Chapter 5.7, Problem 12PSB

To determine

To find:The value of NM.

Answer to Problem 12PSB

The value of NM is 10.

Explanation of Solution

Given Information:

Perimeter of a parallelogram $ABCD$ is 52

Perimeter of a $ABNM$ is 36

  $NC=AM$

Formula used:

Perimeter of a parallelogram $=x+y$

  $x\text{and}y$ are the adjacent sides of the parallelogram

Perimeter of a quadrilateral $=a+b+c+d$

  $a,b,c\text{and}d$ are the sides of the quadrilateral

Calculation:

  $ABCD$ is a parallelogram

  $\Rightarrow AB=DC\text{and}BC=AD$

We know that, perimeter of a parallelogram $=2\left(x+y\right)$

Let $AB=x\text{and}BC=y$

  $\therefore$ Perimeter of parallelogram $ABCD\text{is}\left(P_1\right)=2\left(x+y\right)$

  $\begin{array}{l}\therefore 52=2\left(x+y\right)\\ \therefore x+y=26\mathrm{...}\left(\text{i}\right)\end{array}$

Let $BN=a\text{and}NC=b$

As $BC=BN+NC$

  $\begin{array}{l}\Rightarrow BC=a+b\\ \Rightarrow AD=a+b\mathrm{...}\left(\because BC=AD\right)\end{array}$

  $\begin{array}{l}\text{Since,}NC=AM=b\\ \Rightarrow AD=AM+MD=b+MD\end{array}$

  $\begin{array}{l}\Rightarrow a+b=b+MD\\ \Rightarrow MD=a\end{array}$

Now, we know that Perimeter of a quadrilateral $=$ sum of all its sides

  $\therefore$ Perimeter of quadrilateral $ABNM\text{is}\left(P_2\right)=AB+BN+NM+AM$

  $\begin{array}{l}\therefore P_2=x+a+NM+b\\ \therefore 36=x+a+b+NM\\ \therefore 36=x+y+NM\\ \therefore 36=26+NM\\ \therefore NM=10\end{array}$