12. Herons Formula Mathematics class 9 exercise Exercise 12.2
12. Herons Formula Mathematics class 9 exercise Exercise 12.2 ncert book solution in english-medium
NCERT Books Subjects for class 9th Hindi Medium
Exercise 12.1
Exercise 12.1
Q1. 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:
Perimeter of traffic signal board = 180 cm
Side of traffic signal board
Using equation (1), area of traffic signal board
Q2: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 Rs 5000 per m2 per year. A company hired one of its walls for 3 months. How much rent did it pay?
The sides of the triangle a,b,c are of 122 m, 22 m, and 120 m respectively.
Perimeter of triangle = (122 + 22 + 120) m
2s = 264 m
s = 132 m
By Heron's formula,
Rent of 1 m2 area per year = Rs 5000
Rent of 1 m2 area per month = Rs
Rent of 1320 m2 area for 3 months =
= Rs (5000 × 330) = Rs 1650000
Therefore, the company had to pay Rs 1650000
Q3: There is a slide in the park. One of its side walls has been painted in the same color with a message “KEEP THE PARK GREEN AND CLEAN” (see the given figure). If the sides of the wall are 15m, 11m, and 6m, find the area painted in color.
It can be observed that the area to be painted in colour is a triangle, having its sides as 11 m, 6 m, and 15 m.
Perimeter of such a triangle = (11 + 6 + 15) m
2 s = 32 m
s = 16 m
By Heron's formula
Area of triangle=
Therefore, the area painted in color is
Q4 : Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.
Answer :
Let the third side of the triangle be x.
Perimeter of the given triangle = 42 cm
18 cm + 10 cm + c = 42
c = 14 cm
5. Sides of a triangle are in the ratio of 12: 17: 25 and its perimeter is 540cm. Find its area of triangle.
Solution:
: Let the common ratio between the sides of the given triangle be x.
Therefore, the side of the triangle will be 12x, 17x, and 25x.
Perimeter of this triangle = 540 cm
12x + 17x + 25x = 540 cm
54x = 540 cm
x = 10 cm
Sides of the triangle will be 120 cm, 170 cm, and 250 cm.
Q6 : 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 third side of this triangle be x.
Perimeter of triangle = 30 cm
12 cm + 12 cm + x = 30 cm
x = 6 cm
Exercise 12.2
Q1: A park, in the shape of a quadrilateral ABCD, has ∠ C = 90º, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?
Solution: Let us join BD.
In ΔBCD, applying Pythagoras theorem,
BD2 = BC2 + CD2
= (12)2 + (5)2
= 144 + 25
BD2 = 169
BD = 13 m
Area of ΔBCD
For ΔABD,
By Heron's formula,
Area of Quadrilateral = Area of ΔABD + Area of ΔBCD
= 30m2+ 35.04m2
= 65.04m2
Q2: 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:
In ΔABC
AB = 3cm, BC = 4cm, AC = 5cm
Area of ΔABC
For ΔADC,
Perimeter = 2s = AC + CD + DA = (5 + 4 + 5) cm = 14 cm
s = 7 cm
By Heron's formula,
Area of ABCD = Area of ΔABC + Area of ΔADC
= 6 + 9.166
= 15.166 = 15.2cm2
Q3: Radha made a picture of an aero plane with colored papers as shown in the given figure. Find the total area of the paper used.
Solution:
For triangle I
This triangle is an isosceles triangle.
Perimeter = 2s = (5 + 5 + 1) cm = 11cm
S = = 5.5cm
Area of the triangle
For quadrilateral II
This quadrilateral is a rectangle.
Area = l × b = (6.5 × 1) cm2 = 6.5 cm2
For quadrilateral III
This quadrilateral is a trapezium.
Perpendicular height of parallelogram
Area = Area of parallelogram + Area of equilateral triangle
= 0.866 + 0.433 = 1.299 cm2
Area of triangle (IV) = Area of triangle in (V)
Total area of the paper used = 2.488 + 6.5 + 1.299 + 4.5 × 2
= 19.287 cm2
Q4: 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:
Here, Sides of triangle be A, B, C.
A = 26cm, b = 28cm, c = 30
⇒ s = 42cm
Area of parallelogram = Area of triangle
28cm × h = 336 cm2
h = 336/28 cm
h = 12 cm
The height of the parallelogram is 12 cm.
Q5: 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:
Diagonal AC divides the rhombus ABCD into two congruent triangles of
equal area.
Semi perimeter of ΔABC = (30 + 30 + 48)/2 m = 54 m
Using heron's formula,
Area of field = 2 × area of the ΔABC = (2 × 432)m2 = 864 m2
Thus,
Area of grass field which each cow will be getting = 864/18 m2 = 48 m2
Q6: An umbrella is made by stitching 10 triangular pieces of cloth of two Different colors (see Fig.12.16), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each color is required for the umbrella?
Solution: Semi perimeter of each triangular piece
=120/2 cm = 60cm
Using heron's formula,
Area of the triangular piece
There are 5 colors in umbrella.
Q7: A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in Fig. 12.17. How much paper of each shade has been used in it?
Solution:
As the diagonal of the square bisect each other at right angle.
Area of given kite = ½ (diagonal) 2
= ½ × 32 × 32 = 512 cm2
Area of shade I = Area of shade II
512/2cm2 = 256 cm2
So, area of paper required in each shade = 256 cm2
For the III section,
Length of the sides of triangle = 6cm, 6cm and 8cm
Semi perimeter of triangle = (6 + 6 + 8)/2 cm = 10cm
Using heron's formula,
Q8: 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 Fig. 12.18). Find the cost of polishing the tiles at the rate of 50p per cm2.
Solution:
Here a = 9cm
b = 28cm
c = 35cm
= 36 ´ 2.45
= 88.2cm2
Area of 16 tiles = 16 ´ 88.2
= 1411.2cm2
Cost of polishing the tiles at the rate of 50p per cm2.
= 1411.2cm2 ´ 0.50
= Rs705.60
9. A field is in the shape of a trapezium whose parallel sides are 25 m and10 m. The non-parallel sides are 14 m and 13 m. find the area of the field.
Solution: Let ABCD be the given trapezium with parallel sides AB = 25m and CD = 10m and the non-parallel sides AD = 13m and BC = 14m.
CM ⊥ AB and CE || AD.
In ΔBCE,
BC = 14m, CE = AD = 13 m and
BE = AB - AE = 25 - 10 = 15m
Semi perimeter of the ΔBCE = (15 + 13 + 14)/2 m = 21 m
= 84m2
Also, area of the ΔBCE = 1/2 × BE × CM = 84 m2
⇒ 1/2 × 15 × CM = 84 m2
⇒ CM = 168/15 m2
⇒ CM = 56/5 m2
Area of the parallelogram AECD = Base × Altitude
= AE × CM
= 10 × 84/5
= 112 m2
Area of the trapezium ABCD = Area of AECD + Area of ΔBCE
= (112+ 84) m2
= 196 m2
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Mathematics Chapter List
1. Number Systems
2. Polynomials
3. Coordinate Geometry
4. Linear Equation In Two Variables
5. Introduction To Euclid’s Geometry
6. Lines and Angles
7. Triangles
8. Quadrilaterals
9. Area Parallelograms and Triangles
10. Circles
11. Constructions
12. Herons Formula
13. Surface Areas and Volumes
14. Statistics
15. Probability
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