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Question

In a quadrilateral $$ABCD, AB = 7\ cm, BC = 6\ cm, CD =12\ cm, DA = 15\ cm, AC = 9\ cm$$. Its area is


A
(440+54) cm2
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B
(440+44) cm2
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C
(110+44) cm2
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D
(340+64) cm2
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Solution

The correct option is A $$(\sqrt{440} + 54)$$ $$cm^2$$
By applying heron's formula in triangle ABC having the sides $$7\ cm, 6\ cm$$ and $$9\ cm$$, we have
$$\displaystyle s = \frac{7+6+9}{2} = 11\ cm$$
Area of triangle $$= \sqrt{s(s-a)(s-b)(s-c)}$$
$$ = \sqrt{11(11-7)(11-6)(11-9)}$$
$$ = \sqrt { 440 }\  cm^2 $$
Now, in triangle $$ADC$$ having sides $$15\ cm, 12\ cm$$ and $$9\ cm$$, we have
$$\displaystyle s = \frac{15+12+9}{2} = 18\ cm$$
$$ = \sqrt{18(18-15)(18-12)(18-9)}$$
$$ =  54\ cm^2 $$
There, the area of quadrilateral $$ABCD =(\sqrt { 440 } +54)\ { cm }^{ 2 }$$

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Mathematics

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