NCERT Solutions for Class 9 Maths Chapter 12

NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Exercise 12.1 and Exercise 12.2 in English Medium or प्रश्नावली 12.1 and प्रश्नावली 12.2 in हिंदी मीडियम to study online or in PDF form to free download. All NCERT Solutions as well as Apps are in Hindi Medium and English Medium Free.

 Class 9: Maths – गणित Chapter 12: Heron’s Formula

NCERT Solutions for Class 9 Maths Chapter 12

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9 Maths Exercise 12.1 Solutions

NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Exercise 12.1 is given below. See also Exercise 12.2 in English Medium or प्रश्नावली 12.1 and प्रश्नावली 12.2 in हिंदी मीडियम to study online. Visit to Class 9 Maths main page or move to Top of the page.

9 Maths Exercise 12.2 Solutions

NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Exercise 12.2 is given below. See also Exercise 12.1 in English Medium or प्रश्नावली 12.1 and प्रश्नावली 12.2 in हिंदी मीडियम to study online. Visit to Class 9 Maths main page or move to Top of the page.

9 गणित के प्रश्नावली 12.1 के हल

NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula प्रश्नावली 12.1 is given below. See also प्रश्नावली 12.2 in Hindi Medium or  Exercise 12.1 and Exercise 12.2 in English Medium to study online. Visit to Class 9 Maths main page or move to Top of the page.

9 गणित के प्रश्नावली 12.2 के हल

NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula प्रश्नावली 12.2 is given below. See also  प्रश्नावली 12.1 in Hindi or  Exercise 12.1 and Exercise 12.2 in English Medium to study online. Visit to Class 9 Maths main page or move to Top of the page.

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Important Formulae and Terms on Heron’s Formula
• Area of Triangle: The total space inside the boundary of the triangle is known as area of the triangle.
• Area of triangle = ½ × base × height
• Area of equilateral triangle: Let the side of an equilateral triangle be k. Then, area of an equilateral triangle = (√3/4) k². Square units and altitude = (√3/2) k units.
• Area of an isosceles triangle: Let B be the base and S be the equal sides of an isosceles triangle, then area of an isosceles triangle = [B√(4S² – B²)]/2 square units.
• Perimeter: Perimeter of a triangle is equal to the sum of its three sides. It is denoted by 2s, where s is the semi-perimeter of a triangle.
• Heron’s Formula: The formula given by Heron about the area of a triangle is known as Heron’s formula. According to this formula, area of triangle = √[s (s – a)(s – b)(s – c)], where a, b and c are three sides of the triangle and s is the semi-perimeter. This formula is also used for finding the area of quadrilateral. In quadrilateral, we join one diagonal to divide the quadrilateral into two triangles and then find the area of each triangle separately by Heron’s formula.

There is a slide in a park. One of its side walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN”. If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.

Here, the sides of triangle are
a = 15 m, b = 11 m and c = 6 m.
So, the semi-perimeter of triangle is given by
s = (a + b + c)/2
= (15 + 11 + 6)/2
= 32/2
=16 m
Therefore,
using Heron’s formula, the area of triangle
= √(s(s – a)(s – b)(s – c) )
= √(16(16 – 15)(16 – 11)(16 – 6) )
= √(16(1)(5)(10) )
= √(4 × 4 ×(1)(5)(5 × 2) )
= 4 × 5 √2
= 20 √2 m^2

Hence, the area painted in colour is 20√2 m^2.

एक त्रिभुज और एक समांतर चतुर्भुज का एक ही आधार है और क्षेत्रफल भी एक ही है। यदि त्रिभुज की भुजाएँ 26 cm, 28 cm और 30 cm हैं तथा समांतर चतुर्भुज 28 cm के आधार पर स्थित है, तो उसकी संगत ऊँचाई ज्ञात कीजिए।

यहाँ, त्रिभुज ABE की भुजाएँ
a = 28 cm, b = 26 cm और c = 30 cm हैं।
अत:, त्रिभुज का अर्धपरिमाप
s = (a + b + c)/2
= (28 + 26 + 30)/2
= 84/2
= 42 cm
इसलिए, हीरोन के सूत्र से,
त्रिभुज का क्षेत्रफल
= √(s(s – a)(s – b)(s – c) )
= √(42(42 – 28)(42 – 26)(42 – 30) )
= √(42(14)(16)(12) )
= √112896
= 336 cm^2

हम जानते हैं कि समांतर चतुर्भुज का क्षेत्रफल = आधार × संगत ऊँचाई
प्रश्नानुसार, समांतर चतुर्भुज का क्षेत्रफल = त्रिभुज का क्षेत्रफल
⇒ आधार × संगत ऊँचाई = 336
⇒ 28 × संगत ऊँचाई = 336
⇒ संगत ऊँचाई = 336/28 = 12 cm

एक पार्क चतुर्भुज ABCD के आकार का है, जिसमें ∠C = 90°, AB = 9 cm, BC = 12 cm, CD = 5 cm और AD = 8 cm है। इस पार्क का कितना क्षेत्रफल है?

BD को मिलाया।
त्रिभुज BDC में, पाइथागोरस प्रमेय से
BD^2 = BC^2 + CD^2
〖⇒BD〗^2 =〖12〗^2 + 5^2
= 144 + 25 = 169
⇒ BD =13 cm

त्रिभुज BDC का क्षेत्रफल
= 1/2 × BC × DC
= 1/2 × 12 × 5
= 30 cm^2

यहाँ, त्रिभुज ABD में, त्रिभुज की भुजाएँ
a = 9 cm, b = 8 cm और c = 13 cm हैं।
अत:, त्रिभुज ABD का अर्धपरिमाप
s = (a + b + c)/2
= (9 + 8 + 13)/2
= 30/2
= 15 cm

इसलिए, हीरोन के सूत्र से, त्रिभुज ABD का क्षेत्रफल
= √(s(s – a)(s – b)(s – c) )
= √(15(15 – 9)(15 – 8)(15 – 13) )
= √(15(6)(7)(2) )
= √1260
= 35.5 cm^2 (लगभग)
पार्क का कुल क्षेत्रफल = त्रिभुज BDC का क्षेत्रफल + त्रिभुज ABD का क्षेत्रफल
⇒ पार्क का कुल क्षेत्रफल = 30 + 3535
= 65.5 cm^2

अतः, पार्क का क्षेत्रफल 65.5 cm^2 है।

Find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is 42cm.

Here, the sides of triangle are
a = 18 cm, b = 10 cm and perimeter is 42 cm.
We know that the perimeter of triangle
= a + b + c
⇒ 42 = 18 + 10 + c
⇒ c = 14 cm

So, the semi-perimeter of triangle is given by
s = (a + b + c)/2
= 42/2
=21 cm

Therefore, using Heron’s formula, the area of triangle
= √(s(s – a)(s – b)(s – c) )
= √(21(21 – 18)(21 – 10)(21 – 14) )
= √(21(3)(11)(7) )
= √(7 × 3 × (3)(11)(7) )
= 7 × 3√11
= 21√11 cm^2

Hence, the area of triangle is 21√11 cm^2.

Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540cm. Find its area.

Perimeter of triangle =540 cm
The ratio of sides of triangle = 12:17:25
Let, one of the sides of triangle a = 12x
Therefore, remaining two sides are b = 17x and c = 25x.

We know that the perimeter of triangle
= a + b + c
⇒ 540 = 12x + 17x + 25x
⇒ 540 = 54x
⇒ x =540/54 = 10

So, the sides of triangle are
a = 12 × 10 =120 cm,
b = 17 × 10 =170 cm and
c = 25 × 10 =250 cm.

So, the semi-perimeter of triangle is given by
s = (a + b + c)/2
= 540/2
=270 cm
Therefore, using Heron’s formula, the area of triangle
= √(s(s – a)(s – b)(s – c) )
= √(270(270 – 120)(270 – 170)(270 – 250) )
= √(270(150)(100)(20) )
= √81000000
= 9000 cm^2

Hence, the area of triangle is 9000 cm^2.

An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.

Perimeter of triangle = 30 cm
Two sides of triangle b = 12 cm and c = 12 cm.
Let, the third side = a cm
We know that the perimeter of triangle
= a + b + c
⇒ 30 = a + 12 + 12
⇒ 30 – 24 = a
⇒ a = 6

So, the semi-perimeter of triangle is given by
s = (a + b + c)/2
= 30/2
= 15 cm

Therefore, using Heron’s formula, the area of triangle
= √(s(s – a)(s – b)(s – c) )
= √(15(15 – 6)(15 – 12)(15 – 12) )
= √(15(9)(3)(3) )
= 9√15 cm^2

Hence, the area of triangle is 9√15 cm^2.

1 thought on “NCERT Solutions for Class 9 Maths Chapter 12”

1. ajay prajapati says:

your solution is very easier to understand