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CAT 2021 Slot 2 — QA Question 5

Basics of QuadrilateralsEasy

The sides AB and CD of a trapezium ABCD are parallel, with AB being the smaller side. P is the midpoint of CD and ABPD is a parallelogram. If the difference between the areas of the parallelogram ABPD and the triangle BPC is 10 sq cm, then the area, in sq cm, of the trapezium ABCD is 

Answer & solution

  • A

    20

  • 30

  • C

    40

  • D

    25

Solution

Easy

Since PP is the midpoint of CDCD and ABPDABPD is a parallelogram, triangle BPCBPC shares the same base length (PC=DP=ABPC=DP=AB) and the same height as the parallelogram. That fixes a clean ratio between their areas, which combined with the given difference gives both areas — and the trapezium is just their sum.

D C P A B ABPD BPC
1

Compare the two areas. ABPDABPD is a parallelogram on base DPDP; triangle BPCBPC has base PCPC. As PP is the midpoint of DCDC, DP=PCDP=PC, and both shapes lie between the same parallels (same height). A parallelogram has twice the area of a triangle on an equal base and equal height.

Area(ABPD)=2⋅Area(BPC)\begin{aligned} &\text{Area}(ABPD)=2\cdot\text{Area}(BPC) \end{aligned}
2

Use the given difference. Let Area(BPC)=t\text{Area}(BPC)=t.

Area(ABPD)−Area(BPC)=10⇒ 2t−t=10(from step 1)⇒ t=10\begin{aligned} &\text{Area}(ABPD)-\text{Area}(BPC)=10\\ &\Rightarrow\ 2t-t=10 \quad\text{(from step 1)}\\ &\Rightarrow\ t=10 \end{aligned}

So Area(BPC)=10\text{Area}(BPC)=10 and Area(ABPD)=20\text{Area}(ABPD)=20.

3

Assemble the trapezium. ABCDABCD is exactly parallelogram ABPDABPD plus triangle BPCBPC.

Area(ABCD)=20+10=30\begin{aligned} &\text{Area}(ABCD)=20+10=30 \end{aligned}
Area of trapezium ABCD=30 sq cm\text{Area of trapezium }ABCD=30\ \text{sq cm}
CAT 2021 Slot 2 QA Q5: The sides AB and CD of a trapezium ABCD are parallel, with AB being the smaller side. P is the midpoint of CD — Solution | TheCATExam