# Ex.11.2 Q4 Mensuration Solution - NCERT Maths Class 8

## Question

The diagonal of a quadrilateral shaped field is \(24\,\rm{ m}\) and the perpendiculars dropped on it from the remaining opposite vertices are \(8\,\rm{ m}\) and \(13\,\rm{ m.}\) Find the area of the field.

## Text Solution

**What is Known?**

The shape of the field is quadrilateral, and length of the diagonal and length of the perpendicular drop is given.

**What is unknown?**

Area of the field \(ABCD.\)

**Reasoning:**

A general quadrilateral can be split into two triangles by drawing one of its diagrams .

Area for quadrilateral will be the sum of area of two triangles.

**Steps:**

Area of quadrilateral \(ABCD\)

\[\begin{align} &= \! \text{Area of }\Delta \,\text{ABC} \! + \! \text{Area of }\Delta \,\text{ADC} \\ & = \! \frac{1}{2} \! \times \! (AC \! \times \! BE) \! + \! \frac{1}{2} \! \times \! (AC \! \times \! FD) \\ & = \! \frac{1}{2} \! \times \! AC(BE \! + \! FD) \\ & = \! \frac{1}{2} \! \times \! 24\,\text{m}(13\,\text{m} \! +8 \! \,\text{m}) \\& = \! \frac{1}{2} \! \times \! 24\,\text{m} \! \times \! 21\,\text{m} \\ & = \! 252\,{{\text{m}}^{2}}\end{align}\]

Thus, area of the field is \(252\,{{\rm{m}}^2}\)