TheCATExam
Sign in

CAT 2023 Slot 2 — QA Question 18

Basics of QuadrilateralsEasy

In a rectangle ABCD, AB = 9 cm and BC = 6 cm. P and Q are two points on BC such that the areas of the figures ABP, APQ, and AQCD are in geometric progression. If the area of the figure AQCD is four times the area of triangle ABP, then BP : PQ : QC is:

Answer & solution

Correct answer: 2 : 4 : 1

  • 2 : 4 : 1

  • B

    1 : 2 : 4

  • C

    1 : 1 : 2

  • D

    1 : 2 : 1

Solution

Easy

The three pieces of the rectangle are in GP. The "AQCD=4×ABPAQCD = 4\times ABP" fact pins the common ratio, which gives the area ratio 1:2:41:2:4. Then convert each area into the unknown lengths BP,PQ,QCBP, PQ, QC along BCBC (they share the same "height" AB=9AB=9) and solve.

A B P Q C D

AB=9AB=9 (the common height for all three pieces measured from AA), BC=AD=6BC=AD=6, and BP+PQ+QC=BC=6BP+PQ+QC=BC=6.

1

Find the common ratio of the GP. Let the areas be A, Ar, Ar2A,\,Ar,\,Ar^2. The condition ties the last to the first:

Ar2=4A⇒ r=2⇒ areas=A:Ar:Ar2=1:2:4\begin{aligned} &Ar^2=4A\\ &\Rightarrow\ r=2\\ &\Rightarrow\ \text{areas}=A:Ar:Ar^2=1:2:4 \end{aligned}
2

Express each area through its base on BCBC. Triangles ABPABP and APQAPQ have apex AA and bases BP,PQBP,PQ on BCBC; the trapezium AQCDAQCD has parallel sides QCQC and AD=6AD=6:

[ABP]=12⋅BP⋅9=4.5 BP[APQ]=12⋅PQ⋅9=4.5 PQ[AQCD]=12(QC+6)⋅9=4.5(QC+6)\begin{aligned} &[ABP]=\tfrac12\cdot BP\cdot 9=4.5\,BP\\ &[APQ]=\tfrac12\cdot PQ\cdot 9=4.5\,PQ\\ &[AQCD]=\tfrac12(QC+6)\cdot 9=4.5(QC+6) \end{aligned}
3

Apply the ratios. [APQ]=2[ABP][APQ]=2[ABP] and [AQCD]=4[ABP][AQCD]=4[ABP], with QC=6−BP−PQQC=6-BP-PQ:

4.5 PQ=2(4.5 BP) ⇒ PQ=2BPQC=6−BP−2BP=6−3BP4.5(QC+6)=4(4.5 BP)⇒ (6−3BP)+6=4BP⇒ 12=7BP ⇒ BP=127\begin{aligned} &4.5\,PQ=2(4.5\,BP)\ \Rightarrow\ PQ=2BP\\ &QC=6-BP-2BP=6-3BP\\ &4.5(QC+6)=4(4.5\,BP)\\ &\Rightarrow\ (6-3BP)+6=4BP\\ &\Rightarrow\ 12=7BP\ \Rightarrow\ BP=\tfrac{12}{7} \end{aligned}
4

Get all three lengths and the ratio:

BP=127,PQ=2BP=247,QC=6−3⋅127=67BP:PQ:QC=127:247:67=2:4:1\begin{aligned} &BP=\tfrac{12}{7},\quad PQ=2BP=\tfrac{24}{7},\quad QC=6-3\cdot\tfrac{12}{7}=\tfrac{6}{7}\\ &BP:PQ:QC=\tfrac{12}{7}:\tfrac{24}{7}:\tfrac{6}{7}=2:4:1 \end{aligned}

BP:PQ:QC=2:4:1BP:PQ:QC=\mathbf{2:4:1}

Notice the areas 1:2:41:2:4 are not the length ratio: the two triangles share the same height so their bases scale as the areas (BP:PQ=1:2)\big(BP:PQ=1:2\big), but the trapezium AQCDAQCD does not, so QCQC comes out smaller — giving 2:4:12:4:1, not 1:2:41:2:4.

Related Basics of Quadrilaterals questions

See all Quadrilaterals & Polygons questions →
CAT 2023 Slot 2 QA Q18 — Basics of Quadrilaterals solution | TheCATExam