NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 12.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 12.1. NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 12.1 Ex 12.1 Class 9 Maths 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: Let each side of the equilateral triangle be a. Semi-perimeter of the triangle, Ex 12.1 Class 9 Maths 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 figure). 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: Let the sides of the triangular will be a = 122m, b = 12cm, c = 22m Semi-perimeter, s = a+b+c2 (122+120+224)m = 2642 m = 132m The area of the triangular side wall Rent for 1 year (i.e. 12 months) per m2 = Rs. 5000 ∴ Rent for 3 months per m2 = Rs. 5000 x 312 = Rent for 3 months for 1320 m2 = Rs. 5000 x 312 x 1320 = Rs. 16,50,000. Ex 12.1 Class 9 Maths Question 3. There is a slide in a park. One of its side Company hired one of its walls for 3 months.walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN” (see figure). If the sides of the wall are 15 m, 11 m and 6m, find the area painted in colour. Solution: Let the sides of the wall be a = 15m, b = 11m, c = 6m Semi-perimeter, Thus, the required area painted in colour = 20√2 m2 Ex 12.1 Class 9 Maths Question 4. Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm. Solution: Let the sides of the triangle be a =18 cm, b = 10 cm and c = x cm Since, perimeter of the triangle = 42 cm ∴ 18cm + 10 cm + xcm = 42 x = [42 – (18 + 10)cm = 14cm Now, semi-permimeter, s = 422cm = 21 cm Thus, the required area of the triangle = 2111−−√ cm2 Ex 12.1 Class 9 Maths Question 5. Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area. Solution: Let the sides of the triangle be a = 12x cm, b = 17x cm, c = 25x cm Perimeter of the triangle = 540 cm Now, 12x + 17x + 25x = 540 ⇒ 54x = 54 ⇒ x = 10 ∴ a = (12 x10)cm = 120cm, b = (17 x 10) cm = 170 cm and c = (25 x 10)cm = 250 cm Now, semi-perimeter, s = 5402cm = 270 cm Ex 12.1 Class 9 Maths 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: Let the sides of an isosceles triangle be a = 12cm, b = 12cm,c = x cm Since, perimeter of the triangle = 30 cm ∴ 12cm + 12cm + x cm = 30 cm ⇒ x = (30 – 24) = 6 Now, semi-perimeter, s = 302cm =15 cm Thus, the required area of the triangle = 9√15 cm2 NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 12.2 Ex 12.2 Class 9 Maths Question 1. A park, in the shape of a quadrilateral ABCD, has ∠C = 90°, AB = 9m,BC = 12m,CD = 5m and AD = 8 m. How much area does it occupy? Solution: Given, a quadrilateral ABCD with ZC = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. Let us join B and D, such that ABCD is a right angled triangle. Now, to find the area of ∆ABD, we need the length of BD. In right-angled ∆BCD, by Pythagoras theorem BD2 = 502 + CD2 ⇒ BD2 = 122 + 52 ⇒ BD2 = 144 + 25 = 169 ⇒ BD = 13 m Now, for ∆ABD, we have a = AB = 9 m, b = AD = 8 m, c = BD = 13 m ∴ Area of quadrilateral ABCD = area of ∆BCD + area of ∆ABD = 30 m2 + 35.5 m2 = 65.5 m2 (approx.) Ex 12.2 Class 9 Maths Question 2. Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm. Solution: Given a quadrilateral ABCD with AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm. For ∆ABC, a = AB = 3 cm, b = BC = 4 cm and c = AC = 5 cm Now, area of quadrilateral ABCD = area of ∆ABC + area of ∆ACD = 6 cm2 + 9.2 cm2 = 15.2 cm2 (approx.) Ex 12.2 Class 9 Maths Question 3. Radha made a picture of an aeroplane with coloured paper as shown in figure. Find the total area of the paper used. Solution: For surface I: It is an isosceles triangle whose sides are a = 5 cm, b = 5 cm, c = 1 cm = (0.75 x 3.3) cm2 = 2.475 cm2 (approx.) For surface II: It is a rectangle with length 6.5 cm and breadth 1 cm. ∴ Area of surface II = Length x Breadth = (6.5 x 1) cm2 = 6.5 cm2 For surface III: It is a trapezium whose parallel sides are 1 cm and 2 cm as shown in the figure given below: For surface IV and V: Surface V is a right-angled triangle with base 6cm arid height 1.5 cm. Also, area of surface IV = area of surface V = 12 x base x height = (12 x 6 x 15) cm2 = 4.5 cm2 Thus, the total area of the paper used = (area of surface I) + (area of surface II) + (area of surface III) + (area of surface IV) + (area of surface V) = [2.475 + 6.5 + 1.3 + 4.5 + 4.5] cm2 = 19.275 cm2 = 19.3 cm2 (approx.) Ex 12.2 Class 9 Maths Question 4. A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram Solution: For the given triangle, we have a = 28 cm, b = 30 cm, c = 26 cm Area of the given parallelogram = Area of the given triangle ∴ Area of the parallelogram = 336 cm2 ⇒ base x height = 336 ⇒ 28 x h = 336, where ‘h’ be the height of the parallelogram. ⇒ h = 33628 = 12 Thus, the required height of the parallelogram = 12 cm Ex 12.2 Class 9 Maths Question 5. A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting? Solution: Here, each side of the rhombus = 30 m. Let ABCD be the given rhombus and the diagonal, BD = 48 m Sides ∆ABC are a = AB = 30m, b = AD = 30m, c = BD = 48m S