Question

14. Triangle ABC is inscribed in a semicircle.
AB-12 cm, BC - 16 cm. Find
1. AC 2. the shaded area. (Take = 3.14)

If u don't know please don't answer it is not for free points

Ans for 1. 20cm
2. 61cm2

Can you please tell me process how to solve it??
if u get this answers correct please send me answer

Please friends it is urgent....
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Answer 1

Given :

AB = 12cm

BC = 16 cm

To find :

1. AC
2. Area of shaded region

Formula used :

• Pythagoras theorem ➠ P² + B² = H²
• Area of semi-circle = π r² ÷ 2
• Area of Triangle = (1/2) base × height

Solution :

(1) Angle substended by diameter at any point on circumference is 90°

Angle B = 90°

therefore ∆ABC is right angle triangle,

therefore using Pythagoras theorem,

➠ AB² + BC² = AC²

➠ AC² = (12 cm)² + (16 cn)²

➠ AC² = 144 cm² + 256 cm²

➠ AC² = 400 cm²

➠ AC = √400 cm²

➠ AC = ± 20 cm

As side can't be negative,

➠ AC = 20 cm

(2) Area of shaded region = Area of semi-circle - Area of triangle

◈ AC = Diameter of semi-circle

➠ Radius of semi-circle = (AC)/2 = 20 cm ÷ 2

◈ Area of semi-circle = π(r²) ÷ 2

➠ Area of semi-circle = 3.14 × (10×10) ÷ 2

➠ Area of semi-circle = 314/2

➠ Area of semi-circle = 157 cm²

◈ Area of triangle = (1/2) × 12 × 16 cm²

➠ Area of triangle = 6 × 16 cm²

➠ Area of triangle = 96 cm²

◈ Area of shaded region = Area of semi-circle - Area of triangle

➠ Area of shaded region = 157cm² - 96cm²

➠ Area of shaded region = 61cm²

ANSWER :

1. AC = 20 cm
2. Area of shaded region = 61cm²