Chapter 11, Problem 4PT

To determine

To find: The perimeter and area of the given triangle.

Answer to Problem 4PT

The perimeter and area of the triangle are $\text{55}\text{.3 yd and 58}{\text{.5 yd}}^{\text{2}}$ respectively.

Explanation of Solution

Given information:

The figure of a triangle with dimensions is given as follows:

  

Figure (1)

Formula used:

(i) Pythagoras theorem for right angled triangle:

  ${\text{Hypontenuse}}^2={\text{Adjacent}}^2+{\text{opposite}}^2$ .

(ii) Area of the triangle is $A=\frac{1}{2}bh$ ; where b is the base and h is the height.

(iii) Perimeter of the triangle is $P=a+b+c$ , where a, b and c are the sides of the triangle.

Calculation:

Compute the area of the required triangle whose base $b=13$ and height $h=9$ as follows:

  $\begin{array}{c}A=\frac{1}{2}\left(13\right)\left(9\right)\\ =\frac{1}{2}\left(117\right)\\ =58.5\end{array}$

Thus, the area of the required triangle is $58.5{\text{ yd}}^{\text{2}}$ .

Use the Pythagoras theorem to compute the value of the side c.

  $\begin{array}{c}{c}^2=9^2+5^2\\ c^2=81+25\\ c=\sqrt{106}\\ \simeq 10.296\end{array}$

Now, compute the perimeter of the required triangle whose sides are $a=32,b=13\text{ and }c=10.296$ as follows:

  $\begin{array}{c}P=32+13+10.296\\ =55.296\end{array}$

Round off the perimeter value to the nearest tenth, then $\text{perimeter}=55.3\text{ yd}$ .

Thus, the perimeter of the triangle is $55.3\text{ yd}$ .

Therefore, the perimeter and the area of the triangle are $\text{55}\text{.3 yd and 58}{\text{.5 yd}}^{\text{2}}$ respectively.