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# 4.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.

As per the figure we have area of field=area of ∆ABD+area of ∆BCD

$$=\frac{1}2\times base \times height+\frac{1}2\times base \times height$$

$$=\frac{1}2\times 24 \times 13+\frac{1}2\times 24\times 8$$

$$=12\times 13+12\times 8$$

$$=12\times(13+8)$$

$$=12\times 21$$

$$=252m^2$$

Hence the required area of the given field=$$252m^2$$