If 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 $13 m$ , then the area of the field is

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Solution

In the given figure,
Diagonal $= 24 m$
Height, $h_{1} = 13 m$
Height, $h_{2} = 8 m$
Hence, area of Quadrilateral $= \frac{d (h_{1} + h_{2})}{2} = \frac{24 (8 + 13)}{2} = 252 m^{2}$

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If the diagonal of a quadrilateral shaped field is 24m and the perpendiculars dropped on it from the remaining opposite vertices are 8m and 13m, then the area of the field is

In the given figure,Diagonal =24mHeight, h1=13mHeight, h2=8mHence, area of Quadrilateral =d(h1+h2)2=24(8+13)2=252m2

If 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 $13 m$ , then the area of the field is

In the given figure,
Diagonal $= 24 m$
Height, $h_{1} = 13 m$
Height, $h_{2} = 8 m$
Hence, area of Quadrilateral $= \frac{d (h_{1} + h_{2})}{2} = \frac{24 (8 + 13)}{2} = 252 m^{2}$