Question

The diagonal of a quadrilateral shaped field is 25 m and the perpendiculars dropped on it from the opposite vertices are 18 m and 12m. Find the area of the field.

Answer 1

Given Information:

The diagonal of a quadrilateral shaped field is 25 m and the perpendiculars dropped on it from the opposite vertices are 18 m and 12m.

Solution:

We know that,

$\sf\small\purple{\underbrace{\underline{Total \: Area = Area \: of \:1st\triangle + Area \: of \: 2nd \triangle}}}$

$\sf{ = \frac{1}{2}(12 \times 25) + \frac{1}{2}(18 \times 25)}$

$\sf{ = \frac{1}{2} \times 300 + \frac{1}{2} \times 450}$

$\sf{ = 150 + 225}$

$\sf{ = 375 {m}^{2} }$

$\sf \tt {\pink{\underline{\therefore{The \: area \: of \: the \: field \: is \: 375m^2}}}}$

Answer 2

Given,

For the quadrilateral shaped field,

→ The length of the diagonal of the quadrilateral shaped field , d = 25 m

→ The perpendiculars dropped on it from the two opposite vertices are :

$\sf \: L_1 = 18 \: m \: \: \: \: and \: \: \: \: L_2 = 12 \: m$

We have to find :

• The area of the field = A

Solution :

We know that ,

${\boxed{\boxed{\begin{array}{cc}\sf \: Area \: of \: a \: quadrilateral \: \: A = \frac{1}{2} \times d \times ( L_1 + L_2)\end{array}}}}$

Here

$\sf \: d = length \: of \: the \: diagonal \\ \\ \sf \: L_1 \: \: and \: \: L_2 = \: length \: of \: those \: perpendiculars$

Now,

${\boxed{\boxed{\begin{array}{cc}\sf \: A = \frac{1}{2} \times 25 \times (18 + 12) \\ \\ = \frac{1}{2} \times 25 \times 30 \\ \\ = 25 \times 15 \\ \\ = 375 \: {m}^{2} \end{array}}}}$

So area of the field is 375 m²