Question

In ΔABC, AB = BC, AC = 42 cm and ∠B = 90°. On AC as diameter, a semicircle is described. Find the area of the figure.

Answer 1

Given:

• ΔABC where AB = BC   ...(Isosceles right triangle)
• ∠B = 90°
• AC = 42 cm   ...(Hypotenuse)
• A semicircle is drawn on AC as the diameter.

Step-wise calculation:

1. Let AB = BC = x.

2. Since ∠B = 90°, by Pythagoras theorem:

AB² + BC² = AC²

⇒ x² + x² = 42²

⇒ 2x² = 1764

⇒ x² = 882

3. Area of ΔABC:

= `1/2 xx AB xx BC` 

= `1/2 xx x xx x` 

= `1/2 xx 882` 

= 441 cm²

4. Radius of the semicircle on AC:

= `"AC"/2` 

= `42/2` 

= 21 cm

5. Area of the semicircle:

= `1/2 xx π xx r^2` 

= `1/2 xx 22/7 xx 21 xx 21` 

= `1/2 xx 22/7 xx 441` 

= `1/2 xx 22 xx 63` 

= 11 × 63

= 693 cm²

6. Area of the figure (triangle + semicircle):

= Area of ΔABC + Area of semicircle

= 441 + 693

= 1134 cm²

The area of the figure is 1134 cm².