Question
In given figure $\text{ABCD}$ is a rectangle, in which $BC =2 AB$. A point $E$ lies on $C D$ produced such that $C E=2 B C$. Find $A C: B E$
✓
Answer
$\text{ABCD}$ is a rectangle and each angle of rectangle is $90^{\circ}$.
Let length of $\quad A B=x\quad \quad \ldots \ldots(i)$
It is given that $B C=2 A B$,
$\Rightarrow B C=2 x ($Using$(i)).............(ii)$
$C E=2 B C=2(2 x)=4 x ($Using $(ii))$
Apply Py thagoras Theorem in right angled $\triangle A B C$,
$A C^2=A B^2+B C^2$
$\Rightarrow A C^2=x^2+(2 x)^2$
$\Rightarrow A C^2=5 x^2$
$\Rightarrow A C=\sqrt{5} x$
$($Neglecting the negative value of square root as length cannot be in negative$).....(iii)$
Apply Py thagoras Theorem in right angled $\triangle BCE$,
$B E^2=B C^2+C E^2$
$\Rightarrow B E^2=(2 x)^2+(4 x)^2$
$\Rightarrow B E^2=20 x^2$
$\Rightarrow B E=\sqrt{20} x=2 \sqrt{5} x ($Neglecting the negative value of square root as length cannot be in negative$)......(iv)$
From $(iii)$ and $(iv),$
$A C: B E=\sqrt{5} x: 2 \sqrt{5} x$
Hence, $A C: B E=1: 2$.
Let length of $\quad A B=x\quad \quad \ldots \ldots(i)$
It is given that $B C=2 A B$,
$\Rightarrow B C=2 x ($Using$(i)).............(ii)$
$C E=2 B C=2(2 x)=4 x ($Using $(ii))$
Apply Py thagoras Theorem in right angled $\triangle A B C$,
$A C^2=A B^2+B C^2$
$\Rightarrow A C^2=x^2+(2 x)^2$
$\Rightarrow A C^2=5 x^2$
$\Rightarrow A C=\sqrt{5} x$
$($Neglecting the negative value of square root as length cannot be in negative$).....(iii)$
Apply Py thagoras Theorem in right angled $\triangle BCE$,
$B E^2=B C^2+C E^2$
$\Rightarrow B E^2=(2 x)^2+(4 x)^2$
$\Rightarrow B E^2=20 x^2$
$\Rightarrow B E=\sqrt{20} x=2 \sqrt{5} x ($Neglecting the negative value of square root as length cannot be in negative$)......(iv)$
From $(iii)$ and $(iv),$
$A C: B E=\sqrt{5} x: 2 \sqrt{5} x$
Hence, $A C: B E=1: 2$.
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