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Question

A park, in the shape of a quadrilateral ABCD has C=90o, AB=9 m,BC=12 m,CD=5 m,AD=8 m. How much area does it occupy?

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Solution

Join the diagonal AD of the quadrilateral ABCD.
BD2=BC2+CD2=(12)2+(5)2=169
BD=169=13m
Area of right angle ΔBCD
=12×BD×CD=12×12×5m2=30m2
Now, the sides of the ΔABD are 9 m,13 m and 8 m.
Semiperimeter of ΔABD
S=9m+13m+8m2=15m.
The area of the ΔABD
=15×(159)×(1513)×(158)m2
15×6×2×7m2=5×3×3×2×2×7
6×5.916m2(approx.)=35.495m2(approx.)=35.5(approx.)
Thus, area of the quadrilateral ABCD =ar(ΔBCD)+ar(ΔABD)
=30m2+35.5m2(approx.)=65.5m2(approx.)

300690_243170_ans_0dbe21bbc9c94cdb8c73f3db3acb264d.png

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