Chapter 11.9, Problem 1AGP

x

To determine

To determine the area of the composite figure.

Answer to Problem 1AGP

  $61.3$ square millimeters.

Explanation of Solution

Given information: Figure consists of an isosceles triangle, with common sides $a=16$ millimeters and base $b=6$ millimeters. Semicircle with diameter $d=6$ millimeters.

Formula used: Area of isosceles triangle $A=\frac{b}{2}\times \sqrt{a^2-\frac{b^2}{4}}$ and area of semicircle $S=\frac{\pi \times r^2}{2}$

Calculation:Area of isosceles triangle $A=\frac{b}{2}\times \sqrt{a^2-\frac{b^2}{4}}$

  $=\frac{6}{2}\times \sqrt{{16}^2-\frac{6^2}{4}}$

This can be simplified,

  $=3\times \sqrt{256-9}$

Hence it can be simplified as,

  $=3\times \sqrt{247}$

On taking the square root of the expression.

  $=3\times 15.72$

On multiplying the expression.

  $=47.16$ square millimeters.

Area of the semicircle $S=\frac{\pi \times r^2}{2}$

Also,

  $r=\frac{d}{2}$

  $=\frac{6}{2}$

  $=3$ millimeters.

  $=\frac{\pi \times 3^2}{2}$

Putting the value of radius.

  $=\frac{22\times 9}{7\times 2}$

On simplifying the expression.

  $=\frac{198}{14}$

On converting the fraction to a decimal number.

  $=14.14$ square millimeters

Adding A and S would give the area of the composite figure.

  $=61.3$ square millimeters.