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CAT 2018 Slot 1 — QA Question 32

Similarity of TrianglesEasy

Given an equilateral triangle T1 with side 24 cm, a second triangle T2 is formed by joining the midpoints of the sides of T1. Then a third triangle T3 is formed by joining the midpoints of the sides of T2. If this process of forming triangles is continued, the sum of the areas, in sq cm, of infinitely many such triangles T1, T2, T3,... will be

Answer & solution

Correct answer: 192√3

  • A

    248√3

  • B

    164√3

  • C

    188√3

  • 192√3

Solution

Easy

Joining midpoints halves every side, so each triangle has side half the previous one and area one-quarter of the previous one. The areas therefore form an infinite geometric series with ratio 14\tfrac14; sum it.

T₁ (side 24)
1

Find the first area and the common ratio. The midpoint triangle has half the side, hence (12)2=14\left(\tfrac12\right)^2=\tfrac14 the area.

A(T1)=34⋅242=34⋅576=1443ratio r=14(area scales as side2)\begin{aligned} &A(T_1)=\frac{\sqrt3}{4}\cdot 24^2 = \frac{\sqrt3}{4}\cdot 576 = 144\sqrt3\\ &\text{ratio } r = \frac{1}{4} \quad\text{(area scales as side}^2\text{)} \end{aligned}
2

Sum the infinite geometric series. Use S=a1−rS=\dfrac{a}{1-r} with a=1443a=144\sqrt3, r=14r=\tfrac14.

S=14431−14⇒ S=144334⇒ S=1443×43=1923\begin{aligned} &S = \frac{144\sqrt3}{1-\tfrac14}\\ &\Rightarrow\ S = \frac{144\sqrt3}{\tfrac34}\\ &\Rightarrow\ S = 144\sqrt3 \times \frac{4}{3} = 192\sqrt3 \end{aligned}
Sum of areas=1923 sq cm\text{Sum of areas} = 192\sqrt3 \ \text{sq cm}

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CAT 2018 Slot 1 QA Q32: Given an equilateral triangle T 1 with side 24 cm, a second triangle T 2 is formed by joining the midpoints of — Solution | TheCATExam