Chapter 11.2, Problem 35HP

To determine

To find: x in parallelogram ABCD.

Answer to Problem 35HP

Height( x ) of the parallelogram ABCD is 7.2 unit.

Explanation of Solution

Given information: Sides of the parallelogram $\text{AB}=15\text{ unit}$ and $\text{AD}=9\text{ unit}$ . One of the diagonal of the parallelogram $\text{DB}=12\text{ unit}$ .

Formula used: Area of right angle triangle is given by

  $\text{Area}=\frac{1}{2}\times b\times h$

where, $b\&h$ are base and height of right angle triangle respectively.

Also, Area of parallelogram is given by

  $\text{Area of parallelogram}=b\times h$

where, $b\&h$ are base and height of parallelogram respectively.

Calculation:Since $\Delta \text{DBC}$ is right angle triangle.

So, $\begin{array}{l}\text{Area of }\Delta \text{DBC}=\frac{1}{2}\times BC\times DB\\ =\frac{1}{2}\times 9\times 12\left[\because BC=9\text{ unit as opposite sides of parallelogram are equal}\right]\\ =54\text{ square unit}\end{array}$

As opposite sides of parallelogram are parallel. So, AB & DC are parallel.

Now, AB & DC are parallel and DB is transversal.

So, corresponding angle $\angle \text{ADB}$ and $\angle \text{DBC}$ are equal.

  $\therefore \angle \text{ADB}=90°$

  $\begin{array}{l}\text{Area of }\Delta \text{ADB}=\frac{1}{2}\times AD\times DB\\ =\frac{1}{2}\times 9\times 12\\ =54\text{ square unit}\end{array}$

So, $\begin{array}{l}\text{Area of parallelogram ABCD }=\text{Area of }\Delta \text{ADB}+\text{Area of }\Delta \text{DBC}\\ =54+54\\ =108\text{ square unit}\end{array}$

Now, Base(AB) = 15 unit

Height (x ) = ?

  $\begin{array}{l}\text{Area of parallelogram ABCD }=AB\times x\\ \Rightarrow 108=15\times x\\ \Rightarrow x=7.2\text{ unit}\end{array}$

Thus, Height( x ) of the parallelogram ABCD is 7.2 unit.