Question

in the given figure ABPCis a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter . find the area of the shaded region

Answer 1

HELLO DEAR,



RADIUS (r) OF THE CIRCLE = AB = AC = 14 cm

AREA OF QUADRANT ABPC = 1/4×π×r 2 = (1/4×22/7×14×14) cm2

= 154 cm2

AREA OF ∆ABC = 1/2×AC×AB = (1/2×14×14) cm2

= 98 cm2 .

AREA OF SEGMENT BPC = AREA OF QUADRANT ABPC − AREA OF ∆ABC

                                    = (154− 98) cm2

                                    = 56 cm2        .............(1)

IN A RIGHT-ANGLED TRIANGLE BAC

BC2 = BA2 + AC2     (By Pythagoras theorem)

⇒ BC2  = (142 + 142) cm2
⇒ BC2  = 14×14×2 cm2
⇒ BC  = 14√2 cm

IN THE FIGURE BC IS THE DIAMETER OF THE SEMI-CIRCLE BEC.

d = 2r

THEN THE RADIUS WILL BE =  14√2 / 2 cm = 7√ 2 cm

AREA OF SEMI-CIRCLE BEC


= 1/2×π×r2 = ( 1/2×22/7×7√2×7√2) cm2

= (1/2×22/7×7×7×2) cm2 = 154 cm2 


AREA OF THE SHADED PORTION = AREA OF SEMI-CIRCLE BEC − AREA OF SEGMENT BPC        ( from (1) )

  = 154 cm2 − 56 cm2 = 98 cm2 .




I HOPE ITS HELP YOU DEAR,
THANKS