Chapter 12.3, Problem 56SR

To determine

To find:

The perimeter and area of the given figure.

Answer to Problem 56SR

The perimeter is $73\text{ mm}$ and area is $180{\text{ mm}}^2$ .

Explanation of Solution

Given information:

  

Calculation:

From the given figure.

Apply the Pythagorean theorem to find the height of the triangle.

  $c^2=a^2+b^2$

Substitute $a=13$ , $b=h$ and $c=18$ in the above formula:

  $\begin{array}{c}{c}^2=a^2+b^2\\ {18}^2={13}^2+h^2\\ h^2={18}^2-{13}^2\\ h=\sqrt{{18}^2-{13}^2}\\ =\sqrt{324-169}\\ =\sqrt{155}\\ \approx 12.4\end{array}$

Area of the triangle is:

  $\begin{array}{c}A=\frac{1}{2}bh\\ =\frac{1}{2}\left(29\right)\left(12.4\right)\\ =29\left(6.2\right)\\ =179.8\\ \approx 180{\text{ mm}}^2\end{array}$

Again, apply the Pythagorean theorem to find the third side of the triangle.

  $c^2=a^2+b^2$

Substitute $a=29$ , $b=\sqrt{155}$ and $c=18$ in the above formula:

  $\begin{array}{c}{c}^2=a^2+b^2\\ c^2={29}^2+{\left(\sqrt{155}\right)}^2\\ c^2=841+155\\ c=\sqrt{841+155}\\ =\sqrt{996}\\ \approx 31.6\end{array}$

The perimeter of the given figure is as follows:

  $\begin{array}{c}P=12.4+31.6+29\\ \approx 73\text{ mm}\end{array}$

Hence, the perimeter is $73\text{ mm}$ and area is $180{\text{ mm}}^2$ .