Question

The diagonal of a quadrilateral shaped field is $24m$ and the perpendiculars dropped on it from the remaining opposite vertices are $8m$ and $13m$.

Find the area of the field.

Answer 1

Step 1.Let $ABCD$ is a quadrilateral shaped field, in which $h_1=13m$ is the perpendicular drooped on $AC$ from the vertices $B$ and $h_2=8m$ is the perpendicular drooped on $AC$ from the vertices $D$.

Step 2.Finding the area of $∆ABC$

Here we can see that we have two triangle $∆ABC$ and $∆ADC$

Area of triangle $∆ABC=\frac{1}{2}xACx{h}_1$

$$\begin{gathered} =\frac{1}{2}x24x13 \\ =12x13 \\ =156m^2 \end{gathered}$$

Step 3.Finding the area of $∆ADC$

area of triangle $∆ADC=\frac{1}{2}xACx{h}_2$

$$\begin{gathered} =\frac{1}{2}x24x8 \\ =12x8 \\ =96m^2 \end{gathered}$$

Step 4.Finding the area of field

Therefore area of field $=$Area of triangle $∆ABC$+ area of triangle $∆ADC$

$$\begin{gathered} =\frac{1}{2}xACx{h}_1+\frac{1}{2}xACx{h}_2 \\ =156+96 \\ =252m^2 \end{gathered}$$

Hence the area of the field. is $252m^2$