Question

find the area of a quadrilateral ABCD in which AB = 3 cm,BC=4cm,CD=4cm
DA=5 cm and AC=5 cm.
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Answer 1

Answer:

this is your answer but it is without formula

Answer 2

AnswEr:

Area of Quadilateral ABCD = Area of ∆ ABC + Area of ∆ ACD

We have,

$\qquad\tt\green{AB^2+BC^2=AC^2}$

$\implies$ $\angle$ ABC = 90°.

• $\tt\pink{Area\:of\:\triangle\:ABC}$

$\tt\dfrac{1}{2}(AB\times\:BC)=$

$\tt\dfrac{1}{2}\times\:3\times\:4\:cm^2=6\:cm^2$

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Let 2s be the perimeter of ∆ACD and a = AC = 5 cm , b = CD = 4 cm and c = AD = 5 cm

•°• 2s = a + b + c => 2s = 5 + 4 + 5 => s = 7

•°• $\tt\blue{Area\:of\:\triangle\:ACD=}$

$\tt = \sqrt{s(s - a)(s - b)(s - c)} \\ \\ \tt = \sqrt{7(7 - 5)(7 - 4)(7 - 5)} \: {cm}^{2} \\ \\ \tt = \sqrt{7 \times 2 \times 3 \times 2} \: {cm}^{2} \\ \\ \tt = 2 \sqrt{21} \: {cm}^{2} = 2 \times 4.58 = 9.16$

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•°• $\tt\green{Area\:of\:quadilateral\:ABCD:-}$

$\tt = (6 + 2 \sqrt{21} ) \: {cm}^{2} \\ \\ \tt = 15.16 \: {cm}^{2}$

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