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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 (440+54) cm2
By applying heron's formula in triangle ABC having the sides 7 cm,6 cm and 9 cm, we have
s=7+6+92=11 cm
Area of triangle =s(sa)(sb)(sc)
=11(117)(116)(119)
=440 cm2
Now, in triangle ADC having sides 15 cm,12 cm and 9 cm, we have
s=15+12+92=18 cm
=18(1815)(1812)(189)
=54 cm2
There, the area of quadrilateral ABCD=(440+54) cm2
300593_244738_ans.png

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