Heron's Formula

Heron's formula is a significant concept covered in the CBSE Class 9 Maths syllabus in the first semester. Therefore, grasping the concept is crucial. One of the best methods to do so is to use the NCERT Solutions for Class 9 Maths Chapter 12 Heron's Formula. These answers were developed by experienced teachers based on the current CBSE term-by-term syllabus. The NCERT Solutions for Class 9 include detailed and step-by-step explanations for all of the solutions provided in this chapter's exercises.

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**Question 1. A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side ‘a’. Find the area of the signal board, using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board?**

**Solution: **

Given that the signal board's side = a

The signal board's perimeter is 3a = 180 cm.

Therefore, a = 60 cm

The signal board's semi-perimeter (s)

= 3a/2

Using Heron's formula, you can

The triangle signal board's area will be

= \( \sqrt{s (s-a) (s- b) (s-c)} \)

= \( \sqrt{\frac{3a}{2} (\frac{3a}{2}-a) (\frac{3a}{2}- a) (\frac{3a}{2} - a)} \)

= \( \sqrt{\frac{3a}{2} \times \frac{a}{2} \times \frac{a}{2} \times \frac{a}{2}} \)

= \( \sqrt{\frac{3a^{4}}{16}} \)

= \( \sqrt{3} \frac{a^{2}}{16} \)

= \( \sqrt{3} \frac{a^{2}}{4} \)

= \( \frac{\sqrt{3}}{4} \times 60 \times 60 \)

= 900\( \sqrt{3} \) cm^{2}

**Question 2. The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see Fig. 12.9). The advertisements yield an earning of ₹5000 per m2 per year. A company hired one of its walls for 3 months. How much rent did it pay?**

**Solution:**

The triangle ABC has sides of 122 m, 22 m, and 120 m, respectively.

The perimeter will now be (122 + 22 + 120)

= 264 m.

Additionally, the semi-perimeter (s) = 264/2

= 132 m.

Using Heron's formula as a guide,

The triangle area is

= \( \sqrt{s (s-a) (s-b) (s-c)} \)

= \( \sqrt{132 (132-122) (132-22) (132-120)} ~m^{2} \)

= \( \sqrt{132 \times 10 \times 110 \times 12}~ m^{2} \)

= \( \sqrt{1742400}~m^{2} \)

= 1320 \( m^{2} \)

We know that the yearly advertising rent is approximately 5000 per \( m^{2} \).

As a result, the three-month rent for one wall is Rs. (1320 \( \times \) 5000 \( \times \) 3)/12 = Rs. 16,50,000.

**Question 3. 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” (see Fig. 12.10 ). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.**

**Solution:**

The dimensions of the wall are 15 metres, 11 metres, and 6 metres.

As a result, the semi-perimeter (s) of a triangle wall is (15+11+6)/2 m

= 16 m.

Using Heron's formula as a guide,

Area of the message

= \( \sqrt{s (s-a) (s-b) (s-c)} \)

= \( \sqrt{16 (16-15) (16-11) (16-6)}~ m^{2} \)

= \( \sqrt{16 \times 1 \times 5 \times 10}~m^{2} \)

= \( \sqrt{800}~m^{2} \)

= 20 \( \sqrt{2}~ m^{2} \)

**Question 4. Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42cm.**

Solution: Assume that the triangle's third side is "x."

The triangle's three sides are now 18 cm, 10 cm, and "x" cm.

The triangle's perimeter is calculated to be 42 cm.

As a result, x = 42-18+10 cm

= 14 cm.

As a result, the triangle's semi-perimeter = 42/2

= 21 cm.

Using Heron's formula as a guide,

The triangle's area,

= \( \sqrt{s (s-a) (s-b) (s-c)} \)

= \( \sqrt{21 (21-18) (21-10) (21-14)}~cm^{2} \)

= \( \sqrt{21 \times 3 \times 11 \times 7}~ cm^{2} \)

= 21 \( \sqrt{11}~ cm^{2} \)

**Question 5. Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540cm. Find its area.**

**Solution:**

The sides of the triangle have a ratio of 12: 17: 25.

Let the common ratio between the triangle's sides be "x."

As a result, the sides are 12x, 17x, and 25x, respectively.

It is also assumed that the triangle's circumference is 540 cm.

12x+17x+25x = 540 cm

54 \( \times \) 54 cm = 540 cm

As a result of this,

x = 10.

The triangle's sides are now 120 cm, 170 cm, and 250 cm.

As a result, the triangle's semi-perimeter (s) = 540/2

= 270 cm.

Using Heron's formula as a guide,

Triangle area,

= \( \sqrt{s (s-a) (s-b) (s-c)} \)

= \( \sqrt{270 (270-170) (270-150) (270-250)}~ cm^{2} \)

= \( \sqrt{270 \times 150 \times 100 \times 20}~ cm^{2} \)

= 9000 \( cm^{2} \)

**Question 6. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.**

**Solution: **

Allow the third side to be x at first.

It is assumed that the equal sides' length is 12 cm and the perimeter is 30 cm.

Thus, 30 = 12 + 12 + x

As a result, the third side's length equals 6 cm.

As a result, the isosceles triangle's semiperimeter (s) = 30/2 cm

= 15 cm.

Using Heron's formula as a guide,

Triangle area,

= \( \sqrt{s (s-a) (s-b) (s-c)} \)

= \( \sqrt{15 (15-12) (15-12) (15-6)} cm^{2} \)

= \( \sqrt{15 \times 3 \times 3 \times 9} ~cm^{2} \)

= \( \sqrt{1215}~ cm^{2} \)

= \( 9 \sqrt{15}~ cm^{2} \)

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