Question

find the area of quadrilateral ABCD in which AB equal to 3 cm BC equal to 4 cm CD equal to 4 cm equal to 5 cm and ac equal to 5cm​

Answer 1

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For △ABC

AC² = AB² + BC²

(5)² = (3)² + (4)²

So, △ABC is a right angle triangle, right angled at a point B.

Area of △ABC

$= \: \frac{1}{2} \times AB \times BC \\ \implies \: \frac{1}{2} \times 3 \times 4 \: = {6cm}^{2}$

For △ADC

Perimeter

$= 2s = AC \: + \: CD \: + \: DA \\ 2s = (5 \: + \: 4 \: + \: 5) \:cm \\ 2s = 14\:cm \\ s = \frac{14}{2} \\ s = 7\:cm$

• By Heron's Formula

Area of Triangle

$= \sqrt{s(s - a)(s - b)(s - c)} \:{cm}^{2}$

Area of △

$= \sqrt{7(7 - 5)(7 - 5)(7 \times 4)} \: {cm}^{2} \\ = \sqrt{7 \times 2 \times 2 \times 3} \:{cm}^{2} \\ = 2 \sqrt{21} \:{cm}^{2} \\= (2 \times 4.583) \:{cm}^{2} \\ = 9.166\: {cm}^{2}$

Area of △ABCD = Area of △ABC + Area of △ACD

$=(6 + 9.166) \:\:{cm}^{2} \\ = 15.166\: \:{cm}^{2} \\ = 15.166\: \:{cm}^{2} \\ = \boxed{\red{15.2\: \:{cm}^{2} }}\:\:\:\:(Approximately)$

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Answer 2

Answer:

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