NCERT Solutions for Class 9 Maths. Chapter 7 Triangles. Question Answer.

## NCERT Solutions for Class 9 Maths

### Chapter 7 Triangles

**Exercise 7.1**

**Question 1.**

In quadrilateral ACBD, AC = AD and AB bisects ∠ A (see figure). Show that ∆ABC ≅ ∆ABD. What can you say about BC and BD?

In quadrilateral ACBD, AC = AD and AB bisects ∠ A (see figure). Show that ∆ABC ≅ ∆ABD. What can you say about BC and BD?

Solution:

In quadrilateral ACBD, we have AC = AD and AB being the bisector of ∠A.

Now, In ∆ABC and ∆ABD,

AC = AD (Given)

∠ CAB = ∠ DAB ( AB bisects ∠ CAB)

and AB = AB (Common)

∴ ∆ ABC ≅ ∆ABD (By SAS congruence axiom)

∴ BC = BD (By CPCT)

Now, In ∆ABC and ∆ABD,

AC = AD (Given)

∠ CAB = ∠ DAB ( AB bisects ∠ CAB)

and AB = AB (Common)

∴ ∆ ABC ≅ ∆ABD (By SAS congruence axiom)

∴ BC = BD (By CPCT)

**Question 2.**

ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see figure). Prove thatABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see figure). Prove that

**(i) ∆ABD ≅ ∆BAC****(ii) BD = AC**

(iii) ∠ABD = ∠ BAC

(iii) ∠ABD = ∠ BAC

Solution:

In quadrilateral ACBD, we have AD = BC and ∠ DAB = ∠ CBA

(i) In ∆ ABC and ∆ BAC,

AD = BC (Given)

∠DAB = ∠CBA (Given)

AB = AB (Common)

∴ ∆ ABD ≅ ∆BAC (By SAS congruence)

(ii) Since ∆ABD ≅ ∆BAC

⇒ BD = AC [By C.P.C.T.]

(iii) Since ∆ABD ≅ ∆BAC

⇒ ∠ABD = ∠BAC [By C.P.C.T.]

(i) In ∆ ABC and ∆ BAC,

AD = BC (Given)

∠DAB = ∠CBA (Given)

AB = AB (Common)

∴ ∆ ABD ≅ ∆BAC (By SAS congruence)

(ii) Since ∆ABD ≅ ∆BAC

⇒ BD = AC [By C.P.C.T.]

(iii) Since ∆ABD ≅ ∆BAC

⇒ ∠ABD = ∠BAC [By C.P.C.T.]

**Question 3.**

AD and BC are equal perpendiculars to a line segment AB (see figure). Show that CD bisects AB.AD and BC are equal perpendiculars to a line segment AB (see figure). Show that CD bisects AB.

Solution:

In ∆BOC and ∆AOD, we have

∠BOC = ∠AOD

BC = AD [Given]

∠BOC = ∠AOD [Vertically opposite angles]

∴ ∆OBC ≅ ∆OAD [By AAS congruency]

⇒ OB = OA [By C.P.C.T.]

i.e., O is the mid-point of AB.

Thus, CD bisects AB.

∠BOC = ∠AOD

BC = AD [Given]

∠BOC = ∠AOD [Vertically opposite angles]

∴ ∆OBC ≅ ∆OAD [By AAS congruency]

⇒ OB = OA [By C.P.C.T.]

i.e., O is the mid-point of AB.

Thus, CD bisects AB.

**Question 4.**

l and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that ∆ABC = ∆CDA.l and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that ∆ABC = ∆CDA.

Solution:

∵ p || q and AC is a transversal,

∴ ∠BAC = ∠DCA …(1) [Alternate interior angles]

Also l || m and AC is a transversal,

∴ ∠BCA = ∠DAC …(2)

[Alternate interior angles]

Now, in ∆ABC and ∆CDA, we have

∠BAC = ∠DCA [From (1)]

CA = AC [Common]

∠BCA = ∠DAC [From (2)]

∴ ∆ABC ≅ ∆CDA [By ASA congruency]

∴ ∠BAC = ∠DCA …(1) [Alternate interior angles]

Also l || m and AC is a transversal,

∴ ∠BCA = ∠DAC …(2)

[Alternate interior angles]

Now, in ∆ABC and ∆CDA, we have

∠BAC = ∠DCA [From (1)]

CA = AC [Common]

∠BCA = ∠DAC [From (2)]

∴ ∆ABC ≅ ∆CDA [By ASA congruency]

**Question 5.**

Line l is the bisector of an ∠ A and ∠ B is any point on l. BP and BQ are perpendiculars from B to the arms of LA (see figure). Show that

(i) ∆APB ≅ ∆AQB

(ii) BP = BQ or B is equidistant from the arms ot ∠A.Line l is the bisector of an ∠ A and ∠ B is any point on l. BP and BQ are perpendiculars from B to the arms of LA (see figure). Show that

(i) ∆APB ≅ ∆AQB

(ii) BP = BQ or B is equidistant from the arms ot ∠A.

Solution:

We have, l is the bisector of ∠QAP.

∴ ∠QAB = ∠PAB

∠Q = ∠P [Each 90°]

∠ABQ = ∠ABP

[By angle sum property of A]

Now, in ∆APB and ∆AQB, we have

∠ABP = ∠ABQ [Proved above]

AB = BA [Common]

∠PAB = ∠QAB [Given]

∴ ∆APB ≅ ∆AQB [By ASA congruency]

Since ∆APB ≅ ∆AQB

⇒ BP = BQ [By C.P.C.T.]

i. e., [Perpendicular distance of B from AP]

= [Perpendicular distance of B from AQ]

Thus, the point B is equidistant from the arms of ∠A.

We have, l is the bisector of ∠QAP.

∴ ∠QAB = ∠PAB

∠Q = ∠P [Each 90°]

∠ABQ = ∠ABP

[By angle sum property of A]

Now, in ∆APB and ∆AQB, we have

∠ABP = ∠ABQ [Proved above]

AB = BA [Common]

∠PAB = ∠QAB [Given]

∴ ∆APB ≅ ∆AQB [By ASA congruency]

Since ∆APB ≅ ∆AQB

⇒ BP = BQ [By C.P.C.T.]

i. e., [Perpendicular distance of B from AP]

= [Perpendicular distance of B from AQ]

Thus, the point B is equidistant from the arms of ∠A.

**Question 6.**

In figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.In figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

Solution:

We have, ∠BAD = ∠EAC

Adding ∠DAC on both sides, we have

∠BAD + ∠DAC = ∠EAC + ∠DAC

⇒ ∠BAC = ∠DAE

Now, in ∆ABC and ∆ADE. we have

∠BAC = ∠DAE [Proved above]

AB = AD [Given]

AC = AE [Given]

∴ ∆ABC ≅ ∆ADE [By SAS congruency]

⇒ BC = DE [By C.P.C.T.]

Solution:

We have, P is the mid-point of AB.

∴ AP = BP

∠EPA = ∠DPB [Given]

Adding ∠EPD on both sides, we get

∠EPA + ∠EPD = ∠DPB + ∠EPD

⇒ ∠APD = ∠BPE

(i) Now, in ∆DAP and ∆EBP, we have

∠PAD = ∠PBE [ ∵∠BAD = ∠ABE]

AP = BP [Proved above]

∠DPA = ∠EPB [Proved above]

∴ ∆DAP ≅ ∆EBP [By ASA congruency]

(ii) Since, ∆ DAP ≅ ∆ EBP

⇒ AD = BE [By C.P.C.T.]

Adding ∠DAC on both sides, we have

∠BAD + ∠DAC = ∠EAC + ∠DAC

⇒ ∠BAC = ∠DAE

Now, in ∆ABC and ∆ADE. we have

∠BAC = ∠DAE [Proved above]

AB = AD [Given]

AC = AE [Given]

∴ ∆ABC ≅ ∆ADE [By SAS congruency]

⇒ BC = DE [By C.P.C.T.]

**Question 7.**

AS is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠ BAD = ∠ ABE and ∠ EPA = ∠ DPB. (see figure). Show that

(i) ∆DAP ≅ ∆EBP

(ii) AD = BEAS is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠ BAD = ∠ ABE and ∠ EPA = ∠ DPB. (see figure). Show that

(i) ∆DAP ≅ ∆EBP

(ii) AD = BE

Solution:

We have, P is the mid-point of AB.

∴ AP = BP

∠EPA = ∠DPB [Given]

Adding ∠EPD on both sides, we get

∠EPA + ∠EPD = ∠DPB + ∠EPD

⇒ ∠APD = ∠BPE

(i) Now, in ∆DAP and ∆EBP, we have

∠PAD = ∠PBE [ ∵∠BAD = ∠ABE]

AP = BP [Proved above]

∠DPA = ∠EPB [Proved above]

∴ ∆DAP ≅ ∆EBP [By ASA congruency]

(ii) Since, ∆ DAP ≅ ∆ EBP

⇒ AD = BE [By C.P.C.T.]

**Question 8.**

In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see figure). Show that

(i) ∆AMC ≅ ∆BMD

(ii) ∠DBC is a right angle

(iii) ∆DBC ≅ ∆ACBIn right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see figure). Show that

(i) ∆AMC ≅ ∆BMD

(ii) ∠DBC is a right angle

(iii) ∆DBC ≅ ∆ACB

**(iv) CM = AB**Solution:

Since M is the mid – point of AB.

∴ BM = AM

(i) In ∆AMC and ∆BMD, we have

CM = DM [Given]

∠AMC = ∠BMD [Vertically opposite angles]

AM = BM [Proved above]

∴ ∆AMC ≅ ∆BMD [By SAS congruency]

(ii) Since ∆AMC ≅ ∆BMD

⇒ ∠MAC = ∠MBD [By C.P.C.T.]

But they form a pair of alternate interior angles.

∴ AC || DB

Now, BC is a transversal which intersects parallel lines AC and DB,

∴ ∠BCA + ∠DBC = 180° [Co-interior angles]

But ∠BCA = 90° [∆ABC is right angled at C]

∴ 90° + ∠DBC = 180°

⇒ ∠DBC = 90°

(iii) Again, ∆AMC ≅ ∆BMD [Proved above]

∴ AC = BD [By C.P.C.T.]

Now, in ∆DBC and ∆ACB, we have

BD = CA [Proved above]

∠DBC = ∠ACB [Each 90°]

BC = CB [Common]

∴ ∆DBC ≅ ∆ACB [By SAS congruency]

(iv) As ∆DBC ≅ ∆ACB

DC = AB [By C.P.C.T.]

But DM = CM [Given]

∴ CM = DC = AB

⇒ CM = AB

0°

⇒ 2[∠ACB + ∠ACD] + [∠BAC + ∠CAD] = 360°

⇒ 2∠BCD +180° = 360° [∠BAC and ∠CAD form a linear pair]

⇒ 2∠BCD = 360° – 180° = 180°

⇒ ∠BCD = = 90°

Thus, ∠BCD = 90°

∴ BM = AM

(i) In ∆AMC and ∆BMD, we have

CM = DM [Given]

∠AMC = ∠BMD [Vertically opposite angles]

AM = BM [Proved above]

∴ ∆AMC ≅ ∆BMD [By SAS congruency]

(ii) Since ∆AMC ≅ ∆BMD

⇒ ∠MAC = ∠MBD [By C.P.C.T.]

But they form a pair of alternate interior angles.

∴ AC || DB

Now, BC is a transversal which intersects parallel lines AC and DB,

∴ ∠BCA + ∠DBC = 180° [Co-interior angles]

But ∠BCA = 90° [∆ABC is right angled at C]

∴ 90° + ∠DBC = 180°

⇒ ∠DBC = 90°

(iii) Again, ∆AMC ≅ ∆BMD [Proved above]

∴ AC = BD [By C.P.C.T.]

Now, in ∆DBC and ∆ACB, we have

BD = CA [Proved above]

∠DBC = ∠ACB [Each 90°]

BC = CB [Common]

∴ ∆DBC ≅ ∆ACB [By SAS congruency]

(iv) As ∆DBC ≅ ∆ACB

DC = AB [By C.P.C.T.]

But DM = CM [Given]

∴ CM = DC = AB

⇒ CM = AB

0°

⇒ 2[∠ACB + ∠ACD] + [∠BAC + ∠CAD] = 360°

⇒ 2∠BCD +180° = 360° [∠BAC and ∠CAD form a linear pair]

⇒ 2∠BCD = 360° – 180° = 180°

⇒ ∠BCD = = 90°

Thus, ∠BCD = 90°

**Exercise 7.2****Question 7.**

ABC is a right angled triangle in which ∠A = 90° and AB = AC, find ∠B and ∠C.

ABC is a right angled triangle in which ∠A = 90° and AB = AC, find ∠B and ∠C.

Solution:

In ∆ABC, we have AB = AC [Given]

∴ Their opposite angles are equal.

⇒ ∠ACB = ∠ABC

Now, ∠A + ∠B + ∠C = 180° [Angle sum property of a ∆]

⇒ 90° + ∠B + ∠C = 180° [∠A = 90°(Given)]

⇒ ∠B + ∠C= 180°- 90° = 90°

But ∠B = ∠C

∠B = ∠C = = 45°

Thus, ∠B = 45° and ∠C = 45°

∴ Their opposite angles are equal.

⇒ ∠ACB = ∠ABC

Now, ∠A + ∠B + ∠C = 180° [Angle sum property of a ∆]

⇒ 90° + ∠B + ∠C = 180° [∠A = 90°(Given)]

⇒ ∠B + ∠C= 180°- 90° = 90°

But ∠B = ∠C

∠B = ∠C = = 45°

Thus, ∠B = 45° and ∠C = 45°

**Question 8.**

Show that the angles of an equilateral triangle are 60° each.Show that the angles of an equilateral triangle are 60° each.

Solution:

In ∆ABC, we have

AB = BC = CA

[ABC is an equilateral triangle]

AB = BC

⇒ ∠A = ∠C …(1) [Angles opposite to equal sides of a A are equal]

Similarly, AC = BC

⇒ ∠A = ∠B …(2)

From (1) and (2), we have

∠A = ∠B = ∠C = x (say)

Since, ∠A + ∠B + ∠C = 180° [Angle sum property of a A]

∴ x + x + x = 180o

⇒ 3x = 180°

⇒ x = 60°

∴ ∠A = ∠B = ∠C = 60°

Thus, the angles of an equilateral triangle are 60° each.

AB = BC = CA

[ABC is an equilateral triangle]

AB = BC

⇒ ∠A = ∠C …(1) [Angles opposite to equal sides of a A are equal]

Similarly, AC = BC

⇒ ∠A = ∠B …(2)

From (1) and (2), we have

∠A = ∠B = ∠C = x (say)

Since, ∠A + ∠B + ∠C = 180° [Angle sum property of a A]

∴ x + x + x = 180o

⇒ 3x = 180°

⇒ x = 60°

∴ ∠A = ∠B = ∠C = 60°

Thus, the angles of an equilateral triangle are 60° each.

**Exercise 7.3**

**Question 1.**

**∆ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see figure). If AD is extended to intersect BC at P, show that**

**(i) ∆ABD ≅ ∆ACD**

(ii) ∆ABP ≅ ∆ACP

(iii) AP bisects ∠A as well as ∠D

(iv) AP is the perpendicular bisector of BC

(ii) ∆ABP ≅ ∆ACP

(iii) AP bisects ∠A as well as ∠D

(iv) AP is the perpendicular bisector of BC

Solution:

(i) In ∆ABD and ∆ACD, we have

AB = AC [Given]

AD = DA [Common]

BD = CD [Given]

∴ ∆ABD ≅ ∆ACD [By SSS congruency]

∠BAD = ∠CAD [By C.P.C.T.] …(1)

(ii) In ∆ABP and ∆ACP, we have

AB = AC [Given]

∠BAP = ∠CAP [From (1)]

∴ AP = PA [Common]

∴ ∆ABP ≅ ∆ACP [By SAS congruency]

(iii) Since, ∆ABP ≅ ∆ACP

⇒ ∠BAP = ∠CAP [By C.P.C.T.]

∴ AP is the bisector of ∠A.

Again, in ∆BDP and ∆CDP,

we have BD = CD [Given]

DP = PD [Common]

BP = CP [ ∵ ∆ABP ≅ ∆ACP]

⇒ A BDP = ACDP [By SSS congruency]

∴ ∠BDP = ∠CDP [By C.P.C.T.]

⇒ DP (or AP) is the bisector of ∠BDC

∴ AP is the bisector of ∠A as well as ∠D.

(iv) As, ∆ABP ≅ ∆ACP

⇒ ∠APS = ∠APC, BP = CP [By C.P.C.T.]

But ∠APB + ∠APC = 180° [Linear Pair]

∴ ∠APB = ∠APC = 90°

⇒ AP ⊥ BC, also BP = CP

Hence, AP is the perpendicular bisector of BC.

AB = AC [Given]

AD = DA [Common]

BD = CD [Given]

∴ ∆ABD ≅ ∆ACD [By SSS congruency]

∠BAD = ∠CAD [By C.P.C.T.] …(1)

(ii) In ∆ABP and ∆ACP, we have

AB = AC [Given]

∠BAP = ∠CAP [From (1)]

∴ AP = PA [Common]

∴ ∆ABP ≅ ∆ACP [By SAS congruency]

(iii) Since, ∆ABP ≅ ∆ACP

⇒ ∠BAP = ∠CAP [By C.P.C.T.]

∴ AP is the bisector of ∠A.

Again, in ∆BDP and ∆CDP,

we have BD = CD [Given]

DP = PD [Common]

BP = CP [ ∵ ∆ABP ≅ ∆ACP]

⇒ A BDP = ACDP [By SSS congruency]

∴ ∠BDP = ∠CDP [By C.P.C.T.]

⇒ DP (or AP) is the bisector of ∠BDC

∴ AP is the bisector of ∠A as well as ∠D.

(iv) As, ∆ABP ≅ ∆ACP

⇒ ∠APS = ∠APC, BP = CP [By C.P.C.T.]

But ∠APB + ∠APC = 180° [Linear Pair]

∴ ∠APB = ∠APC = 90°

⇒ AP ⊥ BC, also BP = CP

Hence, AP is the perpendicular bisector of BC.

**Question 2.**

AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that

(i) AD bisects BC

(ii) AD bisects ∠AAD is an altitude of an isosceles triangle ABC in which AB = AC. Show that

(i) AD bisects BC

(ii) AD bisects ∠A

Solution:

(i) In right ∆ABD and ∆ACD, we have

AB =AC [Given]

∠ADB = ∠ADC [Each 90°]

AD = DA [Common]

∴ ∆ABD ≅ ∆ACD [By RHS congruency]

So, BD = CD [By C.P.C.T.]

⇒ D is the mid-point of BC or AD bisects BC.

(ii) Since, ∆ABD ≅ ∆ACD,

⇒ ∠BAD = ∠CAD [By C.P.C.T.]

So, AD bisects ∠A

AB =AC [Given]

∠ADB = ∠ADC [Each 90°]

AD = DA [Common]

∴ ∆ABD ≅ ∆ACD [By RHS congruency]

So, BD = CD [By C.P.C.T.]

⇒ D is the mid-point of BC or AD bisects BC.

(ii) Since, ∆ABD ≅ ∆ACD,

⇒ ∠BAD = ∠CAD [By C.P.C.T.]

So, AD bisects ∠A

**Question 3.**

Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and OR and median PN of ∆PQR (see figure). Show that

(i) ∆ABC ≅ ∆PQR

(ii) ∆ABM ≅ ∆PQNTwo sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and OR and median PN of ∆PQR (see figure). Show that

(i) ∆ABC ≅ ∆PQR

(ii) ∆ABM ≅ ∆PQN

Solution:

In ∆ABC, AM is the median.

∴BM = BC ……(1)

In ∆PQR, PN is the median.

∴ QN = QR …(2)

And BC = QR [Given]

⇒ BC = QR

⇒ BM = QN …(3) [From (1) and (2)]

(i) In ∆ABM and ∆PQN, we have

AB = PQ , [Given]

AM = PN [Given]

BM = QN [From (3)]

∴ ∆ABM ≅ ∆PQN [By SSS congruency]

(ii) Since ∆ABM ≅ ∆PQN

⇒ ∠B = ∠Q …(4) [By C.P.C.T.]

Now, in ∆ABC and ∆PQR, we have

∠B = ∠Q [From (4)]

AB = PQ [Given]

BC = QR [Given]

∴ ∆ABC ≅ ∆PQR [By SAS congruency]

In ∆ABC, AM is the median.

∴BM = BC ……(1)

In ∆PQR, PN is the median.

∴ QN = QR …(2)

And BC = QR [Given]

⇒ BC = QR

⇒ BM = QN …(3) [From (1) and (2)]

(i) In ∆ABM and ∆PQN, we have

AB = PQ , [Given]

AM = PN [Given]

BM = QN [From (3)]

∴ ∆ABM ≅ ∆PQN [By SSS congruency]

(ii) Since ∆ABM ≅ ∆PQN

⇒ ∠B = ∠Q …(4) [By C.P.C.T.]

Now, in ∆ABC and ∆PQR, we have

∠B = ∠Q [From (4)]

AB = PQ [Given]

BC = QR [Given]

∴ ∆ABC ≅ ∆PQR [By SAS congruency]

**Question 4.**

BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

Solution:

Since BE ⊥ AC [Given]

∴ BEC is a right triangle such that ∠BEC = 90°

Similarly, ∠CFB = 90°

Now, in right ∆BEC and ∆CFB, we have

BE = CF [Given]

BC = CB [Common hypotenuse]

∠BEC = ∠CFB [Each 90°]

∴ ∆BEC ≅ ∆CFB [By RHS congruency]

So, ∠BCE = ∠CBF [By C.P.C.T.]

or ∠BCA = ∠CBA

Now, in ∆ABC, ∠BCA = ∠CBA

⇒ AB = AC [Sides opposite to equal angles of a ∆ are equal]

∴ ABC is an isosceles triangle.

Since BE ⊥ AC [Given]

∴ BEC is a right triangle such that ∠BEC = 90°

Similarly, ∠CFB = 90°

Now, in right ∆BEC and ∆CFB, we have

BE = CF [Given]

BC = CB [Common hypotenuse]

∠BEC = ∠CFB [Each 90°]

∴ ∆BEC ≅ ∆CFB [By RHS congruency]

So, ∠BCE = ∠CBF [By C.P.C.T.]

or ∠BCA = ∠CBA

Now, in ∆ABC, ∠BCA = ∠CBA

⇒ AB = AC [Sides opposite to equal angles of a ∆ are equal]

∴ ABC is an isosceles triangle.

**Question 5.**

ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.

Solution:

We have, AP ⊥ BC [Given]

∠APB = 90° and ∠APC = 90°

In ∆ABP and ∆ACP, we have

∠APB = ∠APC [Each 90°]

AB = AC [Given]

AP = AP [Common]

∴ ∆ABP ≅ ∆ACP [By RHS congruency]

So, ∠B = ∠C [By C.P.C.T.]

∠APB = 90° and ∠APC = 90°

In ∆ABP and ∆ACP, we have

∠APB = ∠APC [Each 90°]

AB = AC [Given]

AP = AP [Common]

∴ ∆ABP ≅ ∆ACP [By RHS congruency]

So, ∠B = ∠C [By C.P.C.T.]

**Exercise 7.4**

**Question 1.**

Show that in a right angled triangle, the hypotenuse is the longest side.

Show that in a right angled triangle, the hypotenuse is the longest side.

Solution:

Let us consider ∆ABC such that ∠B = 90°

∴ ∠A + ∠B + ∠C = 180°

⇒ ∠A + 90°-+ ∠C = 180°

⇒ ∠A + ∠C = 90°

⇒∠A + ∠C = ∠B

∴ ∠B > ∠A and ∠B > ∠C

⇒ Side opposite to ∠B is longer than the side opposite to ∠A

i.e., AC > BC.

Similarly, AC > AB.

Therefore, we get AC is the longest side. But AC is the hypotenuse of the triangle. Thus, the hypotenuse is the longest side.

∴ ∠A + ∠B + ∠C = 180°

⇒ ∠A + 90°-+ ∠C = 180°

⇒ ∠A + ∠C = 90°

⇒∠A + ∠C = ∠B

∴ ∠B > ∠A and ∠B > ∠C

⇒ Side opposite to ∠B is longer than the side opposite to ∠A

i.e., AC > BC.

Similarly, AC > AB.

Therefore, we get AC is the longest side. But AC is the hypotenuse of the triangle. Thus, the hypotenuse is the longest side.

**Question 2.**

In figure, sides AB and AC of ∆ABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.In figure, sides AB and AC of ∆ABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.

Solution:

∠ABC + ∠PBC = 180° [Linear pair]

and ∠ACB + ∠QCB = 180° [Linear pair]

But ∠PBC < ∠QCB [Given] ⇒ 180° – ∠PBC > 180° – ∠QCB

⇒ ∠ABC > ∠ACB

The side opposite to ∠ABC > the side opposite to ∠ACB

⇒ AC > AB.

∠ABC + ∠PBC = 180° [Linear pair]

and ∠ACB + ∠QCB = 180° [Linear pair]

But ∠PBC < ∠QCB [Given] ⇒ 180° – ∠PBC > 180° – ∠QCB

⇒ ∠ABC > ∠ACB

The side opposite to ∠ABC > the side opposite to ∠ACB

⇒ AC > AB.

**Question 3.**

In figure, ∠B <∠ A and ∠C <∠ D. Show that AD < BC.In figure, ∠B <∠ A and ∠C <∠ D. Show that AD < BC.

Solution: Since ∠A > ∠B [Given]

∴ OB > OA …(1)

[Side opposite to greater angle is longer]

Similarly, OC > OD …(2)

Adding (1) and (2), we have

OB + OC > OA + OD

⇒ BC > AD

∴ OB > OA …(1)

[Side opposite to greater angle is longer]

Similarly, OC > OD …(2)

Adding (1) and (2), we have

OB + OC > OA + OD

⇒ BC > AD

**Question 4.**

AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure). Show that ∠ A > ∠C and ∠B >∠D.AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure). Show that ∠ A > ∠C and ∠B >∠D.

Solution:

Let us join AC.

Now, in ∆ABC, AB < BC [∵ AB is the smallest side of the quadrilateral ABCD] ⇒ BC > AB

⇒ ∠BAC > ∠BCA …(1)

[Angle opposite to longer side of A is greater]

Again, in ∆ACD, CD > AD

[ CD is the longest side of the quadrilateral ABCD]

⇒ ∠CAD > ∠ACD …(2)

[Angle opposite to longer side of ∆ is greater]

Adding (1) and (2), we get

∠BAC + ∠CAD > ∠BCA + ∠ACD

⇒ ∠A > ∠C

Similarly, by joining BD, we have ∠B > ∠D.

Now, in ∆ABC, AB < BC [∵ AB is the smallest side of the quadrilateral ABCD] ⇒ BC > AB

⇒ ∠BAC > ∠BCA …(1)

[Angle opposite to longer side of A is greater]

Again, in ∆ACD, CD > AD

[ CD is the longest side of the quadrilateral ABCD]

⇒ ∠CAD > ∠ACD …(2)

[Angle opposite to longer side of ∆ is greater]

Adding (1) and (2), we get

∠BAC + ∠CAD > ∠BCA + ∠ACD

⇒ ∠A > ∠C

Similarly, by joining BD, we have ∠B > ∠D.

**Question 5.**

In figure, PR > PQ and PS bisect ∠QPR. Prove that ∠PSR >∠PSQ.In figure, PR > PQ and PS bisect ∠QPR. Prove that ∠PSR >∠PSQ.

Solution:

In ∆PQR, PS bisects ∠QPR [Given]

∴ ∠QPS = ∠RPS

and PR > PQ [Given]

⇒ ∠PQS > ∠PRS [Angle opposite to longer side of A is greater]

⇒ ∠PQS + ∠QPS > ∠PRS + ∠RPS …(1) [∵∠QPS = ∠RPS]

∵ Exterior ∠PSR = [∠PQS + ∠QPS]

and exterior ∠PSQ = [∠PRS + ∠RPS]

[An exterior angle is equal to the sum of interior opposite angles]

Now, from (1), we have

∠PSR = ∠PSQ.

∴ ∠QPS = ∠RPS

and PR > PQ [Given]

⇒ ∠PQS > ∠PRS [Angle opposite to longer side of A is greater]

⇒ ∠PQS + ∠QPS > ∠PRS + ∠RPS …(1) [∵∠QPS = ∠RPS]

∵ Exterior ∠PSR = [∠PQS + ∠QPS]

and exterior ∠PSQ = [∠PRS + ∠RPS]

[An exterior angle is equal to the sum of interior opposite angles]

Now, from (1), we have

∠PSR = ∠PSQ.

**Question 6.**

Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

Solution:

Let us consider the ∆PMN such that ∠M = 90°

Since, ∠M + ∠N+ ∠P = 180°

[Sum of angles of a triangle is 180°]

∵ ∠M = 90° [PM ⊥ l]

So, ∠N + ∠P = ∠M

⇒ ∠N < ∠M

⇒ PM < PN …(1)

Similarly, PM < PN1 …(2)

and PM < PN2 …(3)

From (1), (2) and (3), we have PM is the smallest line segment drawn from P on the line l. Thus, the perpendicular line segment is the shortest line segment drawn on a line from a point not on it.

Let us consider the ∆PMN such that ∠M = 90°

Since, ∠M + ∠N+ ∠P = 180°

[Sum of angles of a triangle is 180°]

∵ ∠M = 90° [PM ⊥ l]

So, ∠N + ∠P = ∠M

⇒ ∠N < ∠M

⇒ PM < PN …(1)

Similarly, PM < PN1 …(2)

and PM < PN2 …(3)

From (1), (2) and (3), we have PM is the smallest line segment drawn from P on the line l. Thus, the perpendicular line segment is the shortest line segment drawn on a line from a point not on it.

**Exercise 7.5****Question 1.**

ABC is a triangle. Locate a point in the interior of ∆ ABC which is equidistant from all the vertices of ∆ ABC.

ABC is a triangle. Locate a point in the interior of ∆ ABC which is equidistant from all the vertices of ∆ ABC.

Solution:

Let us consider a ∆ABC.

Draw l, the perpendicular bisector of AB.

Draw m, the perpendicular bisector of BC.

Let the two perpendicular bisectors l and m meet at O.

O is the required point which is equidistant from A, B and C.

Note: If we draw a circle with centre O and radius OB or OC, then it will pass through A, B and C. The point O is called circumcentre of the triangle.

Let us consider a ∆ABC.

Draw l, the perpendicular bisector of AB.

Draw m, the perpendicular bisector of BC.

Let the two perpendicular bisectors l and m meet at O.

O is the required point which is equidistant from A, B and C.

Note: If we draw a circle with centre O and radius OB or OC, then it will pass through A, B and C. The point O is called circumcentre of the triangle.

**Question 2.**

In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.

Solution:

Let us consider a ∆ABC.

Draw m, the bisector of ∠C.

Let the two bisectors l and m meet at O.

Thus, O is the required point which is equidistant from the sides of ∆ABC.

Note: If we draw OM ⊥ BC and draw a circle with O as centre and OM as radius, then the circle will touch the sides of the triangle. Point O is called incentre of the triangle.

Let us consider a ∆ABC.

Draw m, the bisector of ∠C.

Let the two bisectors l and m meet at O.

Thus, O is the required point which is equidistant from the sides of ∆ABC.

Note: If we draw OM ⊥ BC and draw a circle with O as centre and OM as radius, then the circle will touch the sides of the triangle. Point O is called incentre of the triangle.

**Question 3.**

In a huge park, people are concentrated at three points (see figure)In a huge park, people are concentrated at three points (see figure)

**A: where these are different slides and swings for children.**

B: near which a man-made lake is situated.

C: which is near to a large parking and exist.

Where should an ice-cream parlor be set? up so that maximum number of persons can approach it?

[Hint The parlour should be equidistant from A, B and C.]B: near which a man-made lake is situated.

C: which is near to a large parking and exist.

Where should an ice-cream parlor be set? up so that maximum number of persons can approach it?

[Hint The parlour should be equidistant from A, B and C.]

Solution:

Let us join A and B, and draw l, the perpendicular bisector of AB.

Now, join B and C, and draw m, the perpendicular bisector of BC. Let the perpendicular bisectors l and m meet at O.

The point O is the required point where the ice cream parlour be set up.

Note: If we join A and C and draw the perpendicular bisector, then it will also meet (or pass through) the point O.

Let us join A and B, and draw l, the perpendicular bisector of AB.

Now, join B and C, and draw m, the perpendicular bisector of BC. Let the perpendicular bisectors l and m meet at O.

The point O is the required point where the ice cream parlour be set up.

Note: If we join A and C and draw the perpendicular bisector, then it will also meet (or pass through) the point O.

**Question 4.**

Complete the hexagonal and star shaped Rangolies [see Fig. (i) and (ii)] by filling them with as many equilateral triangles of side 1 cm as you can. Count the number of triangles in each case. Which has more triangles?Complete the hexagonal and star shaped Rangolies [see Fig. (i) and (ii)] by filling them with as many equilateral triangles of side 1 cm as you can. Count the number of triangles in each case. Which has more triangles?

Solution:

It is an activity.

We require 150 equilateral triangles of side 1 cm in the Fig. (i) and 300 equilateral triangles in the Fig. (ii).

∴ The Fig. (ii) has more triangles.

It is an activity.

We require 150 equilateral triangles of side 1 cm in the Fig. (i) and 300 equilateral triangles in the Fig. (ii).

∴ The Fig. (ii) has more triangles.