Herons Formula-NCERT Solutions

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?

Ans. For an equilateral triangle with side ‘a’, area

       

       ∴ Each side of the triangle = a cm

       ∴ a + a + a = 180 cm

       ⇒ 3a = 180 cm

       

       Now, s = Semi–perimeter

       ∴ Area of a triangle

           

       ∴ Area of the given triangle

       

       

       Thus, the area of the given triangle

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.). The advertisements yield an earning of Rs. 5000 per m² per year. A company hired one of its walls for 3 months. How much rent did it pay?

Ans. The sides of the triangular wall are

       a = 122 m, b = 120 m, c = 22 m

       

       

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 figure). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.

Ans. ∴ Sides of the wall are 15 m, 11 m and 6 m.

       ∴ a = 15 m, b = 11 m, c = 6 m

       

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

Ans. Let the sides of the triangle be

       a = 18 cm, b = 10 cm and c = ?

       ∴ Perimeter (2s) = 42 cm

       

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

Ans. Perimeter of the triangle = 540 cm

       ⇒ Semi–perimeter of the triangle,

           

       ∴ The sides are in the ratio of 12 : 17 : 25.

       ∴ a = 12x cm, b = 17x cm, c = 25x cm

       ∴ 12x + 17x + 25x = 540

       ⇒ 54x = 540

       

       ∴ a = 12 × 10 = 120 cm

           b = 17 × 10 = 170 cm

           c = 25 × 10 = 250 cm

       ⇒ (s - a) = (270 - 120) cm = 150 cm

           (s - b) = (270 - 170) cm = 100 cm

           (s - c) = (270 - 250) cm = 20 cm

       ∴ Area of the triangle

       

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

Ans. Equal sides of the triangle are 12 cm each.

       Let the third side = x cm.

       Since, perimeter = 30 cm

       ∴ 12 cm + 12 cm + x cm = 30 cm

       ⇒ x = 30 - 12 - 12 cm

       ⇒ x = 6 cm

       Now, semi–perimeter

       ∴ Area of the triangle

       

1.   A park, in the shape of a quadrilateral ABCD, has �C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?

Ans. Let us join B and D, such that BCD is a right–angled triangle.

       We have area of BCD

       

       Now, to find the area of ABD, we need the length of BD.

       ∴ In right, BCD,

       BD² = BC² = CD² [Pythagoras theorem]

       ⇒ BD² = 12² + 5²

       ⇒ BD² = 144 + 25 = 169 = 13²

       

       

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.

Ans. For ΔABC,

       

3.   Radha made a picture of an aeroplane with coloured paper as shown in Fig, Find the total area of the paper used.

Ans. Area of surface I

       It is an isosceles triangle whose sides are

       a = 5 cm, b = 5 cm, c = 1 cm

       

       = 0.75 × 3.3 cm² (approx.) = 2.475 cm² (approx.)

       Area of surface II

       It is a rectangle with length = 6.5 cm and breadth 1 cm.

       ∴ Area of rectangle II = Length × Breadth = 6.5 × 1 = 6.5 cm²

       Area of surface III

       It is a trapezium whose parallel sides are 1 cm and 2 cm as shown in the adjoining figure. Its height is given by

       

       Note: The perpendicular distance between the parallel sides is called the height of the trapezium.

       

       Area of surface IV

       It is a right triangle with base as 6 cm and height as 1.5 cm.

       

       Area of surface V

       ∴ Right triangle V Right triangle IV

       ∴ Area of right triangle V = Area of right triangle IV = 4.5 cm²

       Thus, the total area of the paper used

       = (area I) + (area II) + (area III) + (area IV) + (area V)

       = [2.475 cm² (approx.)] + [6.5 cm²] + [1.3 cm² (approx.)] + [4.5 cm²] + [4.5 cm²]

       = 19.275 cm² = 19.3 cm² (approx.)

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.

Ans. 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

       ∴ [Area of the parallelogram] × [Height of the parallelogram] = 336 cm²

       ∴ [25 cm] × [h cm] = 336 cm²

       

       Thus, the required height of the parallelogram = 12 cm

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?

Ans. Here, each side of the rhombus = 30 m

       One of the diagonal = 48 m

       Since a diagonal divides the rhombus into two congruent triangles.

       Sides of first triangle are

           a = 30 m, b = 30 m, c = 48 m

       

       ∴ Area of second triangle = 432 m²

       ⇒ Total area of both triangles = 432 m² + 432 m² = 864 m²

       ⇒ The area of the rhombus = 864 m²

       Thus, area of grass for 18 cows = 864 m²

       ∴ Area of grass for 1 cow

6.   An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see figure), each piece measuring 20 cm, 50 cm and 50 cm. Now much cloth of each colour is required for the umbrella?

Ans. Sides of each triangular piece are

       a = 20 cm, b = 50 cm, c = 50 cm

       

       Note: Total triangular pieces are 10, i.e. five triangular pieces for each colour.

7.   A kite in the shape of a square with a diagonal 32 an and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in the figure. How much, paper of each shade has been used in it?

Ans. Area of triangle I

       ∴ Diagonal = 32 cm

       

       Area of triangle II

       ∴ Diagonal of a square divides it into two congruent triangles.

       ∴ Area of triangle II = Area of triangle II

       ⇒ Area of triangle II = 256 cm²

       Area of triangle III

       The triangle at the base is having sides as

       a = 8 cm, b = 6 cm, c = 6 cm

       

       

       Thus, the area of different shades are:

       Area of shade I = 256 cm², Area of shade II = 256 cm²

       Area of shade III = 17.92 cm²

8.   A floral design on a floor is made up: of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see figure). Find the cost of polishing the tiles at the rate of 50p per cm².

Ans. There are 16 equal triangular tiles.

       Area of a triangle

       Sides of the triangle are

       a = 9 cm, b = 28 cm, c = 35 cm

       

       Area of 16 triangles:

       Total area of all the triangles

       = 16 × 88.2 cm² (approx.) = 1411.2 cm² (approx.)

       Cost of polishing the tiles

       Rate of polishing = Rs. 0.5 per cm²

       ∴ Cost of polishing all the tiles

       = Rs. 0.5 × 1411.2

       = Rs. = Rs. 705.60 (approx.)

9.   A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non–parallel sides are 14 m and 13 m. Find the area of the field.

Ans. The given field is in th eform of a trapezium ABCD such that parallel sides

       Non–parallel sides are 13 cm and 14 cm.

       We draw BE || AD, such that BE = 13 cm.

       The given field is divided into two shapes: (i) parallelogram ABED (ii) ΔBCE.

       Area of ΔBCE:

       Sides of the triangle are

       a = 13 m, b = 14 m, c = 15 m

       

       Let the height of the BCE corresponding to the side 15 m be 'h' metres

       

       Area of parallelogram ABED

       ∴Area of a parallelogram = base × height

       ∴Area of parallelogram ABED

            = 2 × 56 m² = 112 m²

       Thus, area of the field = Area of BCE + Area of parallelogram ABED = 84 m² + 112 m² = 196 m²