Chapter 11.2, Problem 30WE

To determine

To Find: The area and length of three altitudes.

Answer to Problem 30WE

  $A=12sq.unit$

Length of altitudes are $3,4.8,4.8$

Explanation of Solution

Given: An isosceles triangle has sides that are $5,5,8$ cm long.

  

Using Pythagoras theorem

  $\begin{array}{l}{5}^2=4^2+h^2\\ h^2=25-16\\ h^2=9\\ h=3\end{array}$

Area of triangle $ABC$

  $\begin{array}{l}b=BC=8\\ h=3\\ A=\frac{1}{2}bh\\ A=\frac{1}{2}(8)(3)\\ A=12sq.unit\end{array}$

Now

The length of altitude $BE$

  $\begin{array}{l}b=5\\ h=?\\ A=12\\ A=\frac{1}{2}bh\\ 12=\frac{1}{2}(5)(h)\\ 24=5h\\ h=\frac{24}{5}\\ h=4.8\end{array}$

Similarly

The length of altitude $CF$ is $4.8$

Hence, $A=12sq.unit$

Length of altitudes are $3,4.8,4.8$