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01 Aug 2019, 00:36
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Official CAT 2018 Questions; Section: QA
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
1) 248 \(\sqrt{3}\)
2) 192 \(\sqrt{3}\)
3) 188 \(\sqrt{3}\)
4) 164 \(\sqrt{3}\)
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
1) 248 \(\sqrt{3}\)
2) 192 \(\sqrt{3}\)
3) 188 \(\sqrt{3}\)
4) 164 \(\sqrt{3}\)
02 Aug 2019, 20:02
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Narenn
Official CAT 2018 Questions; Section: QA
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
1) 248 \(\sqrt{3}\)
2) 192 \(\sqrt{3}\)
3) 188 \(\sqrt{3}\)
4) 164 \(\sqrt{3}\)
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
1) 248 \(\sqrt{3}\)
2) 192 \(\sqrt{3}\)
3) 188 \(\sqrt{3}\)
4) 164 \(\sqrt{3}\)
Area of T1 = √3/4*24^2 = 144√3
T2 is formed by joining the midpoints of the sides of T1.
—> Side of T2 = half of T1 = 12.
Similarly, side of T3 = half of T2 = 6
and so on ...
—> Area of T2 = √3/4*12^2 = 36√3
—> Area of T3 = √3/4*6^2 = 9√3
So, series is 144√3, 36√3, 9√3, . . . . GP with ratio, r = 1/4.
Sum of infinite GP series when 0<r<1 is a/(1 - r)
—> Sum = 144√3/(1 - 1/4)
= 144√3/(3/4)
= 144*4/3*√3
= 192√3
IMO Option (2)
Pls Hit kudos if you like the solution
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02 Aug 2019, 21:22
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This question tests knowledge of Geometry, Infinite Geometric Series, and Ratios:
Given- Equilateral Triangle with side 24 cm
Therefore, Area of the triangle with side 24 cm = \(\frac{\sqrt{3}}{4} * 24^2\) = \(144 * \sqrt{3}\) \(cm^2\)
Consider figure A:
ABC is an equilateral triangle.
Triangle PQR is formed with the midpoints of ABC, and hence will be a similar triangle. Triangle PQR will also be an equilateral triangle.
If we observe, Triangles ARQ, RBP, PCQ, and PQR are all Congruent (as they will have the same sides and angles).
Therefore, area of Triangle PQR is \(\frac{1}{4}\) the area of triangle ABC.
Similarly, the area of Triangle XYZ will be \(\frac{1}{4}\) the area of triangle PQR and this will continue infinitely.
Thus, all we need to find is: \(Area ABC + \frac{1}{4}( Area ABC) + (\frac{1}{4})^2(Area ABC) + (\frac{1}{4})^3(Area ABC) + …Infinite terms.\)
The above is a geometric progression with:
\(First Term (a)\) = \(144\sqrt{3}\) , \(Common ratio (r)\) = \(\frac{1}{4}\)
Sum of infinite geometric series: \(\frac{a}{1 – r}\) => \(\frac{144\sqrt{3}}{1 – \frac{1}{4}}\)= \(192\sqrt{3}\)
Answer (2)
Given- Equilateral Triangle with side 24 cm
Therefore, Area of the triangle with side 24 cm = \(\frac{\sqrt{3}}{4} * 24^2\) = \(144 * \sqrt{3}\) \(cm^2\)
Consider figure A:
ABC is an equilateral triangle.
Triangle PQR is formed with the midpoints of ABC, and hence will be a similar triangle. Triangle PQR will also be an equilateral triangle.
If we observe, Triangles ARQ, RBP, PCQ, and PQR are all Congruent (as they will have the same sides and angles).
Therefore, area of Triangle PQR is \(\frac{1}{4}\) the area of triangle ABC.
Similarly, the area of Triangle XYZ will be \(\frac{1}{4}\) the area of triangle PQR and this will continue infinitely.
Thus, all we need to find is: \(Area ABC + \frac{1}{4}( Area ABC) + (\frac{1}{4})^2(Area ABC) + (\frac{1}{4})^3(Area ABC) + …Infinite terms.\)
The above is a geometric progression with:
\(First Term (a)\) = \(144\sqrt{3}\) , \(Common ratio (r)\) = \(\frac{1}{4}\)
Sum of infinite geometric series: \(\frac{a}{1 – r}\) => \(\frac{144\sqrt{3}}{1 – \frac{1}{4}}\)= \(192\sqrt{3}\)
Answer (2)
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