Question

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Answer 1

Answer is

Step by step explanation :-

In ∆BCD ,

angle C = 90°,

${BD^{2}={BC}^{2}+{DC}^{2}$

$= \sqrt{ {24}^{2} + {7}^{2} } \\ = \sqrt{625} = 25m$

Area of ∆BCD :-

$= \frac{1}{2} \times base \times height \\ \\ = \frac{1}{2} \times 24 \times 7 \\ \\ = 84 {cm}^{2}$

In ∆ABD,

Semi perimeter s = $\frac{a+b+c}{2}$

$= \frac{25 + 26 + 27}{2} = 39m$

Area of ∆ABD :-

$= \sqrt{s(s - a)(s - b)(s - c)} \\ \\ = \sqrt{39 \times 14 \times 13 \times 12} \\ \\ = \sqrt{13 \times 3 \times 2 \times 7 \times 13 \times 6 \times 2} \\ \\ = 13 \times 2 \sqrt{7 \times 6 \times 13} \\ \\ = 291 {cm}^{2}$

Area of quadrilateral ABCD :-

= area of ∆BCD + area of ∆ABD

= 291 + 84

= $</strong><strong>{</strong><strong>3</strong><strong>7</strong><strong>5</strong><strong>c</strong><strong>m</strong><strong>}</strong><strong>^</strong><strong>{</strong><strong>2</strong><strong>}</strong><strong>$

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Answer 2

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