Question

11. Find the area of the shaded region in
triangle.
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Answer 1

Answer:

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Answer 2

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384cm²

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Triangle ABD is a right angle triangle...

so, AD² + BD² = AB²

$\bold{ \tt\implies \: { AB}^{2} = } {(12)}^{2} + {(16)}^{2} \\ \\ \bold{ \tt \implies \: AB^{2} = 144 + 256 } \: \: \: \\ \\ \bold{ \tt \implies \: AB^{2} = 400 } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ \tt \implies \: AB = \sqrt{400} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ \\ \bold{ \tt \implies \: AB = 20cm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }$

so, length of side AB is 20cm

now find area of triangle ABC by heron's formula

$\pink{\bold{ \tt \: if \: a, \: b \: and \: c \: are \: sides \: of \: \triangle} } \: \: \: \: \: \: \: \: \\ \\ \pink{ \bold{ \tt area\: = \sqrt{s(s - a)(s - b)(s - c)} }} \\ \\ \pink{\bold{ \tt and \: \: \: \: \: \: \: s = \frac{a + b + c}{2} }} \: \: \: \: \: \: \:$

here,

$\bold{ \tt \: s = \frac{AB + BC + CA}{2} } \: \: \: \: \: \: \: \: \: \\ \\ \bold{ \tt \implies \: s = \frac{20 + 48 + 52}{2} } \\ \\ \bold{ \tt \implies \: s = \frac{120}{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ \tt \implies \: s = 60} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

so area of triangle ABC ,

$\bold{ \tt \: a1 = \sqrt{60(60 - 20)(60 - 48)(60 - 52)} } \\ \\ \bold{ \tt \implies \:a1\: = \sqrt{60 \times 40 \times 12 \times 8} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ \tt \: \implies a1 = 10 \sqrt{6 \times 4 \times 2 \times 6 \times 2 \times 4} } \: \: \: \: \: \: \: \\ \\ \bold{ \tt \implies \: a1 =10 \times 6 \times 2 \times 4 } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \blue{ \bold{ \tt \implies \:a1 = 480 {cm}^{2} }} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

now , find area of triangle ABD

we know that..

$\pink{\bold{ \tt \: area \: of \: \triangle = \frac{1}{2} \times b \times h }}$

so area of triangle ABD,

$\bold{ \tt \: a2 = \frac{1}{2} \times \: AD \times BD} \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ \tt \implies \: \: a2 = \frac{1}{2} \times 12 \times 16} \\ \\ \bold{ \tt \implies \: a2 = 6 \times 16} \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \blue{\bold{ \tt \implies \: a2 = 96 {cm}^{2} }} \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

and area of shaded region,

A = area of ∆ABC - area of ∆ABD

$\bold{ \tt \implies A = \: a1 - a2 } \\ \\ \bold{ \tt \implies A =480 -96 } \\ \\ \red{ \bold{ \tt \implies A = 384 {cm}^{2} } \: \: \: \: }$

hence, area of shaded region is 384cm²