Question

30. A rectangle PQRS is inscribed in a semicircle of
centre O and diameter MN. M, S, R, N are collinear.
RN = 2 cm and QR = 4 cm, then what is the area of
the semicircle not overlapped by the rectangle PQRS.
(a) (12.5pi+ 12) cm² (b) (12.5pi - 12) cm?
(c) (12.5pi+ 24) cm (d) (12.5pi- 24) cm?​

Answer 1

Answer:

I don't know this answer ok

Answer 2

Answer:

(12.5 pi - 24)cm²

Step-by-step explanation:

We know that O is the centre of the semi-circle.

MN is the diameter

So ON must be radius.

RN = 2

ON = 'r'

or we can write

OR = (r-2)

Now using Pythagoras Theorem

OR² + QR² = OQ²

or

(r-2)² +4² = r²

r² + 16 -4r + 16 = r²

(using (a-b)²)

20/4=r

r=5

In Rectangle PQRS

OR = 5-2

OR = 3

SR = 6

Calculating req Area

Area of semicircle - Area of rectangle.

πr²/2 - lb

12.5π - (6×4)

12.5π -24