Chapter 11.3, Problem 34WE

To determine

To find: The area of $▱$ MNOP.

Answer to Problem 34WE

Area of $▱$ MNOP is 122 square units.

Explanation of Solution

Given information: MR = 10 units, RS = 8 units, ST = 3 units and NS = 16 units.

Formula used: Area of trapezoid is given by

  $\text{Area of trapezoid}=\frac{1}{2}\text{(Sum of parallel sides)}\times \text{Height}$

Calculation: First we find the area of trapezium RMNS.

  

Height of trapezium (RS) = 8 units

Length of parallel sides are 10 units and 16 units.

  $\text{Area of trapezoid RMNS}=\frac{1}{2}\text{(Sum of parallel sides)}\times \text{Height}$

  $\begin{array}{l}=\frac{1}{2}\times \left(10+16\right)\times 8\\ =4\times 26\\ =104{\text{ unit}}^2\end{array}$

Draw a perpendicular from M on NS.

Using Pythagoras Theorem,

  $\begin{array}{l}{\text{MN}}^2={\text{MA}}^2+{\text{AN}}^2\\ ={(8)}^2+{(6)}^2\\ =64+36\\ =100\end{array}$

So, MN = 10 units.

Since MN = OP as MNOP is a parallelogram.

So, difference between SN and MR is equal to difference between OV and PT.

  $\begin{array}{l}\text{SN}-\text{MR}=\text{OV}-\text{PT}\\ 16-10=9-\text{PT}\\ 6=9-\text{PT}\\ \text{PT}=9-6\\ \text{PT}=3\text{ units}\end{array}$

Let TV = x units.

Now, we draw a perpendicular from P on OV.

  

Using Pythagoras Theorem,

  $\begin{array}{l}{\text{OP}}^2={\text{PB}}^2+{\text{OB}}^2\\ {(10)}^2={(x)}^2+{(6)}^2\\ 100=x^2+36\\ x^2=100-36\\ x^2=64\\ x=8\end{array}$

So, TV = 8 units.

Now, we find the area of trapezium NOVS, MPTR and OPTV.

  $\text{Area of trapezoid NOVS}=\frac{1}{2}\text{(Sum of parallel sides)}\times \text{Height}$

  $\begin{array}{l}=\frac{1}{2}\times \left(16+9\right)\times 11\\ =\frac{25\times 11}{2}\\ =\frac{275}{2}{\text{ unit}}^2\end{array}$

  $\text{Area of trapezoid MPTR}=\frac{1}{2}\text{(Sum of parallel sides)}\times \text{Height}$

  $\begin{array}{l}=\frac{1}{2}\times \left(10+3\right)\times 11\\ =\frac{13\times 11}{2}\\ =\frac{143}{2}{\text{ unit}}^2\end{array}$

  $\text{Area of trapezoid OPTV}=\frac{1}{2}\text{(Sum of parallel sides)}\times \text{Height}$

  $\begin{array}{l}=\frac{1}{2}\times \left(9+3\right)\times 8\\ =4\times 12\\ =48{\text{ unit}}^2\end{array}$

Area of $▱$ MNOP = Area of trapezium MNSR + Area of trapezium NOVS − Area of trapezium MPTR − Area of trapezium OPTV

  $\begin{array}{l}\text{Area of }▱\text{MNOP }=104+\frac{275}{2}-\frac{143}{2}-48\\ =56+\frac{132}{2}\\ =56+66\\ =122{\text{ unit}}^2\end{array}$

Thus, the area of $▱$ MNOP is 122 square units.