Question

using the measures given in the figure, find the area of ABCD ​

Answer 1

$In \: Right \: triangle \: BCD , \angle C = 90\degree$

$\underline { \blue { By \: Pythagoras \: Theorem : }}$

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

$\implies BD^{2} = 12^{2} + 5^{2}\\= 144 + 25 \\= 169$

$\implies BD = \sqrt{13^{2} }$

$\implies BD = 13\:cm \: ---(1)$

$\red {Area \: of \: ABCD } \\= Area \:of \:\triangle ABD + Area \:of \: \triangle {BDC} \\= \frac{1}{2} \times BD \times AM + \frac{1}{2} \times BC \times CD$

$= \frac{1}{2} \times 13 \times 6+ \frac{1}{2} \times 5 \times 12\\= 13 \times 3 + 5 \times 6\\= 39 + 30 \\= 69 \:cm^{2}$

Therefore.,

$\red {Area \: of \: ABCD }\green { = 69 \:cm^{2}}$

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Answer 2

Refers to attachment....✌✌

hope IT HELPs.....