = EF.

Answer:

$We have B F = D C \Rightarrow B F + C F = D C + C F \Rightarrow B C = D F . . . (i)$

$In △ ABC and △ EDF, we have : BA = DE (Given) ∠ BAC = ∠ DEF (90 ° each) BC = DF [From (i)] Thus, △ ABC ≅ △ EDF (RHS criterion) ∴ AC = EF (CPCT)$
Hence, proved.

Page No 188:

Question 17:

In the given figure, if x = y and AB = CB, then prove that AE = CD.

Answer:

Consider the triangles AEB and CDB.

$∠ E B A = ∠ D B C$ (Common angle) ...(i)

Further, we have:

$∠ BEA = 180 - y ∠ BDC = 180 - x Since x = y, we have : 180 - x = 180 - y \Rightarrow ∠ BEA = ∠ BDC . . . (ii)$

AB = CB (Given) ...(iii)
From (i), (ii) and (iii), we have:
$△ B D C ≅ △ B E A$ (AAS criterion)
∴ AE = CD (CPCT)
Hence, proved.

Page No 189:

Question 18:

ABC is a triangle in which AB = AC. If the bisectors of ∠B and ∠C meet AC and AB in D and E respectively, prove that BD = CE.

Answer:

In triangle ABC, we have:
AB = AC (Given)
$\Rightarrow ∠ B = ∠ C \Rightarrow \frac{1}{2} ∠ B = \frac{1}{2} ∠ C ∴ ∠ ABD = ∠ ACE . . . (i)$

$Now, in △ ADB and △ ACE, we have : ∠ ABD = ∠ ACE [From (i)] AB = BC (Given) ∠ BAD = ∠ CAE (Common angle) ∴ △ ADB ≅ △ AEC (ASA criterion)$

∴ $B D = C E (Corresponding sides of congruent triangles)$
Hence, proved.

Page No 189:

Question 19:

In the adjoining figure, AD is a median of ∆ABC. If BL and CM are drawn perpendiculars on AD and AD produced, prove that BL = CM.

Answer:

$In △ BLD and △ CMD, we have : ∠ BLD = ∠ CMD (90 ° each) BD = CD (AD is median, i . e ., D is the midpoint of BC) ∠ BDL = ∠ CDM (Vertically opposite angle) △ BLD ≅ △ CMD (AAS criterion) ∴ BL = CM (Corresponding sides of congruent triangles)$
Hence, proved.

Page No 189:

Question 20:

In ∆ABC, D is the midpoint of BC. If DL ⊥ AB and DM ⊥ AC such that DL = DM, prove that AB = AC.

Answer:

$In △ BDL and △ CDM, we have : BD = CD (D is midpoint) DL = DM (Given) ∠ BLD = ∠ CMD (90 ° each) Thus, △ BDL ≅ △ CDM (RHS criterion)$
$\Rightarrow BL = MC (CPCT)$
∴ $∠ B = ∠ C$
This makes triangle ABC an isosceles triangle.
Or AB = AC
Hence, proved.

Page No 189:

Question 21:

In ∆ABC, AB = AC and the bisectors of ∠B and ∠C meet at a point O. Prove that BO = CO and the ray AO is the bisector of ∠A.

Answer:

In triangle ABC, we have:
AB = AC (Given)
$\Rightarrow ∠ B = ∠ C \Rightarrow \frac{1}{2} ∠ B = \frac{1}{2} ∠ C \Rightarrow ∠ O B C = ∠ O C B \Rightarrow B O = C O$

$Now, in △ A O B and △ A O C, we have : O B = O C (Proved) A B = A C (Given) A O = A O (Common) ∴ △ A O B ≅ △ A O C (SSS criterion)$
i.e., $∠ B A O = ∠ C A O (Corresponding angles of congruent triangles)$

So, it shows that ray AO is the bisector of $∠ A$ .
Hence, proved.

Page No 189:

Question 22:

In the given figure, PQR is an equilateral triangle and QRST is a square. Prove that (i) PT = PS, (ii) ∠PSR = 15°.

Answer:

Given:
Triangle PQR is an equilateral triangle, i.e., each angle is $60 °$ .
QRST is a square, i.e., each angle is $90 °$ .

$∠ P Q T = ∠ P Q R + ∠ T Q R = 60 + 90 = 150 °$
Similarly, $∠ P R S = 150 °$
Consider the triangles PQT and PRS.

$PQ = PR (Sides of an equilateral triangle) QT = RS (Sides of a square) ∠ PQT = ∠ PRS = 150 ° (Proved) △ PQT ≅ △ PRS (SAS criterion) PT = PS (CPCT)$

Further, the side of the square coincides with the triangle, PR = RS, which makes PRS isosceles.
Let $∠ P S R = ∠ S P R = x$ .
Using angle sum property of a triangle, we get:
$150 ° + x + x = 150 ° + 2 x = 180 ° \Rightarrow 2 x = 30 ° \Rightarrow x = 15 °$

∴ $∠ P S R = 15 °$
Hence, proved.

Page No 189:

Question 23:

In the given figure,, ABC is a triangle, right angled at B. If BCDE is a square on side BC and ACFG is a square on AC, prove that AD = BF.

Answer:

$In △ ACD and △ FCB, we have : ∠ ACD = 90 ° + ∠ BCA and ∠ FCB = 90 ° + ∠ BCA$
∴ $∠ A C D = ∠ F C B$
$CF = CA (Sides of square) CD = CB (Sides of square) △ ACD ≅ △ FCB (SAS criterion)$
AD = BF (CPCT)
Hence, proved.

Page No 190:

Question 24:

Prove that the median from the vertex of an isosceles triangle is the bisector of the vertical angle.

Answer:

The triangle ABC shown in the figure is an isosceles triangle.
∴ AB = AC and $∠ A B C = ∠ A C B$

AD is the median drawn from vertex A to BC.
i.e., BD = DC

To prove: $∠ B A D = ∠ C A D$

Consider the triangles BAD and CAD.

$AB = AC (△ ABC is isosceles) ∠ ABD = ∠ ACD (△ ABC is isosceles) BD = CD (AD is a median)$

∴ $△ A B D ≅ △ A C D$ (SAS criterion)

Since the corresponding angles of the congruent triangles are congruent, $∠ B A D = ∠ C A D$ .

∴ The median from the vertex of an isosceles triangle bisects that angle.

Page No 190:

Question 25:

In the given figure, ABCD is a quadrilateral in which AB || DC and P is the midpoint of BC. On producing, AP and DC meet at Q. Prove that (i) AB = CQ, (ii) DQ = DC + AB.

Answer:

$In △ ABP and △ QCP, we have : ∠ BPA = ∠ CPQ (Vertically opposite angle) ∠ PAB = ∠ PQC (Alternate angles) PB = PC (P is the midpoint) △ ABP ≅ △ QCP (AAS criterion)$

∴ AB = CQ (CPCT)

$DQ = DC + CQ \Rightarrow DQ = DC + AB (Proved, AB = CQ)$
Hence, proved.

Page No 190:

Question 26:

In the given figure, OA = OB and OP = OQ. Prove that (i) PX = QX, (ii) AX = BX.

Answer:

Proof:
$In △ OQA and △ OPB, we have : OQ = OP (Given) OA = OB (Given) ∠ AOQ = ∠ BOP (Common) △ OQA ≅ OPB (SAS criterion)$
$∠ O A Q = ∠ O B P (Corresponding angles of congruent triangles)$
Now, consider triangles BQX and APX.
$Given : OA = OB OP = OQ ∴ OA - OP = OB - OQ \Rightarrow AP = BQ$
Further, $∠ BXQ = ∠ AXP (Vertically opposite angles)$
Also, we have proven that $∠ Q B X = ∠ P A X$ .
$∆ BQX ≅ ∆ APX (AAS criterion)$

$∴ PX = QX (corresponding sides of congruent triangles)$
$Also, AX = BX (corresponding sides of congruent triangles)$
Hence, proved.

Page No 190:

Question 27:

In the given figure, ABCD is a square and P is a point inside it such that PB = PD. Prove that CPA is a straight line.

Answer:

$In △ PAD and △ PAB, we have : AD = AB (Side of a square) AP = AP (Common) PD = PB (Given) △ PAD ≅ PAB (SSS criterion) \Rightarrow ∠ APD = ∠ APB$

$In △ CPD and △ CPB, we have : CD = CB (Sides of square) CP = CP (Common) PD = PB (Given) △ CPD ≅ △ CPB (SSS test) \Rightarrow ∠ CPD = ∠ CPB$

$∴ ∠ APD + ∠ CPD = ∠ APB + ∠ CPB But ∠ APD + ∠ CPD + ∠ APB + ∠ CPB = 360 ° \Rightarrow ∠ APD + ∠ CPD = 180 °$

So, CPA is a straight line.
Hence, proved.

Page No 190:

Question 28:

In the given figure, ABC is an equilateral triangle; PQ || AC and AC is produced to R such that CR = BP. Prove that QR bisects PC.

Answer:

$Since ∆ ABC is an equilateral ∆, then ∠ ABC = ∠ BCA = ∠ CAB = 60 ° Since PQ ∥ CA and PC is a transversal, then ∠ BPQ = ∠ BCA = 60 ° (Corresponding angles) Since PQ ∥ CA and QA is a transversal, then ∠ BQP = ∠ BAC = 60 ° (Corresponding angles) Further, ∠ B = 60 ° In ∆ BPQ, ∠ B = ∠ BPQ = ∠ BQP = 60 ° ∴ △ BPQ is an equilateral triangle . i . e ., BP = PQ = BQ Now, BP = CR \Rightarrow PQ = CR . . . . (1) Considering △ MPQ and △ MCR, we get : ∠ PQM = ∠ MRC (Alternate interior angles) ∠ PMQ = ∠ CMR (Vertically opposite angles) PQ = CR [using (1)] ∆ MPQ ≅ ∆ MCR (AAS criterion) \Rightarrow MP = MC [Corresponding parts of congruent triangles are equal] \Rightarrow QR bisects PC .$

Page No 190:

Question 29:

In the given figure, ABCD is a quadrilateral in which AB = AD and BC = DC. Prove that (i) AC bisects ∠A and ∠C, (ii) AC is the perpendicular bisector of BD.

Answer:

Consider the triangles ABC and ADC.
$AB = AD (Given) BC = BD (Given) AC = AC (Common) ∴ △ ABC ≅ △ ADC (SSS criterion) \Rightarrow ∠ BAC = ∠ DAC (Corresponding angles of congruent triangles) \Rightarrow ∠ BCA = ∠ DCA (Corresponding angles of congruent triangles)$

Thus, AC bisects angles A and C.

Let AC intersect BD at O. Consider the triangles AOB and AOD.
$∠ BAO = ∠ DAO (Proved) AB = AD (Given) Thus, △ ABD is an isosceles traingle . i . e ., ∠ ABO = ∠ ADO \Rightarrow △ AOB ≅ △ AOD (AAS criterion) \Rightarrow BO = OD$
Thus, AC bisects BD at O.

$Further, ∠ BOA = ∠ DOA (Corresponding angles of congruent triangles) Also, ∠ BOA + ∠ DOA = 180 ° \Rightarrow ∠ BOA = ∠ DOA = 90 °$

Thus, AC is the perpendicular bisector of BD.

Page No 190:

Question 30:

In the given figure, the bisectors of ∠B and ∠C of ∆ABC meet at I. If IP ⊥ BC, IQ ⊥ CA and IR ⊥ AB, prove that (i) IP = IQ = IR, (ii) IA bisects ∠A.

Answer:

$In △ IPC and △ IQC, we have : ∠ IPC = ∠ IQC . . . (90 ° each)$

$∠ PCI = ∠ QCI (CI is the bisector of angle C) IC = IC (Common side) Thus, ∆ IPC ≅ △ IQC (AAS criterion)$

$\Rightarrow IP = IQ (Corresponding parts of congruent triangles) . . . . . . (1)$

$Similarly, in △ IPB and △ IRB, we have : ∠ IPB = ∠ IRB (90 ° each)$

$∠ PBI = ∠ RBI (BI is the bisector of angle B) IB = IB (Common side) Thus, ∆ IPB ≅ △ IRB (AAS criterion)$

$\Rightarrow IP = IR (Corresponding parts of congruent triangles) . . . . . (2)$

From (1) and (2), we have,

IP = IQ = IR

Now consider the right triangles, ARI and AQI.

$IR = IQ (Proved above) AI = AI (Common side) ∠ ARI = ∠ AQI (90 ° each) ∆ ARI ≅ ∆ AQI (RHS criterion) \Rightarrow ∠ RAI = ∠ QAI (Corresponding parts of congruent triangles)$

Thus, IA bisects angle A.

Page No 191:

Question 31:

In the adjoining figure, P is a point in the interior of ∠AOB. If PL ⊥ OA and PM ⊥ OB such that PL = PM, show that OP is the bisector of ∠AOB.

Answer:

$In △ OMP and △ OLP, we have : OP = OP (Common) PL = PM (Given) ∠ BMP = ∠ PLA = 90 ° \Rightarrow △ OMP ≅ ∆ OPL (RHS criterion)$

i.e., $∠ P O M = ∠ P O L$

Thus, OP is the bisector of $∠ A O B$ .

Page No 191:

Question 32:

In the given figure, ABCD is a square, M is the midpoint of AB and PQ ⊥ CM meets AD at P and CB produced at Q. Prove that (i) PA = BQ, (ii) CP = AB + PA.

Answer:

Proof:
$In △ AMP and △ BMQ, we have : MA = MB (M is midpoint) ∠ AMP = ∠ BMQ (Vertically opposite angle s) ∠ PAM = ∠ QBM (90 ° each) ∴ △ AMP ≅ △ BMQ (ASA criterion)$
$\Rightarrow P A = B Q (CPCT) . . . (i)$
and $M P = M Q$ (CPCT)

$Now, in △ CMP and △ CMQ, we have : CM = CM (Common) ∠ CMP = ∠ CMQ (90 ° each) MP = MQ (Proved) ∴ △ CMP ≅ △ CMQ (SAS criterion)$

Thus, $CP = CQ = CB + BQ (Corresponding sides of congruent triangles)$
But $C B = A B (Sides of a square)$
$B Q = P A [From (i)]$
∴ $C P = A B + P A$
Hence, proved.

Page No 191:

Question 33:

In the adjoining figure, explain how one can find the breadth of the river without crossing it.

Answer:

Let AB be the breadth of the river.
M is any point situated on the bank of the river.
Let O be the mid point of BM.
Moving along perpendicular to point such that A,O and N are in a straight line.
Then MN is the required breadth of the river.

$In △ OBA and △ OMN, we have : OB = OM (O is midpoint) ∠ OBA = ∠ OMN (Each 90 °) ∠ AOB = ∠ NOM (Vertically opposite angle) ∴ △ OBA ≅ △ OMN (ASA criterion)$

Thus, MN = AB (CPCT)
If MN is known, one can measure the width of the river without actually crossing it.

Page No 191:

Question 34:

In ∆ABC, if ∠A = 36° and ∠B = 64°, name the longest and shortest sides of the triangle.

Answer:

In triangle ABC, $∠ A = 36 °, ∠ B = 64 °$ .

$Now, ∠ A + ∠ B + ∠ C = 180 ° (Angle sum property) \Rightarrow 36 ° + 64 ° + ∠ C = 180 ° \Rightarrow ∠ C = 180 ° - 100 ° \Rightarrow ∠ C = 80 °$

The side opposite the largest angle is the greatest and the side opposite the smallest angle is the shortest.
The side opposite to $∠ A$ , BC, is the shortest side of the triangle and the side opposite to $∠ C$ , AB, the longest side of the triangle.

Page No 191:

Question 35:

In ∆ABC, if ∠A = 90°, which is the longest side?

Answer:

$I n ∆ A B C,$ $∠ A = 90 °$ .
The sum of the other two angles must be such that
$∠ A + ∠ B + ∠ C = 180 ° \Rightarrow ∠ B + ∠ C = 90 ° (∵ ∠ A = 90 °)$

Thus, angles B and C both measure less than 90 $°$ .
Angle A is the largest in the triangle and the side opposite to it, i.e., BC is the longest.
∴ BC is the hypotenuse of the right triangle ABC.

Page No 191:

Question 36:

In ∆ABC, if ∠A = ∠B = 45° name the longest side.

Answer:

In $∆ A B C$ , we have:
$∠ A = ∠ B = 45 ° (Given)$

$Further, ∠ A + ∠ B + ∠ C = 180 ° \Rightarrow 45 ° + 45 ° + ∠ C = 180 ° \Rightarrow ∠ C = 180 ° - 45 ° - 45 ° = 90 °$

Thus, the side opposite to $∠ C$ , i.e., AB, is the longest side.

The side AB is the hypotenuse of triangle ABC.

Page No 191:

Question 37:

In ∆ABC, side AB is produced to D such that BD = BC. If ∠B = 60° and ∠A = 70°, prove that (i) AD > CD and (ii) AD > AC.

Answer:

In triangle CBA, CBD is an exterior angle.

$i . e ., ∠ C B A + ∠ C B D = 180 ° \Rightarrow 60 ° + ∠ C B D = 180 ° \Rightarrow ∠ C B D = 120 °$

Triangle BCD is isosceles and BC = BD.
Let $∠ B C D = ∠ B D C = x °$ .
$In △ C B D, we have : \Rightarrow ∠ B C D + ∠ C B D + ∠ C D B = 180 ° \Rightarrow x + 120 ° + x = 180 \Rightarrow 2 x = 60 ° \Rightarrow x = 30 ° ∴ ∠ B C D = ∠ B D C = 30 °$

In triangle ADC, $∠ C = ∠ A C B + ∠ B C D = 50 ° + 30 ° = 80 °$
$∠ A = 70 ° a n d ∠ D = 30 °$

$∴ ∠ C > ∠ A \Rightarrow A D > C D . . . (1)$

$Also, ∠ C > ∠ D \Rightarrow A D > A C . . . (2)$

Page No 191:

Question 38:

In ∆ABC, ∠B = 35°, ∠C = 65° and the bisector of ∠BAC meets BC in X. Arrange AX, BX and CX in descending order.

Answer:

To arrange AX, BX and CX in descending order, we have to find the measure of the angle opposite of which these sides lie.

In triangle ABC, we have:

$∠ A + ∠ B + ∠ C = 180 ° \Rightarrow ∠ A + 35 ° + 65 ° = 180 ° \Rightarrow ∠ A = 80 °$

Since AX bisects angle BAC, we have:
$∠ B A X = 40 °$

So, in triangle AXC, we have:
$∠ A X C + 40 ° + 65 ° = 180 ° \Rightarrow ∠ A X C = 75 °$
Thus, AX > CX ...(i)

In triangle AXB, we have:
$∠ A X B + 35 ° + 40 ° = 180 ° \Rightarrow ∠ A X B = 105 °$
Thus, BX > AX ...(ii)

From (i) and (ii), we get:
$B X > A X > C X$

Page No 191:

Question 39:

In ∆ABC, if AD is the bisector of ∠A, show that AB > BD and AC > DC.

Answer:

$∠ ADB = ∠ CAD + ∠ ACD (Exterior angle property) \Rightarrow ∠ ADB > ∠ CAD \Rightarrow ∠ ADB > ∠ BAD [∵ ∠ CAD = ∠ BAD]$
Thus, AB > BD

$Similarly, ∠ A D C > ∠ B A D \Rightarrow ∠ A D C > ∠ C A D$
Thus, AC > DC
Hence, proved.

Page No 191:

Question 40:

In the given figure, ABC is a triangle in which AB = AC. If D be a point on BC produced, prove that AD > AC.

Answer:

$A B = A C ∠ A B C = ∠ A C B$

$As ∠ A C B > ∠ A D C, we have : ∠ A B C > ∠ A D B \Rightarrow ∠ A B D > ∠ A D B$

∴ AD > AB

AB =AC (Given)
∴ AD > AC
Hence, proved.

Page No 192:

Question 41:

In the adjoining figure, AC > AB and AD is the bisector of ∠A. Show that ∠ADC > ∠ADB.

Answer:

Given:
$A C > A B \Rightarrow ∠ B > ∠ C ∴ ∠ B + ∠ B A D > ∠ C + ∠ C A D (∵ ∠ B A D = ∠ C A D) . . . (i)$

In triangles ABD and ADC, the sums of the respective angles will be 180 $°$ .

$∠ B + ∠ B A D + ∠ B D A = ∠ C + ∠ A D C + ∠ D A C = 180 ° \Rightarrow ∠ B D A = 180 ° - ∠ B - ∠ B A D and \Rightarrow ∠ A D C = 180 ° - ∠ C - ∠ C A D$

Using (i), we get:

$∠ B + ∠ B A D > ∠ C + ∠ C A D \Rightarrow 180 ° - ∠ B - ∠ B A D < 180 ° - ∠ C - ∠ C A D \Rightarrow ∠ B D A < ∠ A D C \Rightarrow ∠ A D B < ∠ A D C$

Hence, proved.

Page No 192:

Question 42:

In ∆PQR, if S is any point on the side QR, show that PQ + QR + RP > 2PS.

Answer:

$P Q + Q S > P S . . . (1) [Sum of two sides of a triangle is greater than its third side] P R + S R > P S . . . (2) [Sum of two sides of a triangle is greater than its third side]$

On adding (1) and (2) we get:
$P Q + Q S + P R + S R > 2 P S \Rightarrow P Q + Q R + P R > 2 P S$

Hence, proved.

Page No 192:

Question 43:

In the given figure, O is the centre of the circle and XOY is a diameter. If XZ is any other chord of the circle, show that XY > XZ.

Answer:

$In △ XOZ, we have : OX + OZ > XZ ∴ OX + OY > XZ (∵ OX = OY = OZ) But OX + OY = XY So, XY > XZ$
Hence Proved.

Page No 192:

Question 44:

If O is a point within ∆ABC, show that:
(i) AB + AC > OB + OC
(ii) AB + BC + CA > OA + OB + OC
(iii) $O A + O B + O C > \frac{1}{2} (A B + B C + C A)$

Answer:

Given:
In triangle ABC, O is any interior point.
We know that any segment from a point O inside a triangle to any vertex of the triangle cannot be longer than the two sides adjacent to the vertex.
Thus, OA cannot be longer than both AB and CA (if this is possible, then O is outside the triangle).

(i) OA cannot be longer than both AB and CA.
$A B > O B . . . (1) A C > O C . . . (2) Thus, A B + A C > O B + O C . . . [Adding (1) and (2)]$

$(i i) A B > O A . . . . . . (3) B C > O B . . . . . (4) C A > O C . . . . . (5)$
Adding the above three equations, we get:
$Thus, A B + B C + C A > O A + O B + O C$ ...(6)

OA cannot be longer than both AB and CA.
$A B > O B . . . . . (5) A C > O C . . . . . (6) A B + A C > O B + O C . . . . . . . . . . [On adding (5) and (6)]$
Thus, the first equation to be proved is shown correct.

(iii) Now, consider the triangles OAC, OBA and OBC.
We have:
$O A + O C > A C O A + O B > A B O B + O C > B C Adding the above three equations, we get : O A + O C + O A + O B + O B + O C > A B + A C + B C \Rightarrow 2 (O A + O B + O C) > A B + A C + B C Thus, O A + O B + O C > \frac{1}{2} (A B + B C + C A)$

Page No 192:

Question 45:

Can we draw a triangle ABC with AB = 3 cm, BC = 3.5 cm and CA = 6.5 cm? Why?

Answer:

No.

A triangle can be drawn when the sum of any two sides is greater than the third side.
Here, $A B + B C = A C$
i.e., 3+3.5 = 6.5
So, we cannot draw the triangle ABC.

Page No 195:

Question 1:

Which of the following is not a criterion for congruence of triangles?
(a) SSA
(b) SAS
(c) ASA
(d) SSS

Answer:

(a) SSA
SSA is not a criterion for congruence of triangles.

Page No 195:

Question 2:

If AB = QR, BC = RP and CA = PQ, then which of the following holds?
(a) ∆ABC ≅ ∆PQR
(b) ∆CBA ≅ ∆PQR
(c) ∆CAB ≅ ∆PQR
(d) ∆BCA ≅ ∆PQR

Answer:

Page No 195:

Question 3:

If ∆ABC ≅ ∆ PQR and ∆ABC is not congruent to ∆RPQ, then which of the following is not true?
(a) BC = PQ
(b) AC = PR
(c) BC = QR
(d) AB = PQ

Answer:

(a) BC = PQ

If $∆ A B C ≅ ∆ P Q R$ and $∆ A B C$ is not congruent to $∆ R P Q$ , then
BC $\neq$ PQ
Hence, it is false.

Page No 195:

Question 4:

It is given that ∆ABC ≅ ∆ FDE in which AB = 5 cm, ∠B = 40°, ∠A = 80° and FD = 5 cm. Then, which of the following is true?
(a) ∠D = 60°
(b) ∠E = 60°
(c) ∠F = 60°
(d) ∠D = 80°

Answer:

(b) $∠ E = 60 °$

$∆ A B C ≅ ∆ F D E$
AB = 5cm, $∠ B = 40 °, ∠ A = 80 °$ and FD = 5cm

$Then ∠ A + ∠ B + ∠ C = 180 ° \Rightarrow 80 ° + 40 ° + ∠ C = 180 ° \Rightarrow ∠ C = 60 ° A l s o, ∠ C = ∠ E$
∴ $∠ E = 60 °$

Page No 196:

Question 5:

In ∆ ABC, AB = 2.5 cm and BC = 6 cm. Then, the length of AC cannot be
(a) 3.4 cm
(b) 4 cm
(c) 3.8 cm
(d) 3.6 cm

Answer:

(a) 3.4 cm
In triangle ABC, AB = 2.5 cm and BC = 6 cm.
The length of a side must be less than the sum of the other two sides.
Let the side AC be x cm.
i.e., $x < 2.5 + 6 \Rightarrow x < 8.5$
In addition, the length of a side must be greater then the difference between the other two sides.
i.e., $x > 6 - 2.5 \Rightarrow x > 3.5$
Hence, the limits for the value of x is $3.5 < x < 8.5$ .
Thus, the length of AC cannot be 3.4 cm

Page No 196:

Question 6:

In ∆ABC, ∠A = 40° and ∠B = 60°. Then the longest side of ∆ABC is
(a) BC
(b) AC
(c) AB
(d) cannot be determined

Answer:

(c) AB

In triangle ABC, we have:
$∠ A = 40 °, ∠ B = 60 °$ ...(Given)
$Here, ∠ A + ∠ B + ∠ C = 180 ° \Rightarrow 60 ° + 40 ° + ∠ C = 180 ° \Rightarrow ∠ C = 80 °$
∴ The side opposite to $∠ C$ , i.e., AB, is the longest side of triangle ABC.

Page No 196:

Question 7:

In ∆ABC, ∠B = 35°, ∠C = 65° and the bisector AD of ∠BAC meets BC at D. Then, which of the following is true?
(a) AD > BD > CD
(b) BD > AD > CD
(c) AD > CD > BD
(d) None of these

Answer:

(b) $B D > A D > C D$

In $∆ A B C$ , we have:
$∠ B = 35 °, ∠ C = 65 °$ and the bisector AD of $∠ B A C$ meets BC at D.
Then we have:
$∠ A + ∠ B + ∠ C = 180 ° \Rightarrow ∠ A + 35 ° + 65 ° = 180 ° \Rightarrow ∠ A = 80 °$

AD is the angle bisector of $∠ B A C$
∴ $∠ B A D = ∠ C A D = 40 °$

Now, in triangle ABD, we have:
$∠ B A D > ∠ A B D$
$\Rightarrow B D > A D$
Also, in triangle ACD, we have:
$∠ A C D > ∠ C A D$
$\Rightarrow A D > C D$
$∴ B D > A D > C D$

Page No 196:

Question 8:

In the given figure, AB > AC. Then which of the following is true?
(a) AB < AD
(b) AB = AD
(c) AB > AD
(d) Cannot be determined

Answer:

(c) AB > AD

$A B > A C$ is given.

∴ $∠ A C B > ∠ A B C$

$Now, ∠ A D B > ∠ A C D (exterior angle) \Rightarrow ∠ A D B > ∠ A C B > ∠ A B C \Rightarrow ∠ A D B > ∠ A B D \Rightarrow A B > A D$

Page No 196:

Question 9:

In the given figure, AB > AC. If BO and CO are the bisectors of ∠B and ∠C respectively, then
(a) OB < OC
(b) OB = OC
(c) OB > OC

Answer:

(c) OB > OC

AB >AC (Given)
$\Rightarrow ∠ C > ∠ B$
$\Rightarrow \frac{1}{2} ∠ C > \frac{1}{2} ∠ B$
$\Rightarrow ∠ O C B > ∠ O B C$ (Given)
$\Rightarrow O B > O C$

Page No 196:

Question 10:

In the given figure, AB = AC and OB = OC. Then, ∠ABO : ∠ACO = ?
(a) 1 : 1
(b) 2 : 1
(c) 1 : 2
(d) None of these

Answer:

(a) 1:1

In $∆ O A B and ∆ O A C, we have :$
$AB = AC (Given) OB = OC (Given) OA = OA (Common side)$
$Thus, ∆ OAB ≅ OAC (SSS criterion)$
i.e., $∠ A B O = ∠ A C O$
∴ $∠ A B O : ∠ A C O = 1 : 1$

Page No 196:

Question 11:

In ∆ABC, if ∠C > ∠B, then
(a) BC > AC
(b) AB > AC
(c) AB > AC
(d) BC > AC

Answer:

(b) AB > AC

In $∆ A B C$ , we have:
$∠ C > ∠ B$
The side opposite to the greater angle is larger.
$∴ A B > A C$

Page No 197:

Question 12:

O is any point in the interior of ∆ABC. Then, which of the following is true?
(a) $(O A + O B + O C) > (A B + B C + C A)$
(b) $(O A + O B + O C) > \frac{1}{2} (A B + B C + C A)$
(c) $(O A + O B + O C) < \frac{1}{2} (A B + B C + C A)$
(d) None of these

Answer:

(c) $(O A + O B + O C) > \frac{1}{2} (A B + B C + C A)$

In $△ O A B, △ O B C and △ O C A, we have :$
$O A + O B > A B, O B + O C > B C and O C + O A > A C$
Adding them, we get:

$2 (O A + O B + O C) > (A B + B C + C A) \Rightarrow (O A + O B + O C) > \frac{1}{2} (A B + B C + C A)$

Page No 197:

Question 13:

If the altitudes from two vertices of a triangle to the opposite sides are equal, then the triangle is
(a) equilateral
(b) isosceles
(c) scalene
(d) right-angled

Answer:

(b) isosceles

In $∆ A B C, BL ∥ AC .$
$CM ∥ AB such that BL = CM .$

To prove: AB = AC

$In △ ABL and △ ACM, BL = CM (Given) ∠ BAL = ∠ CAM (Common angle) ∠ ALB = ∠ AMC (Each 90 °) △ ABL ≅ △ ACM (AAS criterion) ∴ AB = AC (CPCT)$

Page No 197:

Question 14:

In the given figure, AE = DB, CB = EF and ∠ABC = ∠FED. Then, which of the following is true?
(a) ∆ABC ≅ ∆DEF
(b) ∆ABC ≅ ∆EFD
(c) ∆ABC ≅ ∆FED
(d) ∆ABC ≅ ∆EDF

Answer:

(a) $△ A B C ≅ △ D E F$

AE = DB ...(Given)
CB = EF ...(Given)

$A B = (A D - D B) = (A D - A E) a n d D E = (A D - A E)$
$In △ ABC and △ DEF, we have : AB = DE (Proved) CB = EF . . . (Given) ∠ ABC = ∠ FED . . . (Given) △ ABC ≅ △ DEF (SAS criterion)$

Page No 197:

Question 15:

In the given figure, BE ⊥ CA and CF ⊥ BA such that BE = CF. Then, which of the following is true?
(a) ∆ABE ≅ ∆ACF
(b) ∆ABE ≅ ∆AFC
(c) ∆ABE ≅ ∆CAF
(d) ∆ABE ≅ ∆FAC

Answer:

(a) $△ A B E ≅ △ A C F$

$In △ ABE and △ ACF, we have : BE = CF (Given) ∠ BEA = ∠ CFA = 90 ° ∠ A = ∠ A (Common) △ ABE ≅ △ ACF (AAS criterion)$

Page No 197:

Question 16:

In the given figure, D is the midpoint of BC, DE ⊥ AB and DF ⊥ AC such that DE = DF. Then, which of the following is true?
(a) AB = AC
(b) AC = BC
(c) AB = BC
(d) None of these

Answer:

(a) AB = AC

$In △ BED and △ CFD, we get : DE = DF (Given) BD = CD (Given) ∠ E = ∠ F = 90 ° (Given) ∴ △ BED ≅ △ CFD (RHS criterion) \Rightarrow ∠ B = ∠ C \Rightarrow AC = AB$

Page No 198:

Question 17:

In ∆ABC and ∆DEF, it is given that AB = DE and BC = EF. In order that ∆ABC ≅ ∆DEF, we must have
(a) ∠A = ∠D
(b) ∠B = ∠E
(c) ∠C = ∠F
(d) none of these

Answer:

(b) $∠ B = ∠ E$

In $△ A B C and △ D E F, we have :$
AB = DE (Given)
BC = EF (Given)
In order that $△ A B C ≅ D E F$ , we must have $∠ B = ∠ E$ .

Page No 198:

Question 18:

In ∆ABC and ∆DEF, it is given that ∠B = ∠E and ∠C = ∠F. In order that ∆ABC ≅ ∆DEF, we must have
(a) AB = DF
(b) AC = DE
(c) BC = EF
(d) ∠A = ∠D

Answer:

Page No 198:

Question 19:

In ∆ABC and ∆PQR, it is given that AB = AC, ∠C = ∠P and ∠B = ∠Q. Then, the two triangles are
(a) isosceles but not congruent
(b) isosceles and congruent
(c) congruent but not isosceles
(d) neither congruent nor isosceles

Answer:

(a) isosceles but not congruent

$A B = A C \Rightarrow ∠ C = ∠ B \Rightarrow ∠ P = ∠ Q [∵ ∠ C = ∠ P and ∠ B = ∠ Q]$
Thus, both the triangles are isosceles but not congruent.

Page No 198:

Question 20:

Which is true?
(a) A triangle can have two right angles.
(b) A triangle can have two obtuse angles.
(c) A triangle can have two acute angles.
(d) An exterior angle of a triangle is less than either of the interior opposite angles.

Answer:

(c) A triangle can have two acute angles.
The sum of two acute angles is always less than 180^o, thus satisfying the angle sum property of a triangle.
Therefore, a triangle can have two acute angles.

Page No 198:

Question 21:

Three statements are given below:
I. In a ∆ABC in which AB = AC, the altitude AD bisects BC.
II. If the altitudes AD, BE and CF of ∆ABC are equal, then ∆ABC is equilateral.
III. If D is the midpoint of the hypotenuse AC of a right ∆ABC, then BD = AC.
Which is true?
(a) I only
(b) II only
(c) I and II
(d) II and III

Answer:

(c) I and II are true
(I)

$In △ ADB and △ ADC, we get : AB = AC (Given) AD = AD (Common) ∠ ADB = ∠ ADC = 90 ° (Given) △ ADB ≅ △ ADC (RHS criterion) \Rightarrow BD = DC (CPCT)$

∴ AD bisects BC, which is true.

(II)

$In △ ABE and △ ACF, we have : ∠ AEB = ∠ AFC (Each 90 °) ∠ BAE = ∠ CAF (Common) BE = CF (Given) ∆ ABL ≅ △ ACM (AAS criterion) \Rightarrow AB = AC (CPCT)$
Similarly, if AD is perpendicular to BC and AD = BE, then we can prove that BC = AC.
Therefore, triangle ABC is an equilateral triangle.
Hence, (II) is true.

(III)

$In △ ABC, ∠ ABC = 90 ° (BD perpendicular AC)$
Then we know that AD = BD = CD
Therefore, BD = AC is not true.
So, only (I) and (II) are true.

Page No 199:

Question 22:

Assertion: If AD is a median of ∆ABC, then AB + AC > 2AD.
Reason: The angles opposite to equal sides of a triangle are equal.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.

In triangle ABC, AD is a median.
We know that the sum of two sides of a triangle is greater than the third side.

$A B + D B > A D . . . . (i) A C + D C > A D . . . . . . . . . (i i) \Rightarrow A B + A C + (D B + D C) > 2 A D \Rightarrow A B + A C + B C > 2 A D$

In any triangle, angles opposite to equal sides are equal.
Clearly, the assertion and the reason are both true, but reason (R) does not give assertion (A).

Page No 199:

Question 23:

Assertion: In a quadrilateral ABCD, we have (AB + BC + CD + DA) > 2AC.

Reason: The sum of any two sides of a triangle is greater than the third side.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
We know that sum of any two sides of a triangle is greater than the third side.
Therefore, $(A B + B C) > A C$ and $(C D + D A) > A C$ .
Adding them, we get:
$(A B + B C + C D + D A) > 2 A C$
Thus, the assertion is true, and the reason gives the assertion.

Page No 199:

Question 24:

Assertion: ∆ABC and ∆DBC are two isosceles triangles on the same base BC. Then, ∠ABD = ∠ACD.

Reason: The angles opposite to equal sides of a triangle are equal.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

We know that the angles opposite to the equal sides of a triangle are equal.
So, the reason is true.

$AB = AC \Rightarrow ∠ ABC = ∠ ACB DB = DC \Rightarrow ∠ DBC = ∠ DCB Subtracting them, we get : \Rightarrow ∠ ABC - ∠ DBC = ∠ ACB - ∠ DCB ∴ ∠ ABD = ∠ ACD$

So, the assertion (A) is true and, clearly, the reason (R) gives the assertion.

Page No 200:

Question 25:

Assertion: It is always possible to draw a triangle whose sides measure 4 cm, 5 cm and 10 cm respectively.
Reason: In an isosceles ∆ABC with AB = AC, if BD and CE are bisectors of ∠B and ∠C respectively, then BD = CE.

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(d) Assertion is false and Reason is true.
Reason (R):

$A B = A C \Rightarrow ∠ C = ∠ B \Rightarrow \frac{1}{2} ∠ C = \frac{1}{2} ∠ B \Rightarrow ∠ B C E = ∠ C B D$

$In △ BDC and △ CEB, we have : ∠ CBD = ∠ BCE (Proved) ∠ DCB = ∠ EBC (Given) BC = CB (Common) ∴ △ BDC ≅ CEB \Rightarrow BD = CE (CPCT)$
Thus, the reason (R) is true.
The assertion (A) is false, as the sum of two sides of a triangle is always greater than the third side.
Here, $(4 + 5) < 10$
So, the reason (R) is true, but the assertion (A) is false.

Page No 200:

Question 26:

Assertion: In the given figure, ∆ABC is given with AB = AC and BA is produced to D such that AB = AD. Then, ∠BCD = 90°.

Reason: In the given figure AB = AC and D is a point on BC produced. Then, AB > AD.

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(c) Assertion is true and Reason is false.
Reason (R):
$A B = A C \Rightarrow ∠ A C B = ∠ A B C . . . (i)$
The side BC of triangle ABC is produced to D.
$∠ A C D > ∠ A C B = ∠ A B C But in ∆ ADC, side DC is produced to B . ∠ A C B > ∠ A D C . . . (i i) \Rightarrow ∠ A B C > ∠ A D C [From (i) and (ii)] \Rightarrow A D > A B$
Hence, the reason (R) is false.
Clearly, the assertion is true, but the reason (R) is false.

Page No 200:

Question 27:

Match the following columns:

Column I	Column II
(a) ∆ABC, if AB = AC and ∠A = 50°, then ∠C = ...... .	(p) its perimeter
(b) The vertical angle of an isosceles triangle is 130°. Then, each base angle is ...... .	(q) 15°
(c) The sum of three altitudes of a ∆ABC is less than ....... .	(r) 65°
(d) In the given figure, ABCD is a square and ∆EDC is an equilateral triangle. Then, ∠EBC is ...... .	(s) 25°

(a) ......,
(b) ......,
(c) ......,
(d) ......,

Answer:

(a) $\Rightarrow$ (r)
In $△ A B C,$ if AB = AC and $∠ A = 50 °$ , then
$∠ B = ∠ C B y a n g l e s u m p r o p e r t y, ∠ A + ∠ B + ∠ C = 180 ° \Rightarrow 50 ° + 2 ∠ B = 180 ° \Rightarrow 2 ∠ B = 130 ° \Rightarrow ∠ B = 65 ° = ∠ C$

$∠ C = 65 °$ .

(b) $\Rightarrow$ (s)
If the vertical angle of an isosceles triangle is $130 °$ , then
$L e t t h e v e r t i c a l a n g l e b e ∠ A . o r, ∠ B = ∠ C B y a n g l e s u m p r o p e r t y, ∠ A + ∠ B + ∠ C = 180 ° \Rightarrow 130 ° + 2 ∠ B = 180 ° \Rightarrow 2 ∠ B = 50 ° \Rightarrow ∠ B = 25 ° = ∠ C$

Each base angle is $25 °$ .

(c) $\Rightarrow$ (p)
The sum of three altitudes of a $△ A B C$ is less than its perimeter.

(d) $\Rightarrow$ (q)
In the given figure, ABCD is a square and $△ E D C$ is an equilateral triangle.
Then,
$E D = D C = E C = A D = A B = B C In △ E C B, E C = C B \Rightarrow ∠ C = ∠ B Let ∠ C = ∠ B = x Also, ∠ E C D = 60 ° a n d ∠ D C B = 90 ° Thus, ∠ E C B = 60 ° + 90 ° = 150 ° o r, x + x + 150 ° = 180 ° o r, 2 x = 30 ° o r, x = 15 °$
Therefore, $∠ E B C = 15 °$ .

Page No 201:

Question 28:

Fill in the blanks with < or >.
(a) (Sum of any two sides of a triangle) ...... (the third side)
(b) (Difference of any two sides of a triangle) ...... (the third side)
(c) (Sum of three altitudes of a triangle) ...... (sum of its three side)
(d) (Sum of any two sides of a triangle) ...... (twice the median to the 3rd side)
(e) (Perimenter of a triangle) ...... (sum of its three medians)

Answer:

a) Sum of any two sides of a triangle > the third side

b) Difference of any two sides of a triangle < the third side

c) Sum of three altitudes of a triangle < sum of its three side

d) Sum of any two sides of a triangle > twice the median to the 3rd side

e) Perimeter of a triangle > sum of its three medians

Page No 201:

Question 29:

Fill in the blanks.
(a) Each angle of an equilateral triangle measures ...... .
(b) Medians of an equilateral triangle are ...... .
(c) In a right triangle the hypotenuse is the ...... side.
(d) Drawing a ∆ABC with AB = 3 cm, BC = 4 cm and CA = 7 cm is ...... .

Answer:

a) Each angle of an equilateral triangle measures $60 °$ .

b) Medians of an equilateral triangle are equal.

c) In a right triangle, the hypotenuse is the longest side.

d) Drawing a $△ A B C$ with AB = 3cm, BC = 4cm and CA = 7cm is not possible.

Page No 205:

Question 1:

In an equilateral ∆ABC, find ∠A.

Answer:

In an equilateral $△ A B C$ , all the angles are equal.
Let the three angles $∠ A, ∠ B and ∠ C b e x °$ .
$Now, x ° + x ° + x ° = 180 ° (sum of all the angles of triangle is 180 °) \Rightarrow 3 x ° = 180 ° \Rightarrow x ° = 60 ° ∴ ∠ A = 60 °$

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Question 2:

In a ∆ABC, if AB = AC and ∠B = 65°, find ∠A.

Answer:

In $△ A B C$ , we have:
AB = AC
$∠ B = 65 °$
Since ABC is an isosceles triangle, we have:
$∠ C = ∠ B$
$∠ C = 65 °$
In triangle ABC, we have:
$∠ A + ∠ B + ∠ C = 180 ° \Rightarrow ∠ A + 65 + 65 = 180 ° \Rightarrow ∠ A = 180 ° - 130 ° ∴ ∠ A = 50 °$

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Question 3:

In a right ∆ABC, ∠B = 90°. Find the longest side.

Answer:

In right $∆ A B C$ , we have:
$∠ B = 90 °$
$∠ A + ∠ C = 90 ° ∴ ∠ A, ∠ C < ∠ B$
The side opposite to $∠ B$ , i.e., side AC, is the longest.

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Answer:

In a triangle, the side opposite the greater angle is longer. In triangle ABC, $∠ B > ∠ C$ .
Therefore, AC is longer than AB.

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Question 5:

Can we construct a ∆ABC in which AB = 5 cm, BC = 4 cm and AC = 9 cm? Why?

Answer:

No, the sum of two side must be greater than the third side.
AC should be greater than (AB + BC).

In this case, AB + BC = 5 + 4 = 9 cm
AC = 9 cm
Thus, the sum of two sides is not greater than the third side. So, triangle ABC cannot be constructed.

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Question 6:

Find the measure of each exterior angle of an equilateral triangle.

Answer:

Here, exterior angle is $∠ A O D$ .
$∠ A O D + ∠ A O B = 180 ° \Rightarrow 60 ° + ∠ A O B = 180 ° \Rightarrow ∠ A O B = 120 °$
∴ The measure of each exterior angle of an equilateral triangle is $120 °$ .

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Question 7:

Show that the difference of any two sides of a triangle is less than the third side.

Answer:

$Construction : Let AC > AB . Then along AC, set off AD = AB . Join BD . Proof : AB = AD \Rightarrow ∠ 1 = ∠ 2 Side CD of ∆ BCD has been produced to A . ∴ ∠ 2 > ∠ 4 [∵ ext angle > each int . opposite angle] Again, side AD of △ ABD has been produced to C . ∴ ∠ 3 > ∠ 1 [∵ ext angle > each int . opposite angle] Consequently, ∠ 3 > ∠ 2 [∵ ∠ 1 = ∠ 2] Now, ∠ 3 > ∠ 2 and ∠ 2 > ∠ 4 \Rightarrow ∠ 3 > ∠ 4 ∴ BC > CD [side opposite to the greater angle is longer] \Rightarrow CD < BC \Rightarrow AC - AD < BC \Rightarrow AC - AB < BC [∵ AD = AB] Hence, AC - AB < BC Similarly, BC - AC < AB and BC - AB < AC .$

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Question 8:

In a right ∆ABC, ∠B = 90° and D is the mid-point of AC. Prove that $B D = \frac{1}{2} A C$ .

Answer:

AD = DC     (∵ D is the midpoint)
$∠ A D B = ∠ B D C$    (Altitude = BD)
BD = BD     (Common side)
∴ $△ A D B ≅ △ C D B$      (By SAS criterion)
⇒ $∠ A = ∠ C$     (CPCT)
Now, we know that $∠ B = 90 °$ .
$\Rightarrow ∠ A = ∠ ABD = 45 °$     (Using sum angle property)
Similarly, $∠ B D C = 90 °$     (BD is altitude)
$∠ C = 45 °$     (Proved)
$∠ D B C = 45 °$
$\Rightarrow ∠ A B D = 45 °$
$\Rightarrow B D = C D a n d B D = A D$     (Isosceles triangle property)
As AD+DC =AC
$\Rightarrow B D + B D = A C \Rightarrow 2 B D = A C \Rightarrow B D = \frac{1}{2} A C$
Hence proved.

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Question 9:

Prove that the perimeter of a triangle is greater than the sum of its three medians.

Answer:

Let ABC be the triangle and D, E and F be the midpoints of BC, CA and AB, respectively.
Since the sum of two sides of a triangle is greater than twice the median bisecting the third side, we have:
$A B + A C > 2 A D Similarly, B C + A C > 2 C F Also, B C + A B > 2 B E$
On adding all opf the above inequalities, we get:
$(A B + A C) + (B C + A C) + (B C + A B) > 2 A D + 2 C D + 2 B E \Rightarrow 2 (A B + B C + A C) > 2 (A D + B E + C F) ∴ A B + B C + A C > A D + B E + C F$

Thus, the perimeter of triangle is greater than the sum of the medians.

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Question 10:

Which is true?
(a) A triangle can have two acute angles.
(b) A triangle can have two right angles.
(c) A triangle can have two obtuse angles.
(d) An exterior angles of a triangle is always less than either of the interior opposite angles.

Answer:

(a) A triangle can have two acute angles.

This is because the sum of two acute angles is always less than 180^o, thus satisfying the angle sum property of a triangle.
Therefore, a triangle can have two acute angles.

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Question 11:

In ∆ABC, BD ⊥ AC and CE ⊥ AB such that BE = CD. Prove that BD = CE.

Answer:

In $△ B D C a n d △ C E B,$ we have:
$BE = CD (Given) ∠ BEC = ∠ CDB = 90 ° \Rightarrow BC = BC (Common) △ BDC ≅ △ CEB (RHS criterion)$
∴ $B D = C E$ (CPCT)
Hence proved.

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Question 12:

In ∆ABC, AB = AC. Side BA is produced to D such that AD = AB. Prove that ∠BCD = 90°.

Answer:

In triangle ABC, we have:

$AB = AC (Given) \Rightarrow ∠ ACB = ∠ ABC (Angles opposite to equal sides of a triangle are also equal)$
In triangle ACD, we have:
$AC = AD \Rightarrow ∠ ADC = ∠ ACD (Angles opposite to equal side of a triangle are also equal)$
In triangle BCD, we have:
$∠ A B C + ∠ B C D + ∠ A D C = 180 ° (Angle sum property of triangle) \Rightarrow ∠ A C B + ∠ A C B + ∠ A C D + A C D = 180 ° \Rightarrow 2 (∠ A C B + ∠ A C D) = 180 ° \Rightarrow 2 (∠ B C D) = 180 ° ∴ ∠ B C D = 90 °$
Hence proved.

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Question 13:

In the given figure, it is given that AD = BC and AC = BD. Prove that ∠CAD = ∠CBD and ∠ADC = ∠BCD.

Answer:

Consider the triangles DAC and CBD.

$\begin{array}{r} A D = B C A C = B D \\ D C = D C \end{array}$

By SSS criterion of congruence, $∆ A D C ≅ B C D$ .

By CPCT, we have:

$\begin{array}{r} ∠ C A D = ∠ C B D ∠ A D C = ∠ B C D \\ ∠ A C D = ∠ B D C \end{array}$

Hence proved.

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Question 14:

Prove that the angles opposite to equal sides of a triangle are equal.

Answer:

In triangle PQR, $P Q = Q R .$
To prove: $∠ Q = ∠ R$
Proof:
Draw a bisector of $∠ P$ and let S be the point of intersection.

$In △ QPS and △ RPS, we have : PQ = PR (Given) ∠ QPS = ∠ RPS (By construction) PS = PS (Common) △ QPS ≅ △ RPS (SAS criterion) ∠ PQS = ∠ PRS (CPCT)$
∴ $∠ Q = ∠ R$
Hence Proved.

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Question 15:

In an isosceles ∆ABC, AB = AC and the bisectors of are joint.
Prove that: (i) OB = OC (ii) ∠OAB = ∠OAC

Answer:

In $△ A B O a n d △ A C O$ , we have:
$AB = AC (Given) AO = AO (Common) ∠ ABO = ∠ ACO (In an isoscles triangle, opposite angles are equal) △ ABO ≅ △ ACO (SAS criterion)$
⇒ OB = OC (CPCT)
⇒ $∠ O A B = ∠ O A C (CPCT)$
Hence proved.

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Question 16:

Prove that, of all line segments that can be drawn to a given line, from a point, not lying on it, the perpendicular line segment is the shortest.

Answer:

Given: l is a straight line and A is a point not lying on l.
AB is perpendicular to line l.
C is a point on l.
To prove: $A B < A C$
Proof:
$∠ B = 90 ° I n △ A B C, we have : ∠ C < 90 ° [∠ A + ∠ B + ∠ C = 180 \Rightarrow ∠ A + ∠ C = 90 ° \Rightarrow ∠ C < 90 °] \Rightarrow ∠ C < ∠ B \Rightarrow A B < A C (Greater angle has a greater side opposite to it)$
Since, C can lie anywhere on l, AB is the shortest of all line segments drawn from A to l.

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Question 17:

Assertion: Each angle of an equilateral triangle is 60°.
Reason: Angles opposite to equal sides of a triangle are equal.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(a) The Assertion and Reason are both true and the Reason is a correct explanation of the Assertion.
We know that each angle of an equilateral triangle is 60°.
Also, angles opposite to the equal sides of a triangle are equal.
Clearly, both the Assertion and the Reason are true .

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Question 18:

Assertion: If AD is a median of ∆ABC, then AB + AC > 2AD.

Reason: In a triangle the sum of two sides is greater than the third side.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(a) The Assertion and Reason are both true and the Reason is a correct explanation of the Assertion.
So, the Reason is obviously true. Let us consider the Assertion.

In triangle ABD, $A B + B D > A D$ .

Similarly, in triangle ADC, $A C + C D > A D$ .

Adding the above two expressions, we get:
$A B + A C + B D + C D > A D + A D \Rightarrow A B + A C + B D + D C > 2 A D \Rightarrow A B + A C + B C > 2 A D$

Thus, the Assertion is true and the Reason is true. The Reason is an explanation of the Assertion.

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Question 19:

Match the following columns:

Column I	Column II
(a) In ∆ABC, if AB = AC and ∠A = 70°, then ∠C = ...... .	(p) less
(b) The vertical angle of an isosceles triangle is 120°. Each base angle is ...... .	(q) greater
(c) The sum of three medians of a triangle is ...... than the perimeter.	(r) 30°
(d) In a triangle, the sum of any two sides is always ...... than the third side.	(s) 55°

(a) ......,
(b) ......,
(c) ......,
(d) ......,

Answer:

(a) $\Rightarrow$ (s)
In $△ ABC, if AB = AC and ∠ A = 70 °, then ∠ C = 55 ° .$

(b) $\Rightarrow$ (r)
The vertical angle of an isosceles triangle is $120 °$ . Each base angle is $30 °$ .

(c) $\Rightarrow$ (p)
The sum of three medians of a triangle is less than the perimeter.

(d) $\Rightarrow$ (q)
In a triangle, the sum of any two sides is always greater than the third side.

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Question 20:

In the given figure, PQ > PR and QS and RS are the bisectors of ∠Q and ∠R respectively. Show that SQ > SR.

Answer:

Since the angle opposite to the longer side is greater, we have:

$P Q > P R \Rightarrow ∠ R > ∠ Q \Rightarrow \frac{1}{2} ∠ R > \frac{1}{2} ∠ Q \Rightarrow ∠ S R Q > ∠ R Q S \Rightarrow Q S > S R$

∴ $ S Q > S R$

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Question 21:

In the given figure, ABC is a triangle right-angled at B such that ∠BCA = 2 ∠BAC. Show that AC = 2BC.

Answer:

Produce CB to D such that BD = BC.
Join AD.
$In △ ABC and △ ABD, we have : BC = BD (By construction) AB = AB (Common) ∠ ABC = ∠ ABD = 90 ° ∴ △ ABC ≅ △ ABD (SAS criterion) \Rightarrow ∠ CAB = ∠ DAB (CPCT) \Rightarrow AC = AD (CPCT) ∴ ∠ CAD = ∠ CAB + ∠ BAD = x ° + x ° = 2 x °$
But $A C = A D$
$i . e ., ∠ A C D = ∠ A D B = 2 x ° (∵ ∠ B C A = 2 ∠ B A C)$
Therefore, triangle ACD is an equilateral triangle.
$Then A C = C D \Rightarrow A C = 2 B C$

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Question 22:

S is any point in the interior of ∆PQR. Show that ABCD AB + BC + CD + DA > AC + BD.
Figure

Answer:

Produce QS so that it meets PR at T.

$In △ P Q T, we have : P Q + P T > Q T \Rightarrow P Q + P T > S Q + S T . . . (i) In △ S T R, we have : S T + T R > S R . . . . (i i) Adding (i) and (ii), we get : P Q + P T + S T + T R > S Q + S T + S R \Rightarrow P Q + P T + T R > S Q + S R \Rightarrow P Q + P R > S Q + S R \Rightarrow S Q + S R < P Q + P R$

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Question 23:

Show that in a quadrilateral ABCD AB + BC + CD + DA > AC + BD.
Figure

Answer:

ABCD is a quadrilateral.
AC and BD are diagonals.

$We know that the sum of two sides of a triangle is greater than the third side . In △ A C B, we have : A B + B C > A C . . . (1) In △ B D C, we have : B C + C D > B D . . . (2) In △ A C D, we have : A D + D C > A C . . . (3) In △ B A D, we have : A B + A D > B D . . . (4) Adding (1), (2), (3) & (4), we get : A B + B C + B C + C D + A D + D C + A B + A D > A C + B D + A C + B D \Rightarrow 2 A B + 2 B C + 2 C D + 2 A D > 2 A C + 2 B D \Rightarrow A B + B C + C D + A D > A C + B D$

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