Question

The altitude of an isosceles triangle drawn to its base is $3cm$, and its perimeter is $18cm$. Find the length of its base.

Short answer

The length of its base is $8cm$.

Step-by-step solution

Step 1. Given information.

Assume b to be the base of the isosceles triangle and one of the equal sides isa.

Height of the isosceles triangle,

$$\begin{gathered} \frac{\sqrt{4a^2-b^2}}{2}=3 \\ \sqrt{4a^2-b^2}=6 \\ 4a^2-b^2-36=0......\left(i\right) \end{gathered}$$

Perimeter of the isosceles triangle,

$$\begin{gathered} 2a+b=18 \\ 2a+b-18=0......\left(ii\right) \end{gathered}$$

Solve equation (ii) for b,

$b=18-2a......\left(iii\right)$

Step 2. Substitute the value of b in equation (i).

Substitute the value of bin equation (i),

$$\begin{gathered} 4a^2-{\left(18-2a\right)}^2-36=0 \\ 4a^2-\left(324-72a+4a^2\right)-36=0 \\ 72a-360=0 \\ a=5 \end{gathered}$$

Substitute the value of a in equation (iii),

role="math" localid="1646850774031" $$\begin{gathered} b=18-2\left(5\right) \\ b=18-10 \\ b=8 \end{gathered}$$