Chapter 11.9, Problem 24HP

To determine

Tofind:the area of the floweras shown. Round up to nearest tenth.

Answer to Problem 24HP

  $59.8\text{sq}\text{inch}$ .

Explanation of Solution

Given:

Reena created the flower at the right by placing semicircles around a regular pentagongiven figure is

  

Calculation:

Consider the diameter of the semicircle is4 inch.

The objective is to find the area of the figure.

Now, divide the given figure in to five semicircle, 2 right angle triangles and an isosceles triangle.

The area A of a semicircle is given by the formula $A=\frac{1}{2}\pi r^2$ where r is the radius.

The area A of the triangle is given by the formula $A=\frac{1}{2}ba$ where b is the base and a is the altitude.

Thus, area A of a semicircle is given by

  $\begin{array}{l}A=\frac{1}{2}\pi r^2\\ =\frac{1}{2}\pi {\left(2\right)}^2\\ =2\pi \{\because \pi =3.14\}\\ =6.28\end{array}$

Therefore, area of five semicircle is $5\left(6.28\right)\text{square}\text{inch}$ , that is $31.4\text{square}\text{inch}$ .

And the area A of the triangle is given

  $\begin{array}{l}A=\frac{1}{2}ba\\ =\frac{1}{2}\left(4\right)\left(4\right)=8\text{sq}\text{inch}\end{array}$

Therefore, area A of the two triangle is $2\left(8\right)\text{sq}\text{inch}$ , that is $16\text{sq}\text{inch}$ .

Now find the altitude so that to find the area of the isosceles triangle:

Now use Pythagoras theorem to find a.

  $\begin{array}{l}{6.5}^2=a^2+2^2\\ 42.25=a^2+4\\ \therefore a^2=42.25-4=38.25\\ \therefore a=\sqrt{38.25}\approx 6.2\end{array}$ .

Now area of the isosceles triangle is

  $\begin{array}{l}A=\frac{1}{2}ba\\ =\frac{1}{2}\left(4\right)\left(6.2\right)=12.4\text{sq}\text{inch}\end{array}$ .

Thus, the required area of the flower is the sum of the area of 6 semicircle, 2 right triangles, and one isosceles triangle.

Therefore, $A=31.4+16+12.4=59.8\text{sq}\text{inch}$

Hence, the required area of flower is $59.8\text{sq}\text{inch}$ .