# Ex.11.2 Q1 Mensuration Solution - NCERT Maths Class 8

## Question

The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are \(1\,\rm{ m}\) and \(1.2\,\rm{ m}\) and perpendicular distance between them is \(0.8\rm{ m}.\)

## Text Solution

**What is Known?**

Shape of the table is trapezium and dimensions of the table.

**What is unknown?**

Area of the table.

**Reasoning:**

The table is rectangular in the middle and triangle at the end. Usually the area of the problem is sum of areas of two right angle triangle and rectangle.

**Steps:**

Base of the triangles

\[\begin{align}(CE+FD) &= (CD-EF)\\&= 1.2\;\rm{cm} – 1\,\rm{m} \\&= 0.2\,\rm{m}\end{align}\]

Base of one triangle \(\begin{align} = {\rm{CE}} = {\rm{FD}} = \frac{{0.2}}{2} = 0.1{\rm{m}} \end{align}\)

Height of the triangle \(= AE = BF = 0.3\,\rm{m}\)

Area of the triangle \(ACE\) \(=\) Area of the triangle \(BDF\)

\[\begin{align} &= \frac{1}{2} \times \rm{base} \times \rm{height}\\& = \frac{1}{2} \times 0.1{\rm{m}} \times 0.8{\rm{m}} = 0.04{{\rm{m}}^2}\end{align}\]

Area of the rectangle \(ABEF\)

\[\begin{align} &= 1{\rm{m}} \times 0.8{\rm{m}}\\ &= 0.8\,\rm{m^2}\end{align}\]

Therefore,

Area of the table \(=\) Area of the triangle \(ACE\) \(+\) Area of the rectangle \( ABEF\) \(+\) Area of the triangle \( BFD\)

\[\begin{align}&= 0.04\,{{\rm{m}}^2} \times 0.8\,{{\rm{m}}^2} + 0.04\,{{\rm{m}}^2}\\&= 0.88\,{{\rm{m}}^2}\end{align}\]

- Live one on one classroom and doubt clearing
- Practice worksheets in and after class for conceptual clarity
- Personalized curriculum to keep up with school