Chapter 11.6, Problem 21PSC

a.

To determine

To calculate: The area of shaded region which contains triangle andsemicircles.

Answer to Problem 21PSC

The area of shaded region is $120$ .

Explanation of Solution

Given information:

The sides of a triangle are 24 and 10.

Formula used:

Area of triangle: $A=\frac{1}{2}\ast b\ast h$

  $\begin{array}{l}b=baseoftriangle\\ h=heightoftriangle\end{array}$

Area of sector $=\left(\frac{measureofarc}{360^o}\right)\pi r^2$

r = radius of circle.

Area of a circle: $A=\pi r^2$

r = radius of circle.

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

  

In right angle triangle,

  $a^2+b^2\text{= }{c}^2$

Calculation:

  

The shaded area is formed by adding the triangle and two smaller semicircles then subtracting the larger semicircle.

  $r=\frac{1}{2}(10)=5$

  $\begin{array}{l}{A}_{triangle}=\frac{1}{2}\ast b\ast h\\ =\frac{1}{2}\ast 24\ast 10\\ =120\end{array}$

  $\begin{array}{l}{A}_{sector}=\left(\frac{measureofarc}{360^o}\right)\pi r^2\\ $ =(\frac{180^o}{360^o})\pi {(5)}^2$=\frac{1}{2}25\pi \\ =12.5\pi \end{array}$

  $r=\frac{1}{2}(24)=12$

  $\begin{array}{l}{A}_{sector}=\left(\frac{measureofarc}{360^o}\right)\pi r^2\\ $ =(\frac{180^o}{360^o})\pi {(12)}^2$=\frac{1}{2}144\pi \\ =72\pi \end{array}$

Triangle is a Pythagorean triple $\left(10,24,26\right)$

  $r=\frac{1}{2}(26)=13$

  $\begin{array}{l}{A}_{longsemicircle}=\frac{1}{2}\pi r^2\\ $ =\frac{1}{2}\pi {(13)}^2$=\frac{1}{2}169\pi \\ A_{shadedarea}=A_{triangle}+A_{sector}+A_{sector}-A_{longsemicircle}\\ =120+\frac{144\pi }{2}+\frac{25\pi }{2}-\frac{169\pi }{2}\\ =120\end{array}$

b.

To determine

To find: The area of triangle.

Answer to Problem 21PSC

The area of triangle is $120$ .

Explanation of Solution

Given information:

  $\begin{array}{l}Base=10,\\ Height=24.\end{array}$

Formula used:

Area of triangle: $A=\frac{1}{2}\ast b\ast h$

  $\begin{array}{l}b=baseoftriangle\\ h=heightoftriangle\end{array}$

Calculation:

  

  $\begin{array}{l}{A}_{triangle}=\frac{1}{2}\ast b\ast h\\ =\frac{1}{2}\ast 24\ast 10\\ =120\end{array}$