Chapter 11, Problem 18PT

To determine

To find: The perimeter and area of the figure.

Answer to Problem 18PT

The perimeter of the figure is $30\mathrm{m}$ .

The area of the figure is $51.57{\mathrm{m}}^2$ .

Explanation of Solution

Given Information: In figure sides of triangle and square are given.

Calculation: In figure sides of triangle and square are given.

Perimeter of figure :

  $\begin{array}{l}=6+6+6+6+6\\ =30\mathrm{m}\end{array}$

Area of figure:

  $\begin{array}{l}=areaoftriangle+areaofsquare\\ $ =\frac{\sqrt{3}}{4}{(6)}^2+6\times 6$=\frac{\sqrt{3}}{4}(36)+36\\ =1.73\times 9+36\\ =15.57+36\\ =51.57{\mathrm{m}}^2\end{array}$

Thus, the perimeter of the triangle is $30\mathrm{m}$ and the area of the triangle is $51.57{\mathrm{m}}^2$ .