CAT 2018 Set-1 | Question: 76

Question

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

• A. $164\sqrt 3$
• B. $188\sqrt 3$
• C. $248\sqrt 3$
• D. $192\sqrt 3$

Answer 1

We can draw the diagram,

Given that, side of $T_{1} = 24 \; \text{cm}$

So, side of $T_{2} = \frac{1}{2} \; (\text{side of}  \;T_{1}) = \frac{1}{2} \times 24 = 12 \; \text{cm}$

Side of $T_{3} = \frac {1}{2} \; ( \text{side of} \; T_{2})= \frac{1}{2} \times 12 = 6 \; \text{cm}$

The area of equilateral triangle $ = \frac{\sqrt{3}}{4} \; (\text{side})^{2}$

The sum of the areas of infinitely many triangles $ = \text{area}(T_{1}) + \text{area}(T_{2}) + \text{area}(T_{3}) +\cdots $

$\qquad  = \frac{\sqrt{3}}{4}(24)^{2} + \frac{\sqrt{3}}{4}(12)^{2} + \frac{\sqrt{3}}{4}(6)^{2} + \cdots $

$\qquad = \frac{\sqrt{3}}{4} \left(24^{2} + 12^{2} + 6^{2} + \cdots \right)$

The sum of infinite $\text{GP}$ series $ = \dfrac{a}{(1-r)};$ where $ a = $ first term, and $r = $ common ratio.

Here, $ \text{a} = 24^{2} = 576, \text{r} = \left(\frac{12}{24}\right)^{2} =  \left(\frac{1}{2}\right)^{2} = \frac{1}{4}$

Now, the sum of the areas of infinitely many triangles $ = \frac{\sqrt{3}}{4} \left[ \frac{24^{2}}{\left(1-\frac{1}{4}\right)} \right]$

$ \qquad = \frac{\sqrt{3}}{4} \left[ \frac{576}{\left(\frac{4-1}{4}\right)} \right] $

$\qquad = \frac{\sqrt{3}}{4} \left(\frac{576}{\frac{3}{4}} \right) $

$\qquad = \frac{\sqrt{3}}{4} \left(\frac{576 \times 4}{3}\right) $

$\qquad = 192 \sqrt{3} \; \text{cm}^{2}$

Correct Answer $: \text{D}$