Question

In rhombus $ABCD$, the diagonals $AC$ and $BD$ intersect at $E$. If $AE=x$, $BE=(x+7)$, and $AB=(x+8)$, find the lengths of the diagonals $AC$ and $BD$.

Answer 1

It is given that $AE=x,BE=(x+7)$ and $AB=(x+8)$.

In $△AEB$, $\angle E={90}^0$

Using pythagoras theorem, we have

$A{B}^2=A{E}^2+B{E}^2\Rightarrow (x+8{)}^2=x^2+(x+7{)}^2\Rightarrow x^2+16x+64=x^2+x^2+14x+49(\because (a+b{)}^2=a^2+b^2+2ab)\Rightarrow x^2-2x^2+16x-14x+64-49=0\Rightarrow -x^2+2x+15=0\Rightarrow x^2-2x-15=0\Rightarrow x^2-5x+3x-15=0\Rightarrow x(x-5)+3(x-5)=0\Rightarrow (x+3)=0,(x-5)=0\Rightarrow x=-3,x=5$

Since the length of the rhombus cannot be negative thus, $x=5$.

Therefore, $AE=5$ cm, $BE=5+7=12$ cm,

$AC=2\times AE=2\times 5=10$ cm and

$BD=2\times BE=2\times 12=24$ cm

Hence, the length of the diagonals are $AC=10$ cm and $BD=24$ cm.