Calculate the area of the shaded region in Fig. 12.1.
Solution:
Given, the figure represents two triangles
We have to find the area of the shaded region.
Consider triangles ABC and BDC
Area of shaded region = area of triangle ABC - area of triangle BDC
By Heron’s formula,
Area of triangle = √s(s - a)(s - b)(s - c)
Where s= semiperimeter
s = (a + b + c)/2
In triangle ABC,
a = 120m
b = 122 m
c = 22 m
So, s = (120 + 122 + 22)/2
= 264/2
s = 132 m
Now, area = √[132(132 - 120)(132 - 122)(132 - 22)]
= √[132(12)(10)(110)]
= √[11 × 12 × 12 × 10 × 10 × 11]
= 11 × 12 × 10
Area of triangle ABC = 1320 m²
In triangle BDC,
a = 24 m
b = 26 m
c = 22 m
s = (22 + 24 + 26)/2
= 72/2
s = 36 m
Area = √36(36 - 24)(36 - 26)(36 - 22)
= √36(12)(10)(14)
= √12 × 3 × 12 × 5 × 2 × 7 × 2
= (12 × 2)√5 × 7 × 3
= 24√105
= 24(10.25)
Area of triangle BDC = 246 m²
Area of shaded region = 1320 - 246
= 1074 m²
Therefore, the area of shaded region is 1074 m²
✦ Try This: In the given figure AB=16 cm, then find the area of the shaded region
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 12
NCERT Exemplar Class 9 Maths Exercise 12.3 Sample Problem 2
Calculate the area of the shaded region in Fig. 12.1.
Summary:
The area of the shaded region in Fig 12.1 is 1074 m²
☛ Related Questions:
- The cost of levelling the ground in the form of a triangle having the sides 51 m, 37 m and 20 m at t . . . .
- In a triangle, the sides are given as 11 cm, 12 cm and 13 cm. The length of the altitude is 10.25 cm . . . .
- The sides of a triangular field are 41 m, 40 m and 9 m. Find the number of rose beds that can be pre . . . .