Question

$ABCD$ is a trapezium in which $AB$ is parallel to $CD$. The sides $AD$ and $BC$ when extended, intersect at point $E$. If $AB = 2	ext{ cm}$, $CD = 1	ext{ cm}$, and perimeter of $ABCD$ is $6	ext{ cm}$, then the perimeter, in $	ext{cm}$, of $	riangle AEB$ is

Accepted Answer

C. Since $CD \parallel AB$, $	riangle EDC \sim 	riangle EAB$. The ratio of similarity is $\frac{CD}{AB} = \frac{1}{2}$. Hence: $$\frac{EC}{EB} = \frac{1}{2} \implies EC = CB$$ $$\frac{ED}{EA} = \frac{1}{2} \implies ED = DA$$ Perimeter of $ABCD = AB + BC + CD + DA = 2 + BC + 1 + DA = 6$. $$3 + BC + DA = 6 \implies BC + DA = 3$$ Perimeter of $	riangle AEB = AB + BE + EA = AB + 2BC + 2DA = 2 + 2(BC + DA) = 2 + 2(3) = 8	ext{ cm}$.

Perimeter of Triangle Extended from Trapezium

Explanation

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