Chapter 12, Problem 9STP

To determine

To find the perimeter of given isosceles triangle

Answer to Problem 9STP

Perimeter, $P=84.7cm$

Explanation of Solution

Given information:

For the given isosceles triangleABC with equal sides AB and AC ,

Length of base (b) $BC=20$ cm $\angle ABC=\angle ACB={72}^{°}$

Formulas used:

• Perimeter of isosceles triangle $=2a+b$ where a is the length of equal sides and b is the base.
• Cosine of an angle is the ratio of adjacent side to hypotenuse

Draw a perpendicular bisector from point A to side BCas shown below:

That is, $BD=DC=10cm$

To find side AB, apply cosine ratio at angle ABD on the right triangle ABD

$\angle ABD={72}^{°}$

That is, $\cos B=\frac{\text{adjascent side}}{\text{hypotenuse}}$

$\cos B=\frac{\text{BD}}{\text{AB}}$

$$

$\cos 72=\frac{10}{\text{AB}}$

$AB=\frac{10}{\cos 72}$

$=32.36$

∴$AB=AC=32.36cm$

Perimeter of isosceles triangle ABC , $$

$\begin{array}{c}P=2a+b\\ =2\left(32.36\right)+20\\ =84.72cm\end{array}$