Question

71. In the given figure, ABC is a quadrant of a
circle of radius 10 cm and a semi-circle is
drawn with BC as diameter. Find the area
of shaded region.
CBSE 2008, 08C
B
10 cm
AF 10 cm​

Answer 1

Given :

• ABC is a quadrant of a circle of radius 10cm.
• ABC is a quadrant of a circle of radius 10cm.A semi-circle is drawn with BC as diameter.

To find :

• The area of shaded portion. A

Solution :

_____________________

Case I

• Finding the area of quadrant of the circle which is of radius 10

Using formula : Area of sector = θ/360πr²

We have,

• θ = 90° [ As the angle is a quadrant of circle.
• r = 10cm

Putting the values

Area of sector = θ/360πr²

$→90 \div 360 \times 3.14 \times 10 {}^{2} \\ →1 \div 4 \times 3.14 \times 100 \: \: \: \: \: \: \\ →1 \div 4 \times 314 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ →78.5cm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Case II

• Finding area of ∆ABC

Using formula : Area of triangle = 1/2 × base × height

We have,

• Base = 10cm
• height = 10cm

Putting the values

Area of triangle = 1/2 × base × height

$→ \frac{1}{2} \times 10 \times 10 \\ → \frac{1}{2} \times 100 \: \: \: \: \: \: \: \: \\ →50cm {}^{2} \: \: \: \: \: \: \: \: \: \:$

Case III

• Finding length of BC

Using Pythagoras theorem = BC² = AB² + AC²

We have,

• AB = 10cm
• AC = 10cm

Putting the values

BC² = AB² + AC²

BC² = 10² + 10²

BC² = 100 + 100

BC² = 200

BC = √200

BC = 14.14 cm

Case IV :

• Finding the area of semi circle ia drawn with BC as diameter

Using Formula : Area of semi circle = 1/2πr²

We have,

• r = 14.14/2 = 7.07 cm

Putting the values

Area of semi circle = 1/2πr²

$→ \frac{1}{2} \times 3.14 \times 7.07 {}^{2} \\ →\frac{1}{2} \times 3.14 \times49.98 \\ → \frac{1}{2} \times 156.93 \: \: \: \: \: \: \: \: \: \: \\ →78.46cm {}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \:$

Case V :

• Finding area of shaded portion

Formula : Area of shaded portion : [ Area of semi circle - ( area of sector - area of triangle)]

We have,

• Area of semi circle = 78.46cm²
• Area of sector = 78.5cm²
• Area of triangle = 50cm²

Putting the values

Area of shaded portion : [ Area of semi circle - ( area of sector - area of triangle)]

$→78.46 - (78.5 - 50) \\ →78.46 - 28.5 \: \: \: \: \: \: \: \: \: \: \: \: \\ →49.96$

Hence, the area of shaded portion = 49.96cm²