Question

In the given figure ABC is an equilateral triangle with side 10 cm and DBC is right angled triangle with ∠D= 90°. If BD = 6 cm, then find the area of the shaded portion

Answer 1

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A/Q ___

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 =

$> \large \: \frac{1}{2} \times b \: \times h \: \\ \\ > \frac{1}{2} \times 6 \times 8 = 24 \: {cm}^{2}$

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

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Answer 2

Answer:

19.3cm2

Step-by-step explanation:

In ABC

Side =10cm

area =√3/4×a2

√3/4×10×10=√3/4×100 CM2

43.3CM2

In DBC

D= 90°

BC^2= BD^2+ DC^2

100= 36+DC^2

DC=√100-36

DC=√64

DC=8

AREA OF TRIANGLE DBC=1/2 ×B×H

1/2×6×8 =24CM2

SHADED PORTION= 43.3-24=19.3 CM2