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Question

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

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Solution

Given -: In quadrilateral ABCD where C=90°, AB=9m, BC=12m, CD=5m and AD=8m. Join BD

Step 1: Find area of BCD

By applying Pythagoras theorem in BCD, we get

BD2=BC2+CD2

BD2=122+52

BD2=144+25

BD2=169

BD=169=13m

Now, the area of ΔBCD =12×12×5=6×5=30m2 ( area of triangle = 12×base×height )

Step 2: Find area of ΔABD

The semi perimeter of ΔABD is s=a+b+c2

s=9+13+82=302=15m

Using Heron’s formula,

Area of ΔABD =ss-as-bs-c

=1515-915-1315-8

=15×6×2×7

=3×5×2×3×2×7

=2×35×7

=635m2=35.5m2(approx)

Step 3: Find area of quadrilateral ABCD

∴ The area of quadrilateral ABCD = Area of BCD+Area of ΔABD

=30m2+35.5m2=65.5m2

Hence, the area of quadrilateral ABCD=65.5m2


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