Question

find the perimeter and the area of quadrilateral ABCD in which
AB = 9 CM, AD = 12 CM, BD = 15 CM, CD = 17 CM, and ∠CBD =90°

Answer 1

$\huge\bold\red{Answer~:}$
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In ΔBCD,

$\angle BCD=90°$

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

$= \sqrt{ {17}^{2} - {15}^{2} } \\ \\ = \sqrt{64} \\ \\ = 8 \: cm$

Area of quadrilateral = ar(ΔBCD+ΔABD)

Area of ΔBCD :

$= \frac{1}{2} \times base \times height \\ \\ = \frac{1}{2} \times 15 \times 8 \\ \\ = 60 { \: cm}^{2}$

Area of ΔABD :

semi perimeter :

$= \frac{a + b + c}{2} \\ \\ = \frac{15 + 12 + 9}{2} \\ \\ = 18 \: cm$

Area :

$= \sqrt{s(s - a)(s - b)(s - c)} \\ \\ = \sqrt{18 \times 3 \times 6 \times 9} \\ \\ = 54 \: {cm}^{2}$

Area of quadrilateral :

$= (54 + 60) { \: cm}^{2} \\ \\ = 114 { \: cm}^{2}$

$\huge\bold\red{Answer~is~{114cm}^{2}}$

Answer 2

Step-by-step explanation:

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