Chapter 9.7, Problem 16PSB

To determine

To find: The perimeter of isosceles trapezoid JMOP.

Answer to Problem 16PSB

The perimeter of isosceles trapezoid JMOP is $=16+12\sqrt{3}\text{ units}$

Explanation of Solution

Given information :

PK is an altitude of isosceles trapezoid JMOP = 6 units

PO = 8 units and $\angle J=45°$

Formula used :

Triangle $45°-45°-90°$ has sides ratio as $1:\sqrt{3}:2$

Perimeter is sum of the sides

Calculation :

In triangle PKJ ,

JK = x (say), PK = $x\sqrt{3}$ = 6 …given

  $\therefore$ x = $\frac{6}{\sqrt{3}}=2\sqrt{3}$

And JK $=2\sqrt{3}$ units, PJ = 2x = $4\sqrt{3}$

JM = $8+2\times JK=8+2\times 2\sqrt{3}=8+4\sqrt{3}$

Now,

Perimeter

  $\begin{array}{l}=JP+PO+OM+MJ\\ =4\sqrt{3}+8+4\sqrt{3}+8+4\sqrt{3}\\ =16+12\sqrt{3}\text{ units}\end{array}$