In Fig. 11.7, AB is the diameter of the circle, AC = 6 cm and BC = 8 cm. Find the area of the shaded region (Use π = 3.14)
Solution:
Given, AB is the diameter of the circle.
The sides of triangle AC = 6 cm and BC = 8 cm
We have to find the area of the shaded region.
We know that a triangle in a semicircle is a right angled triangle with hypotenuse as diameter.
So, ACB is a right triangle with C at right angle
AB² = AC² + BC²
AB² = (6)² + (8)²
AB² = 36 + 64
AB² = 100
Taking square root,
AB = 10 cm
Area of triangle = (1/2) × base × height
Area of triangle ABC = (1/2) × AC × BC
= (1/2) × 6 × 8
= 48/2
= 24 cm²
Diameter of circle = AB = 10 cm
Radius = 10/2 = 5 cm
Area of circle = πr²
= (3.14)(5)(5)
= (3.14)(25)
= 78.5 cm²
Area of shaded region = area of circle - area of triangle
= 78.5 - 24
= 54.5 cm²
Therefore, the area of the shaded region is 48.5 cm².
✦ Try This: In Fig, AB is the diameter of the circle, AC = 9 cm and BC = 12 cm. Find the area of the shaded region (Use π = 3.14).
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 12
NCERT Exemplar Class 10 Maths Exercise 11.3 Problem 7
In Fig. 11.7, AB is the diameter of the circle, AC = 6 cm and BC = 8 cm. Find the area of the shaded region (Use π = 3.14)
Summary:
In Fig. 11.7, AB is the diameter of the circle, AC = 6 cm and BC = 8 cm. The area of the shaded region is 54.5 cm²
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