# NCERT Solutions for Class 9 Maths Chapter 12: Heron’s Formula

## NCERT Solutions for Class 9 Mathematics Chapter 12 Free PDF Download

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Q. The perimeter of a triangular field is 540 m and the ratio of its sides is 12 : 17 : 25. Find the area of the field.

Ans.

$$\text{Given : Ratio of sides of the field = 12 : 17 : 25}\\\text{Let the lengths of sides be 12x, 17x and 25x,}\\ respectively.\\ Perimeter of triangle = 540m\\ ⇒ 12x + 17x + 25x = 540\\ ⇒ 54x = 540\\ ⇒ x = 10m\\ \text{ Lengths of sides are}\\ \text{12 x = 120m, 17 x = 170m and 25 x = 250 m.}\\ Semi-perimeter, s =\frac{a+b+c}{2}=\frac{120+170+250}{2}\\ =\frac{540}{2}\\ = 270m\\ \text{Now, Area of triangle} =\sqrt{\mathstrut s(s-a)(s-b)(s-c)}\\ =\sqrt{\mathstrut 270(270-120)(270-170)(270-250)}\\ =\sqrt{\mathstrut 270(150)(100)(20)}\\ =\sqrt{\mathstrut 100 × 100 × 30 × 270}\\ = 100 × 10 × 9\\ = 9000\\ Hence, the area of the field is 9,000 m^2.$$

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

Ans.

$$\text{Let the base of the isosceles triangle be a cm.}\\\text{Now, Perimeter = 30 cm}\\⇒ 12 + 12 + a = 30\\ ⇒ a = 30 – 24 = 6 cm\\ Also, s =\frac{Perimeter}{2}=\frac{30}{2}=15cm\\ \text{Now, Area of triangle} =\sqrt{\mathstrut s(s-a)(s-b)(s-c)}\\ =\sqrt{\mathstrut 15(15-12)(15-12)(15-5)}\\ =\sqrt{\mathstrut 15×3×3×9}\\ =9\sqrt{\mathstrut 15}cm^2$$

Q. What is the area of an isosceles triangle having base 2 cm and length of one of its equal sides as 4 cm ?

Ans.

$$\text{By Heron’s formula,}\\ s=\frac{a+b+c}{2}\\ =\frac{4+4+2}{2}=5$$

$$A =\sqrt{\mathstrut s(s-a)(s-b)(s-c)}\\ =\sqrt{\mathstrut 5(5-2)(5-4)(5-4)}\\ =\sqrt{\mathstrut 5×3}\\ ={\mathstrut 15}cm^2$$

Q. The edges of a triangular board are 6cm, 8cm and 10cm. What is the cost of painting it at the rate of a 9 paise cm2 ?

Ans.

$$\text{Area of the triangular board, by Heron’s formula,}\\ Here, a = 6, b = 8, c = 10\\ S =\frac{a+b+c}{2}=\frac{6+8+10}{2}=12\\ Area =\sqrt{\mathstrut s(s-a)(s-b)(s-c)}\\ =\sqrt{\mathstrut 12(12-6)(12-8)(12-10)}\\ =\sqrt{\mathstrut 12×6×4×2}\\ =\sqrt{\mathstrut 2×2×3×2×2×2×2}\\ = 2 × 2 × 2 × 3 = 24 cm^2\\ \text{Cost = Area × Rate}$$

Q. 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 colours.

Ans.

$$\text{Area of the triangular board, by Heron’s formula,}\\ Here, a = 6, b = 8, c = 10\\ S =\frac{a+b+c}{2}=\frac{11+6+15}{2}=12\\ Area =\sqrt{\mathstrut s(s-a)(s-b)(s-c)}\\ =\sqrt{\mathstrut 16(16-15)(16-11)(16-6)}\\ =\sqrt{\mathstrut 16×1×5×10}\\ =\sqrt{\mathstrut 4×4×5×5×2}\\ =20\sqrt{\mathstrut 2}\\ \text{Hence, the area painted in colour is}=20\sqrt{\mathstrut 2}m^2$$

Q. A traffic signal board, indicating “School Ahead” is an equilateral triangle with its perimeter 180 cm. What will be the area of the signal board?

Ans.

$$\text{Perimeter of signal board = 180 cm}\\ \text{3 × Side = 180 cm}\\ ⇒ Side =\frac{180}{3}=60cm\\ \text{Area of signal board}=\frac{\sqrt{\mathstrut 3}}{4}×(side)^2\\=\frac{\sqrt{\mathstrut 3}}{4}60×60\\=900\sqrt{\mathstrut 3}cm^2$$

Q. 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. The advertisements yield on earning of ₹5000 per m2 per year. A company hired both walls for 3 months. How much rent did it pay?

Ans. Consider DABC as one of the side walls of the flyover. So, in DABC, a = 122 m, b = 120 m and c = 22 m

$$⇒ s =\frac{a+b+c}{2}\\ =\frac{122+120+22}{2}\\ =\frac{264}{2}=132 m\\ ar(△ABC)={\sqrt{\mathstrut s(s-a)(s-b)(s-c)}}\\ ={\sqrt{\mathstrut 132(132-122)(132-120)(132-22)}}\\ ={\sqrt{\mathstrut 132×10×12×110}}\\ ={\sqrt{\mathstrut 12×11×10×12×11×10}}\\ = 10×11×12\\ =132m^2\\ \text{Area of both walls = 2 × ar(△ABC)}\\ = 2 × 1320\\ = 2640 m^2\\ \text{Rate of rent }= ₹ 5000 per m^2 per year\\ \text{Rate paid for 3 months = Rate × Area × Time}\\ = ₹ 5000 × 2640 ×\frac{3}{12}\\ = ₹ 5000 × 220 × 3\\ = ₹ 33,00,000$$