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CAT 2021 Slot 2 — QA Question 20

Similarity of TrianglesEasy

Let D and E be points on sides AB and AC, respectively, of a triangle ABC, such that AD : BD = 2 : 1 and AE : CE = 2 : 3. If the area of the triangle ADE is 8 sq cm, then the area of the triangle ABC, in sq cm, is 

Answer & solution

Answer: 30

Solution

Medium

Use the "same-height" ratio idea: triangles sharing the same apex split in the ratio of their bases. Step from △ADE\triangle ADE up to △ABE\triangle ABE (dividing along ABAB at DD), then up to △ABC\triangle ABC (dividing along ACAC at EE). Each step multiplies the area by a known ratio.

A B C D E
1

From △ADE\triangle ADE to △ABE\triangle ABE. Both share apex EE and have bases AD, DBAD,\ DB on line ABAB, so their areas are in ratio AD:DB=2:1AD:DB=2:1. Thus [BDE]=12[ADE]=4[BDE]=\tfrac12[ADE]=4.

[ADE]=8,[BDE]=12×8=4⇒ [ABE]=[ADE]+[BDE]=8+4=12\begin{aligned} &[ADE]=8,\quad [BDE]=\frac{1}{2}\times 8=4\\ &\Rightarrow\ [ABE]=[ADE]+[BDE]=8+4=12 \end{aligned}
2

From △ABE\triangle ABE to △CBE\triangle CBE. Both share apex BB with bases AE, ECAE,\ EC on line ACAC, so areas are in ratio AE:EC=2:3AE:EC=2:3.

[ABE][CBE]=AEEC=23⇒ [CBE]=32×12=18\begin{aligned} &\frac{[ABE]}{[CBE]}=\frac{AE}{EC}=\frac{2}{3}\\ &\Rightarrow\ [CBE]=\frac{3}{2}\times 12=18 \end{aligned}
3

Add the pieces. △ABC\triangle ABC is △ABE\triangle ABE plus △CBE\triangle CBE.

[ABC]=[ABE]+[CBE]=12+18=30\begin{aligned} &[ABC]=[ABE]+[CBE]=12+18=30 \end{aligned}

Direct ratio: [ADE][ABC]=ADAB⋅AEAC=23⋅25=415\dfrac{[ADE]}{[ABC]}=\dfrac{AD}{AB}\cdot\dfrac{AE}{AC}=\dfrac{2}{3}\cdot\dfrac{2}{5}=\dfrac{4}{15}, so [ABC]=8⋅154=30[ABC]=8\cdot\dfrac{15}{4}=30.

[ABC]=30 sq cm[ABC]=30\text{ sq cm}
CAT 2021 Slot 2 QA Q20: Let D and E be points on sides AB and AC, respectively, of a triangle ABC, such that AD : BD = 2 : 1 and AE : — Solution | TheCATExam