Question 15
There are two triangles in the diagram which are similar to each other. Divide DE by BC, equate to AD divided by AB and solve for $x$.
$\frac{DE}{BC}=\frac{AD}{AB}$
$\frac{x+3}{x}=\frac{x-4+4}{4}$
$x^{2}=4x+12$
$x^2-4x-12=0$
$(x-6)(x+2)=0$
$x\neq-2$
$DE=9cm$
There are two triangles in the diagram which are similar to each other. Divide DE by BC, equate to AD divided by AB and solve for $x$.
$\frac{DE}{BC}=\frac{AD}{AB}$
$\frac{x+3}{x}=\frac{x-4+4}{4}$
$x^{2}=4x+12$
$x^2-4x-12=0$
$(x-6)(x+2)=0$
$x\neq-2$
$DE=9cm$