Question

Please answer this question​

Answer 1

For Area of the Shaded region,

We can make Formula like below,

Area of shaded part = Area of ∆ ABC - Area of ∆ BDC

So, First we need to find the Area of ∆ ABC = $\sf \frac{1}{2} × Base × Height$

First Refer the attachment, then

Where,

Base = BE

Height = AE

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Now, We have ∆ AEB , ∠E = 90°

So, By using Pythagoras theorem,

AB² = AE² + EB²

(10)² = AE² + (5)²

100 = AE² + 25

100 - 25 = AE²

AE = $\sf \sqrt{75}$

AE = $\sf \sqrt{25×3}$

AE = [/tex] \sf 5\sqrt{3} [/tex] ($\sf \sqrt{3} = 1.732$ given in question)

AE = 5(1.732)

AE = 8.66

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Now, In ∆ ABC,

Area of ∆ ABE = $\sf \frac{1}{2} × Base × Height$

Area of ∆ ABE = $\sf \frac{1}{2} × BE × AE$

Area of ∆ ABE = $\sf \frac{1}{2} × 5 × 8.66$

Area of ∆ ABE = $\sf \frac{1}{2} × 5 × 8.66$

Area of ∆ ABE = $\sf \frac{1}{\cancel2} × 5 × \cancel{8.66}^{\: \: \cancel2 × 4.33}$

Area of ∆ ABE = $\sf 5 × 4.33$

Area of ∆ ABE = 21.65

For Area of ∆ ABC = 2 × Area of ∆ ABE

Area of ∆ ABC = 2 × 21.65

Hence,

Area of ∆ ABC = 43.3 cm²

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For Area of ∆ BDC,

Area of ∆ BDC = $\sf \frac{1}{2} × Base × Height$

Area of ∆ BDC = $\sf \frac{1}{2} × BD × DC$

Where,

Base = BD

Height = DC

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For BD, We can use Pythagoras theorem

BC² = BD² + DC²

(10)² = BD² + (8)²

100 = BD² + 64

100 - 64 = BD²

36 = BD²

BD = $\sf \sqrt{36}$

BD = 6

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Now we have to put in,

Area of ∆ BDC = $\sf \frac{1}{2} × Base × Height$

Area of ∆ BDC = $\sf \frac{1}{2} × BD × DC$

Area of ∆ BDC = $\sf \frac{1}{2} × 6 × 8$

Area of ∆ BDC = $\sf \frac{1}{\cancel2} × \cancel6^{\: \: \cancel2 × 3 } × 8$

Area of ∆ BDC = $\sf 3×8$

Area of ∆ BDC = 24 cm²

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Finally,

For Shaded region

Area of shaded part = Area of ∆ ABC - Area of ∆ BDC

Area of shaded part = 43.3 - 24

Area of shaded part = 19.3 cm²

Hence, Option (A) is right

I hope it helps you...