Question

Hey guyzz...
supp??

Solve the question no.46.

Most satisfying answer will be marked as Brainliest answer...

Do ur best...

waise to maths solve ni krta aaj man Kiya to solve hi ni ho rha.....
Lol. :-P

Answer 1

BC = √ 24^2 + 7^2 = √576 +49= √625 = 25

so radius = 25/2 = 12.5

Area of semi circle = pie R^2/2 = 3.14 × (12.5)^2/2
= 245.3

area of right angled = 1/2 × AB× AC = 1/2×7× 24

= 12× 7 = 84

area of Sector BOD = pier^2/360 × 90

= pie r^2/4

= 3.14× 12.5× 12.5 /4
= 122.6

Now area of shaded = area of semi circle - area of right angled + area of sector

= 245.3 - 84 + 122.6

= 283.9

Answer 2

Given,
AC= 24 cm
AB= 7 cm
angle BOD= 90°
Now,
CB is the diameter
So,
In right-angled ∆ BOD, we have:-
CB= √[(AC)²+(AB)³]
= √[(24)²+(7)²]
= √[576+49]
= √625
= 25 cm
So,
Radius= CB/2= 25/2 cm
Now,
Area of shaded region=
Area of Circle-[Ar.(∆BOD)+ Ar.(quad.COD)
So,
Ar.(circle)= πr²= 22/7*(25/2)²
= 22/7*625/4
= 6875/14 cm²
Now,
Ar.(∆BOD)= 1/2*AC*AB
= 1/2*24*7
= 84 cm²
Again,
Ar.(quadCOD)= 1/4*πr²
= 1/4*6875/14
= 6875/56 cm²
Now,
Ar.(shaded region)
= 6875/14-[84+6875/56]
= 6875/14-[(4704+6875)/56]
= 6875/14-[11579/56]
= (27500-11579)/56
= 15921/56
= 283.303
= 283 cm² (approx)

Hope this helps you!