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CAT 2024 Slot 2 — QA Question 9

Basics of QuadrilateralsEasy

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 cm, CD = 1 cm, and perimeter of ABCD is 6 cm, then the perimeter, in cm, of ∆AEB is

Answer & solution

Correct answer: 8

  • 8

  • B

    9

  • C

    7

  • D

    10

Solution

Easy

Extending the non-parallel sides creates similar triangles △EDC∼△EAB\triangle EDC\sim\triangle EAB with ratio CD:AB=1:2CD:AB=1:2. Use the perimeter of the trapezium to get AD+BCAD+BC, then express the perimeter of △AEB\triangle AEB through that similarity.

E D C A B 1 2
1

Use the trapezium perimeter. AB+BC+CD+DA=6AB+BC+CD+DA=6 with AB=2, CD=1AB=2,\,CD=1.

2+1+BC+AD=6⇒ AD+BC=3\begin{aligned} &2+1+BC+AD=6\\ &\Rightarrow\ AD+BC=3 \end{aligned}
2

Set up the similarity. Since CD∥ABCD\parallel AB, △EDC∼△EAB\triangle EDC\sim\triangle EAB with ratio CDAB=12\dfrac{CD}{AB}=\dfrac12. Hence ED=12EAED=\tfrac12 EA and EC=12EBEC=\tfrac12 EB, so D,CD,C are midpoints of EA,EBEA,EB.

AD=EA−ED=12EA=EDBC=EB−EC=12EB=EC\begin{aligned} &AD=EA-ED=\tfrac12 EA=ED\\ &BC=EB-EC=\tfrac12 EB=EC \end{aligned}
3

Express the perimeter of △AEB\triangle AEB. EA=2 ADEA=2\,AD and EB=2 BCEB=2\,BC, so:

Perim(△AEB)=EA+EB+AB=2 AD+2 BC+AB⇒ =2(AD+BC)+AB=2(3)+2(from step 1)⇒ =8\begin{aligned} &\text{Perim}(\triangle AEB)=EA+EB+AB=2\,AD+2\,BC+AB\\ &\Rightarrow\ =2(AD+BC)+AB=2(3)+2 \quad\text{(from step 1)}\\ &\Rightarrow\ =8 \end{aligned}
Perimeter of △AEB=8 cm\text{Perimeter of }\triangle AEB=8\text{ cm}

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CAT 2024 Slot 2 QA Q9 — Basics of Quadrilaterals solution | TheCATExam