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Q.
Diagonal $A C$ of a rectangle $A B C D$ is produced to the point $E$ such that $A C: C E=2: 1 . A B=8 \,cm$ and $B C=6 cm$. Find the length of $D E$.
Geometry
Solution:
Given, $A B=8 cm$ and $B C=6 \,cm$
$\therefore A C=\sqrt{8^2+6^2}=10 \,cm$
And also given $A C: C E=2: 1$
Produce $B C$ to meet $D E$ at the point $P$
As $C P$ is parallel to $A D$,
$\triangle E C P \sim \triangle E A D$
$\therefore \frac{C P}{A D}=\frac{C E}{A E}=\frac{C P}{6}=\frac{1}{3}$
$\Rightarrow C P=2 \, cm .$
$\triangle C P D$ is a right triangle.
$\therefore D P=\sqrt{C D^2+C P^2}=\sqrt{68}=2 \sqrt{17} \,cm$
But $D P: P E=2: 1$ (from (1))
$P E=\sqrt{17} \,cm $
$\therefore D E=D P+P E$
$=2 \sqrt{17}+\sqrt{17}=3 \sqrt{17} \, cm .$
