Chapter 11.6, Problem 5P

To determine

To find: the altitude of the isosceles triangle.

Answer to Problem 5P

  $22.248$ cm.

Explanation of Solution

Given:

The base of the isosceles triangle is18cm and equal sides are 24 cm.

Calculation:

According to the question, length of the right triangle is consecutive even integers in feet.

Let the first side be x.

Therefore, second and third side will be $x+2andx+4$ .

  

Here, ABE is an isosceles triangle where AE is base. Draw altitude AD, then AD bisect BE in to two equal parts. And ABD is a right-angle triangle. Therefore,

By using Pythagoras theorem, $A{D}^2=A{B}^2+B{D}^2$ Thus,

  $\begin{array}{l}A{B}^2=A{D}^2+B{D}^2\\ {\left(24\right)}^2={\left(AD\right)}^2+9^2\left\{\text{substitutes the given value}\right\}\\ 576={\left(AD\right)}^2+81\\ \therefore {\left(AD\right)}^2=576-81\\ \Rightarrow {\left(AD\right)}^2=495\end{array}$

  $\begin{array}{l}AD=\sqrt{495}\left\{\text{taking square root on both the sides}\right\}\\ AD=22.248\{\text{round up to hundred place}}\end{array}$

Hence, the altitude of the isosceles triangle is $22.248$ cm.