Chapter 9.7, Problem 22PSB

To determine

To calculate: The Altitude to the base.

Answer to Problem 22PSB

The altitude to the base is $3\sqrt{3}\text{ units}$

Explanation of Solution

Given information:Here, we have given an isosceles triangle with one of its angles $120°$ and base $18$ units.

Formula used: Perpendicular drawn from vertex bisects both the vertex angle as well as the base.

Calculation: Let the triangle be $\Delta ABC$ . Draw a perpendicular from $\angle A$ to the base $BC$ such that $AD\perp BC$ .

Now, perpendicular drawn bisects both the vertex angle as well as the base of the triangle.

  $\Rightarrow \angle DAC=\frac{120°}{2}=60°$ and $DC=\frac{BC}{2}=\frac{18}{2}=9$

Now, in right $\Delta ADC$ ,

Angles are $90°,60°,30°$ ,

  $\therefore$ Hypotenuse will be $2a$ and perpendicular and base will be $a$ and $a\sqrt{3}$ respectively.

Here, base ( $DC$ ) = 9

  $\begin{array}{l}\Rightarrow a\sqrt{3}=9\\ \Rightarrow a=\frac{9}{\sqrt{3}}\\ \Rightarrow a=\frac{9}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}\\ \Rightarrow a=\frac{9\sqrt{3}}{3}\\ \Rightarrow a=3\sqrt{3}\end{array}$

  $\Rightarrow$ Perpendicular (a) = $3\sqrt{3}$ units