Question

An isosceles triangle has sides that are$5cm,5cm$and8cmlong. Find its area and the lengths of the three altitudes.

Short answer

The area of triangle is$12sq.unit$ and the length of altitudes are $3,4.8,4.8$

Step-by-step solution

Step 1. Draw the diagram.

Step 2. Concept used.

We can use Pythagorean Theorem.

Pythagorean Theorem define that in a right angle triangle, the square of the hypotenuse is equal to the sum of squares of other sides.

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

Step 3. Use the concept of area of triangle.

Area of $△ABC$can be found using the formula

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

Step 4. Find the altitudes.

The length of altitude$BE$

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

Similarly, The length of altitude$CF$ is$4.8$

Therefore, the Area is$12sq.unit$

The length of altitudes are$3,4.8,\mathrm{4.8.}$