Question

107.In the given figure ABCD is a rectangle with the length AC =36 m and breadth =24m
In the triangle ABG, FG is perpendicular to AB and FG=15 m. CED is a semicircle
With CD as its diameter.Calculate the area of the shaded region.(june 2011)
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Answer 1

given

AC is equal to 36

AD is equals to 24

as it is a right angle triangle according to Pythagoras formula

${ac}^{2} = {ad}^{2} + {dc}^{2}$

(36)^2 = (24)^2 + (dc)^2

dc=

$\sqrt{720} = 26.8$

area of the triangle is= 1/2× base(AB)× height(AD)

= 201

area of rectangle =length(AB) × breadth(AD)

=643.2

area of the semicircle is equal to

$\pi {r}^{2} \div 2$

r = 26.8/2 = 13.4

area is equale to 282.05

total area of the figure is =

area of the triangle + area of the rectangle + area of the semicircle.

= 201+643.2+282.05

=1126.25

Answer 2

This is the correct answer . Goof luck bro for RIMC .