Heron's Formula RD Sharma Solutions | Mathematics (Maths) Class 9 PDF Download

 Page 1


RD Sharma Class 9 Solution Chapter 12 Heron’s Formula 
Question 1. 
In the figure, the sides BA and CA have been produced such that BA = AD and CA = 
AE. Prove the segment DE || BC.
Solution: 
Given : Sides BA and CA of ?ABC are produced such that BA = AD are CA = AE. ED is 
joined. 
To prove : DE || BC 
Proof: In ?ABC and ?DAE AB=AD (Given) 
AC = AE (Given) 
?BAC = ?DAE (Vertically opposite angles) 
? ?ABC ? ?DAE (SAS axiom) 
? ?ABC = ?ADE (c.p.c.t.) 
But there are alternate angles 
? DE || BC 
Question 2. 
In a ?PQR, if PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP 
respectively. Prove that LN = MN. 
Solution: 
Given : In ?PQR, PQ = QR 
L, M and N are the mid points of the sides PQ, QR and PR respectively 
Page 2


RD Sharma Class 9 Solution Chapter 12 Heron’s Formula 
Question 1. 
In the figure, the sides BA and CA have been produced such that BA = AD and CA = 
AE. Prove the segment DE || BC.
Solution: 
Given : Sides BA and CA of ?ABC are produced such that BA = AD are CA = AE. ED is 
joined. 
To prove : DE || BC 
Proof: In ?ABC and ?DAE AB=AD (Given) 
AC = AE (Given) 
?BAC = ?DAE (Vertically opposite angles) 
? ?ABC ? ?DAE (SAS axiom) 
? ?ABC = ?ADE (c.p.c.t.) 
But there are alternate angles 
? DE || BC 
Question 2. 
In a ?PQR, if PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP 
respectively. Prove that LN = MN. 
Solution: 
Given : In ?PQR, PQ = QR 
L, M and N are the mid points of the sides PQ, QR and PR respectively 
 
To prove : LM = MN 
Proof : In ?LPN and ?MRH 
PN = RN ( ? M is mid point of PR) 
LP = MR (Half of equal sides) 
?P = ?R (Angles opposite to equal sides) 
? ALPN ? AMRH (SAS axiom) 
? LN = MN (c.p.c.t.) 
Question 3. 
Prove that the medians of an equilateral triangle are equal. 
Solution: 
Given : In ?ABC, AD, BE and CF are the medians of triangle and AB = BC = CA 
 
To prove : AD = BE = CF 
Proof : In ?BCE and ?BCF, 
BC = BC (Common side) 
CE = BF (Half of equal sides) 
?C = ?B (Angles opposite to equal sides) 
? ABCE ? ABCF (SAS axiom) 
? BE = CF (c.p.c.t.) …(i) 
Similarly, we can prove that 
? ?CAD ? ?CAF 
Page 3


RD Sharma Class 9 Solution Chapter 12 Heron’s Formula 
Question 1. 
In the figure, the sides BA and CA have been produced such that BA = AD and CA = 
AE. Prove the segment DE || BC.
Solution: 
Given : Sides BA and CA of ?ABC are produced such that BA = AD are CA = AE. ED is 
joined. 
To prove : DE || BC 
Proof: In ?ABC and ?DAE AB=AD (Given) 
AC = AE (Given) 
?BAC = ?DAE (Vertically opposite angles) 
? ?ABC ? ?DAE (SAS axiom) 
? ?ABC = ?ADE (c.p.c.t.) 
But there are alternate angles 
? DE || BC 
Question 2. 
In a ?PQR, if PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP 
respectively. Prove that LN = MN. 
Solution: 
Given : In ?PQR, PQ = QR 
L, M and N are the mid points of the sides PQ, QR and PR respectively 
 
To prove : LM = MN 
Proof : In ?LPN and ?MRH 
PN = RN ( ? M is mid point of PR) 
LP = MR (Half of equal sides) 
?P = ?R (Angles opposite to equal sides) 
? ALPN ? AMRH (SAS axiom) 
? LN = MN (c.p.c.t.) 
Question 3. 
Prove that the medians of an equilateral triangle are equal. 
Solution: 
Given : In ?ABC, AD, BE and CF are the medians of triangle and AB = BC = CA 
 
To prove : AD = BE = CF 
Proof : In ?BCE and ?BCF, 
BC = BC (Common side) 
CE = BF (Half of equal sides) 
?C = ?B (Angles opposite to equal sides) 
? ABCE ? ABCF (SAS axiom) 
? BE = CF (c.p.c.t.) …(i) 
Similarly, we can prove that 
? ?CAD ? ?CAF 
? AD = CF …(ii) 
From (i) and (ii) 
BE = CF = AD 
? AD = BE = CF 
Question 4. 
In a ?ABC, if ?A = 120° and AB = AC. Find ?B and ?C. 
Solution: 
In ?ABC, ?A = 120° and AB = AC 
? ?B = ?C (Angles opposite to equal sides) 
But ?A + ?B + ?C = 180° (Sum of angles of a triangle) 
 
? 120° + ?B + ?B = 180° 
? 2 ?B = 180° – 120° = 60° 
? ?B = 60 °2 = 30° 
and ?C = ?B = 30° 
Hence ?B = 30° and ?C = 30° 
Question 5. 
In a ?ABC, if AB = AC and ?B = 70°, find ?A. 
Solution: 
In ?ABC, ?B = 70° 
AB =AC 
? ?B = ?C (Angles opposite to equal sides) 
 
But ?B = 70° 
? ?C = 70° 
Page 4


RD Sharma Class 9 Solution Chapter 12 Heron’s Formula 
Question 1. 
In the figure, the sides BA and CA have been produced such that BA = AD and CA = 
AE. Prove the segment DE || BC.
Solution: 
Given : Sides BA and CA of ?ABC are produced such that BA = AD are CA = AE. ED is 
joined. 
To prove : DE || BC 
Proof: In ?ABC and ?DAE AB=AD (Given) 
AC = AE (Given) 
?BAC = ?DAE (Vertically opposite angles) 
? ?ABC ? ?DAE (SAS axiom) 
? ?ABC = ?ADE (c.p.c.t.) 
But there are alternate angles 
? DE || BC 
Question 2. 
In a ?PQR, if PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP 
respectively. Prove that LN = MN. 
Solution: 
Given : In ?PQR, PQ = QR 
L, M and N are the mid points of the sides PQ, QR and PR respectively 
 
To prove : LM = MN 
Proof : In ?LPN and ?MRH 
PN = RN ( ? M is mid point of PR) 
LP = MR (Half of equal sides) 
?P = ?R (Angles opposite to equal sides) 
? ALPN ? AMRH (SAS axiom) 
? LN = MN (c.p.c.t.) 
Question 3. 
Prove that the medians of an equilateral triangle are equal. 
Solution: 
Given : In ?ABC, AD, BE and CF are the medians of triangle and AB = BC = CA 
 
To prove : AD = BE = CF 
Proof : In ?BCE and ?BCF, 
BC = BC (Common side) 
CE = BF (Half of equal sides) 
?C = ?B (Angles opposite to equal sides) 
? ABCE ? ABCF (SAS axiom) 
? BE = CF (c.p.c.t.) …(i) 
Similarly, we can prove that 
? ?CAD ? ?CAF 
? AD = CF …(ii) 
From (i) and (ii) 
BE = CF = AD 
? AD = BE = CF 
Question 4. 
In a ?ABC, if ?A = 120° and AB = AC. Find ?B and ?C. 
Solution: 
In ?ABC, ?A = 120° and AB = AC 
? ?B = ?C (Angles opposite to equal sides) 
But ?A + ?B + ?C = 180° (Sum of angles of a triangle) 
 
? 120° + ?B + ?B = 180° 
? 2 ?B = 180° – 120° = 60° 
? ?B = 60 °2 = 30° 
and ?C = ?B = 30° 
Hence ?B = 30° and ?C = 30° 
Question 5. 
In a ?ABC, if AB = AC and ?B = 70°, find ?A. 
Solution: 
In ?ABC, ?B = 70° 
AB =AC 
? ?B = ?C (Angles opposite to equal sides) 
 
But ?B = 70° 
? ?C = 70° 
But ?A + ?B + ?C = 180° (Sum of angles of a triangle) 
? ?A + 70° + 70° = 180° 
? ?A + 140°= 180° 
??A = 180°- 140° = 40° 
Question 6. 
The vertical angle of an isosceles triangle is 100°. Find its base angles. 
Solution: 
In ?ABC, AB = AC and ?A = 100° 
But AB = AC (In isosceles triangle) 
 
? ?C = ?B (Angles opposite to equal sides) 
?A + ?B + ?C = 180° (Sum of angles of a triangle) 
? 100° + ?B + ?B = 180° ( ? ?C = ?B) 
? 2 ?B = 180° – 100° = 80° 
? ?C = ?B = 40° 
Hence ?B = 40°, ?C = 40° 
Question 7. 
In the figure, AB = AC and ?ACD = 105°, find ?BAC. 
 
Solution: 
In ?ABC, AB = AC 
? ?B = ?C (Angles opposite to equal sides) 
But ?ACB + ?ACD = 180° (Linear pair) 
? ?ACB + 105°= 180° 
? ?ACB = 180°-105° = 75° 
Page 5


RD Sharma Class 9 Solution Chapter 12 Heron’s Formula 
Question 1. 
In the figure, the sides BA and CA have been produced such that BA = AD and CA = 
AE. Prove the segment DE || BC.
Solution: 
Given : Sides BA and CA of ?ABC are produced such that BA = AD are CA = AE. ED is 
joined. 
To prove : DE || BC 
Proof: In ?ABC and ?DAE AB=AD (Given) 
AC = AE (Given) 
?BAC = ?DAE (Vertically opposite angles) 
? ?ABC ? ?DAE (SAS axiom) 
? ?ABC = ?ADE (c.p.c.t.) 
But there are alternate angles 
? DE || BC 
Question 2. 
In a ?PQR, if PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP 
respectively. Prove that LN = MN. 
Solution: 
Given : In ?PQR, PQ = QR 
L, M and N are the mid points of the sides PQ, QR and PR respectively 
 
To prove : LM = MN 
Proof : In ?LPN and ?MRH 
PN = RN ( ? M is mid point of PR) 
LP = MR (Half of equal sides) 
?P = ?R (Angles opposite to equal sides) 
? ALPN ? AMRH (SAS axiom) 
? LN = MN (c.p.c.t.) 
Question 3. 
Prove that the medians of an equilateral triangle are equal. 
Solution: 
Given : In ?ABC, AD, BE and CF are the medians of triangle and AB = BC = CA 
 
To prove : AD = BE = CF 
Proof : In ?BCE and ?BCF, 
BC = BC (Common side) 
CE = BF (Half of equal sides) 
?C = ?B (Angles opposite to equal sides) 
? ABCE ? ABCF (SAS axiom) 
? BE = CF (c.p.c.t.) …(i) 
Similarly, we can prove that 
? ?CAD ? ?CAF 
? AD = CF …(ii) 
From (i) and (ii) 
BE = CF = AD 
? AD = BE = CF 
Question 4. 
In a ?ABC, if ?A = 120° and AB = AC. Find ?B and ?C. 
Solution: 
In ?ABC, ?A = 120° and AB = AC 
? ?B = ?C (Angles opposite to equal sides) 
But ?A + ?B + ?C = 180° (Sum of angles of a triangle) 
 
? 120° + ?B + ?B = 180° 
? 2 ?B = 180° – 120° = 60° 
? ?B = 60 °2 = 30° 
and ?C = ?B = 30° 
Hence ?B = 30° and ?C = 30° 
Question 5. 
In a ?ABC, if AB = AC and ?B = 70°, find ?A. 
Solution: 
In ?ABC, ?B = 70° 
AB =AC 
? ?B = ?C (Angles opposite to equal sides) 
 
But ?B = 70° 
? ?C = 70° 
But ?A + ?B + ?C = 180° (Sum of angles of a triangle) 
? ?A + 70° + 70° = 180° 
? ?A + 140°= 180° 
??A = 180°- 140° = 40° 
Question 6. 
The vertical angle of an isosceles triangle is 100°. Find its base angles. 
Solution: 
In ?ABC, AB = AC and ?A = 100° 
But AB = AC (In isosceles triangle) 
 
? ?C = ?B (Angles opposite to equal sides) 
?A + ?B + ?C = 180° (Sum of angles of a triangle) 
? 100° + ?B + ?B = 180° ( ? ?C = ?B) 
? 2 ?B = 180° – 100° = 80° 
? ?C = ?B = 40° 
Hence ?B = 40°, ?C = 40° 
Question 7. 
In the figure, AB = AC and ?ACD = 105°, find ?BAC. 
 
Solution: 
In ?ABC, AB = AC 
? ?B = ?C (Angles opposite to equal sides) 
But ?ACB + ?ACD = 180° (Linear pair) 
? ?ACB + 105°= 180° 
? ?ACB = 180°-105° = 75° 
? ?ABC = ?ACB = 75° 
But ?A + ?B + ?C = 180° (Sum of angles of a triangle) 
? ?A + 75° + 75° = 180° 
? ?A + 150°= 180° 
? ?A= 180°- 150° = 30° 
? ?BAC = 30° 
Question 8. 
Find the measure of each exterior angle of an equilateral triangle. 
Solution: 
In an equilateral triangle, each interior angle is 60° 
 
But interior angle + exterior angle at each vertex = 180° 
? Each exterior angle = 180° – 60° = 120° 
Question 9. 
If the base of an isosceles triangle is produced on both sides, prove that the exterior 
angles so formed are equal to each other. 
Solution: 
Given : In an isosceles ?ABC, AB = AC 
and base BC is produced both ways 
 
To prove : ?ACD = ?ABE