Question

In an isosceles triangle $ABC$, $AB=BC$ and $BD$ is the altitude to base $AC$. If $DC=x$, $BD=2x-1$ and $BC=2x+1$, find the lengths of all three sides of the triangle.

Answer 1

It is given that $DC=x,BD=(2x-1)$ and $BC=(2x+1)$.

In $△BDC$, $\angle D={90}^0$

Using pythagoras theorem, we have

$B{C}^2=D{C}^2+B{D}^2\Rightarrow (2x+1{)}^2=x^2+(2x-1{)}^2\Rightarrow 4x^2+4x+1=x^2+4x^2-4x+1(\because (a+b{)}^2=a^2+b^2+2ab,(a-b{)}^2=a^2+b^2-2ab)\Rightarrow 4x^2-5x^2+4x+4x+1-1=0\Rightarrow -x^2+8x=0\Rightarrow x^2-8x=0\Rightarrow x(x-8)=0\Rightarrow x=0,(x-8)=0\Rightarrow x=0,x=8$

We reject $x=0$ because if $x=0$ then triangle does not exist thus, $x=8$.

Therefore, $DC=8$ cm, $BC=(2\times 8)+1=16+1=17$ cm,

$AC=2\times DC=2\times 8=16$ cm and

$AB=BC=17$ cm

Hence, the sides of the triangle are $16$ cm, $17$ cm and $17$ cm.