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The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order) respectively, and the angle between the first two sides is a right angle. Find its area.

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$\left [ 24\; cm^{2}+24\sqrt{6}cm^{2} \right ]$

Here we have, AB = 6 cm, BC = 8 cm, CD = 12 cm and AD = 14 cm

According to question, the angle between AB and BC is $90^{\circ}$

Join AC

In Right $\Delta$ ABC, [By Pythagoras theorem]

$AC^{2} = (6)^{2} + (8)^{2}$

$AC^{2} = 36 + 64$

$AC = \sqrt{100}$

$AC = 10 cm$

$Area\; of \; quadrilateral \: ABCD = Ar(\Delta ABC) + Ar(\Delta ACD)$

$Now, we\: find \; area\; of \; \Delta ABC = \frac{1}{2} \times base \times height$

$= \frac{1}{2}\times BC\times AB= \frac{1}{2}\times 8\times 6= (4 \times 6) cm^{2}$

$Area of \Delta ABC = 24 cm^{2}$

$In \Delta ACD, AC = a = 10 cm, CD = b = 12 cm, AD = c = 14 cm$

$S= \frac{a+b+c}{2}= \frac{10+12+14}{2}= \frac{36}{2}= 18cm$

$Using\; Heron's \; formula \; area \; of \Delta ACD=\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}$

$= \sqrt{18\left ( 18-10 \right )\left ( 18-12 \right )\left ( 18-14 \right )}$

$= \sqrt{18\times 8\times 6\times 4}$

$= \sqrt{9\times 2\times 4\times 2\times 3\times 2\times 4}$

$= \sqrt{3\times 3\times 2\times 2\times 3\times 2\times 4\times 4}$

$= 3\times 2\times 4\times \sqrt{3\times 2}= 24\sqrt{6}\; cm^{2}$

$Hence, area \; of\; quadrilateral \; ABCD = 24cm^{2}+24\sqrt{6}\; cm^{2}$

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