Question

Q. No. 8
In the below figure, ABCD is a square drawn inside a circle with centre O. The centre of the square coincides with O and the diagonal AC is horizontal. If AP, DQ are vertical and AP = 45cm, DQ = 25 cm, then find :-
(i) the radius of the circle.
(ii) the side of the square.
(iii) the area of shaded region.
(take √2 = 1.41 and π = 3.14)
For figure see the image provided.

Answer 1

OD=Y=OA
radius=r=OD+25
AP=OQ=RADIUS
(I)
PAO ARE RIGHT TRIANGLE
OA^2+AP^2=OP^2
Y^2+45^2=(Y+25)^2
Y^2+2025=Y^2+625+50Y
1400=50Y
Y=28
radius=25+28=53

2.OD=28
BD=DIAGONAL=√2*SIDE
OD=BD/2
28=SIDE/√2
SIDE=28*√2=39.48

3.area of shaded region =area of circle-area of square

=π(53)^2-1558.67
=8820.26-1558.67=7261.59

Answer 2

Let AO = x cm, then OD = x cm and radius of the circle = OP = OQ = OD + DQ = (x + 25) cm.

Since A PAO is right angled at A, by Pythagoras theorem, we get OP2 = AP2 + AO2 (x + 25)2 = 452 + x2

> x2 + 50x + 625 2025 + x2 50x = 2025 -625 50x = 1400 => x = 28,

the radius of the circle = OP = (28 + 25) cm 53 cm. = (ii) From right angled triangle AOD, by Pythagoras theorem, we get AD2 = AO2 + OD 2 = x2 + x2 = 2x2 = 2 x 282

=

AD = 28 /2 cm = 28 x 1-41 cm 39-48 cm. (iii) Area of the shaded region = area of the circle - area of the square

= (n x 532 - (28 J2 )2) cm2 = (3-14 x 2809 – 1568) cm? = (8820-26 - 1568) cm2 = 7252-26 cm2.

A bucket 15 diameter 77

speed of 1-1 raising the The wheel 84 cm, find 10 The circum

(i

(9 Let AO - x m, then OD = x cm and radius of the circle

- OP - Q - ODDO - ( 25) cm.

Since A PAO in right angled at A, by Pythagoras theorem, we get Op: AMP Or 25) - 45 *+ 3r + 625 - 2025 + - 50r - 2025 - 625

50x - 1400 = 1-28

the radius of the circle - OP - (28 + 25) cm - 53 cm. G) From right angled triangle AOD, by Pythagoras theorem, we get

ADSO - OD = + = 2 = 2 x 28° AÐ- 28 7 28 x 1-41 cm 3948 em.

em

() Atca of the shaded region area of the circle - arca of the square

Ex 5- (28 2 -314 x 2809 - 1568 - 8820-26 1568