In Fig. 12.30, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the
(i) quadrant OACB, (ii) shaded region
We use the formula for area of sector of the circle to solve the problem.
Since quadrant means 1/4th of a complete circle, therefore, the angle at the centre of a quadrant of a circle, θ = 360°/4 = 90°
Area of the quadrant OACB = 1/4πr²
We can get the area of quadrant OACB with radius r = 3.5 cm
Area of shaded region = Area of quadrant OACB - Area of ΔBDO
Since ∠BOD = 90°, ΔBDO is a right-angled triangle.
Using formula, Area of triangle = 1/2 × base × height , we can find area of ΔBDO with base = OB = 3.5 cm (radius of quadrant) and height = OD = 2 cm
(i) Area of quadrant OACB = 1/4πr2
= 1/4 × 22/7 × (3.5 cm)2
= 1/4 × 22/7 × 7/2 × 7/2 cm2
= 77/8 cm2
(ii) In ΔBDO, OB = r = 3.5 cm = 7/2 cm and OD = 2 cm
Area of ΔBDO = 1/2 × base × height
= 1/2 × OB × OD
= 1/2 × 7/2 cm × 2 cm
= 7/2 cm2
From figure, it is observed that:
Area of shaded region = Area of Quadrant OACB - Area of ΔBDO
= 77/8 cm2 - 7/2 cm2
= (77 - 28)/8 cm2
= 49/8 cm2
In Fig. 12.30, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the (i) quadrant OACB, (ii) shaded region
NCERT Solutions Class 10 Maths Chapter 12 Exercise 12.3 Question 12
The area of the quadrant OACB of a circle with center O and radius 3.5 cm and the area of the shaded region are 77/8 cm2 and 49/8 cm2 respectively.
☛ Related Questions:
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