Question:
The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.
Solution:
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Given:
Diagonal of a quadrilateral shaped field $=24 \mathrm{~m}$
Perpendiculars dropped on it from the remaining opposite vertices are $8 \mathrm{~m}$ and $13 \mathrm{~m}$.
Now, we know :
Area $=\frac{1}{2} \times \mathrm{d} \times\left(\mathrm{h}_{1}+\mathrm{h}_{2}\right)$
$\therefore$ Area of the field $=\frac{1}{2} \times 24 \times(8+13)$
$=12 \times 21$
$=252 \mathrm{~m}^{2}$