Question

Diagonal $AC$ of a rectangle $ABCD$ is produced to the point $E$ such that $AC:CE=2:1.AB=8cm$ and $BC=6cm$. Find the length of $DE$.

• A. $2\sqrt{19}cm$
• B. $15cm$
• C. $3\sqrt{17}cm$
• D. $13cm$

Answer 1

Answer: (C)

Given, $AB=8cm$ and $BC=6cm$
$\therefore AC=\sqrt{8^2+6^2}=10cm$
And also given $AC:CE=2:1$
Produce $BC$ to meet $DE$ at the point $P$
As $CP$ is parallel to $AD$,
$△ECP\sim △EAD$
$\therefore \frac{CP}{AD}=\frac{CE}{AE}=\frac{CP}{6}=\frac{1}{3}$
$\Rightarrow CP=2cm.$
$△CPD$ is a right triangle.
$\therefore DP=\sqrt{C{D}^2+C{P}^2}=\sqrt{68}=2\sqrt{17}cm$
But $DP:PE=2:1$ (from (1))
$PE=\sqrt{17}cm$
$\therefore DE=DP+PE$
$=2\sqrt{17}+\sqrt{17}=3\sqrt{17}cm.$