Chapter 8.4, Problem 7ST1

To determine

To find: the altitude to the base of the isosceles triangle.

Answer to Problem 7ST1

  $\text{12 units}$

Explanation of Solution

Given:

The length of sides of the isosceles triangle are 13,13 and 10.

Concept Used:

Theorem 4-4: RHS rule states that the right triangles are congruent if any of side and hypotenuse of right triangles are congruent.

Theorem 8-7: In a $30°-60°-90°$ triangle, the longer Side is $\sqrt{3}$ times as long as the shorter Side and the hypotenuse of right triangle are two times as long as the shorter side of triangle.

First draw an isosceles triangle as mentioned in the question:

Here $△ABD$ is an isosceles triangle, so

$AB=AD$ ….. (1)

$m\angle B=m\angle D$ ….. (2)

Consider the right triangles $△ACB$ and $△ACD$

$AB=AD$ [By equation (1)]

$AC=AC$ [Common side]

So, by HL-theorem $△ACB\cong △ACD$

Thus,

  $BC=CD$

Now consider,

  $\begin{array}{c}BD=10\\ BC+CD=10\\ 2BC=10\end{array}$

  $\begin{array}{c}BC=\frac{10}{2}\\ BC=5\end{array}$

Consider the right triangle $△ACB$ .

Now, $\overline{AB}$ is the hypotenuse and has length 13 and the lengths of the side are $BC=5$ and $AC$ .

So, by theorem 8-2,

  $\begin{array}{c}A{B}^2=B{C}^2+A{C}^2\\ {13}^2=5^2+A{C}^2\\ 169=25+A{C}^2\end{array}$

  $\begin{array}{c}169-25=A{C}^2\\ 144=A{C}^2\end{array}$

Now, take square root of both sides,

  $\begin{array}{c}\sqrt{A{C}^2}=\sqrt{144}\\ AC=12\end{array}$

Hence, the length of the altitude to the base of the isosceles triangle is 12 units.