Question

10) In the given figure, O is the
centre and AOC is a diameter
of the circle. Find
(i) the sum of the areas of
the two shaded segments
4 cm
made by the chords AB
and BC.
(ii) the area of the shaded segment made by the
chord PQ. (take, n = 2]​

Answer 1

Answer:

4.5 cm2(Sqr)

Step-by-step explanation:

In △ABC,

AC2(sqr) = AB2(sqr)+BC2(sqr)

AC2(sqr) =42(sqr)+42(sqr)

AC2(sqr) =16+16=32

AC=4√2cm

thus,  OA=OC=4√2 cm/2 =2√2cm

OB =OA=OC each radius of circle

thus OB=2√2 cm

∴ area of triangle =1/2×4√2×2√2=8 cm

Now, area of semi circle=πr2/2 =22/7×2×2√2×2√2=88/7 cm2(sqr)

=12.5 cm2(sqr)

thus, area of shaded portion =area of semicircle −area of triangle =12.5−8 =4.5 cm2(sqr)

hence, the sum of the area of two shaded segments made by chord AB and BC=4.5 cm2(sqr)

Answer 2

your answer is in the attached pictures