Question

The given figure shows an equilateral triangle ABC whose side is 10 cm and a right angled triangle BDC whose side BD is 8 cm and angle D= 90°. Find the area of the shaded portion.
Ans:- 19.3
pls tell the steps to solve​

Answer 1

Step-by-step explanation:

Given side of Equilateral ∆ = √3/4a²

a= 10 cm

Area = √3/4 ×10 ×10 cm²

Area = 100√3/4 cm² = 43.3

Now In ∆ BDC

D = 90°

BC² = BD² + CD²

100 = 36 + CD²

CD = √64 = 8

Area of ∆ BDC =  1/2 * 6* 8 = 24 cm²

Area of shaded region = 43.3 - 24 =19 .3 cm²

Answer 2

Answer:

19.30 cm²

Step-by-step explanation:

BC = AB = 10 cm

So the area of Triangle ABC

√s(s-a)(s-b)(s-c) = √15(15-10)(15-10)(15-10)

= √15(5)(5)(5)

= 25√3 cm²

Firstly we will find the side DC by pythagoras theorem.

DB² + DC² = BC²

8² + DC² = 10²

DC² = 10² - 8²

DC² = 100 - 64

DC² = 36

DC = √36 = 6 cm

Area of DBC = √s(s-a)(s-b)(s-c)

= √12(12-8)(12-10)(12-6)

= √12(4)(2)(6)

= 24 cm²

Area of shaded portion = 25√3 - 24 = 19.30 cm²