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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

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FAQs on Heron's Formula RD Sharma Solutions - Mathematics (Maths) Class 9

1. What is Heron's formula?

Ans. Heron's formula is a mathematical formula used to find the area of a triangle when the lengths of its sides are known. It is named after Hero of Alexandria, a Greek mathematician who derived this formula.

2. How is Heron's formula derived?

Ans. Heron's formula is derived using the concept of semiperimeter of a triangle. The semiperimeter is calculated by adding the lengths of all three sides of the triangle and dividing it by 2. Then, using the semiperimeter, the area of the triangle is calculated using the formula A = √(s(s-a)(s-b)(s-c)), where A is the area, s is the semiperimeter, and a, b, and c are the lengths of the sides of the triangle.

3. Can Heron's formula be used for all types of triangles?

Ans. Yes, Heron's formula can be used to find the area of any type of triangle, whether it is equilateral, scalene, or isosceles. It is a general formula that works for all triangles, as long as the lengths of the sides are known.

4. How is Heron's formula helpful in real-life applications?

Ans. Heron's formula is used in various real-life applications, especially in fields like architecture, engineering, and surveying. It helps in calculating the area of irregularly shaped land or fields, determining the amount of materials required for construction, and in designing structures like bridges and buildings.

5. Are there any limitations or special cases when using Heron's formula?

Ans. Yes, there are a few limitations and special cases to consider when using Heron's formula. One limitation is that the lengths of the sides must form a valid triangle, i.e., the sum of any two sides must be greater than the third side. Additionally, when the lengths of the sides are very large, the calculations involved in the formula may result in rounding errors. Finally, if the triangle is degenerate (having collinear vertices), Heron's formula will not yield a valid area.

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