Chapter 11, Problem 4MCQ

To determine

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

Answer to Problem 4MCQ

The perimeter and area of the parallelogram are ${\text{164 ft and 1448 ft}}^{\text{2}}$ respectively.

Explanation of Solution

Given information:

The figure of a parallelogram 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) The tangent formula for a right angled triangle:

  $\tan \theta =\frac{\text{Opposite side}}{\text{Adjacent side}}$

(iii) Area of the parallelogram is $A=bh$ ; where b is the base and h is the height.

(iv) Perimeter of the parallelogram is $P=2\left(a+b\right)$ , where a is the side and b is the base.

Calculation:

Calculate the value of height h using the tangent formula as follows:

  $\begin{array}{c}\tan 30°=\frac{\text{19}}{h}\\ \frac{1}{\sqrt{3}}=\frac{\text{19}}{h}\\ h=19\sqrt{3}\end{array}$

Use the Pythagoras theorem to compute the value of y.

  $\begin{array}{c}{y}^2={\left(19\sqrt{3}\right)}^2+{19}^2\\ y^2=1083+361\\ y=\sqrt{1444}\\ =38\end{array}$

Thus, the height, length and breadth of the required parallelogram is $19\sqrt{3}\text{ ft, 44 ft and 38 ft}$ respectively.

Now, compute the area of the required parallelogram as follows:

  $\begin{array}{c}A=44\cdot 19\sqrt{3}\\ =1447.994\end{array}$

Round off the area value to the nearest tenth, then $\text{area}=1448{\text{ ft}}^{\text{2}}$ .

Compute the perimeter of the required parallelogram as follows:

  $\begin{array}{c}P=2\left(44+38\right)\\ =2\left(82\right)\\ =164\end{array}$

Thus, the perimeter of the parallelogram is $164\text{ ft}$ .

Therefore, the perimeter and the area of the parallelogram is ${\text{164 ft and 1448 ft}}^{\text{2}}$ respectively.