1 In given figure three ares are drawn by taking vertices, A, B and C , of an equilateral triangle of side 10 cm, to intersect the sides BC, CA and AB at their respective mid point D, E and F. Find the area of shaded region. (Use it = 3.14).
Answers
Answer:
Given that,
ABC is an equilateral triangle, so AB=BC=Ac=10cm & <A=<B=<C=60°
D, E & F are the mid point of BC, AC & AB
Step-by-step explanation:
solution,
AB = BC = AC = 10
=> AF=BF = BD=CD = AE=CE =5
& <A = <B = <C
do, ar(AFE) = ar(BFD) = ar (CED)
now,
ar(AFE) ==>
θ/360 × πr²
=> 60/360 ×3.14× 5²
=> 1/6 ×3.14× 25 cm²
=> 0.52×25 cm²
=> 13 cm²
now,
ar( AFE, BFD & CED) = 3×13 cm²
= 39 cm²
again,
ar(ABC) = √3/4 × side²
= √3/4 × 10² cm²
= √3/4 × 100 cm²
= √3 × 25 cm²
= 25√3 cm²
now,
ar(DFE)= ar(ABC) - ar(AFE, BFD & CED)
= (25√3 - 39) cm²
Answer
39.25 cm^2
I hope it may help you