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11 Jul 2019, 18:12
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B
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Official CAT 2018 Question; 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
A) 248√3
B) 192√3
C) 188√3
D) 164√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
A) 248√3
B) 192√3
C) 188√3
D) 164√3
11 Jul 2019, 18:42
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Narenn
Official CAT 2018 Question; 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
A) 248√3
B) 192√3
C) 188√3
D) 164√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
A) 248√3
B) 192√3
C) 188√3
D) 164√3
Area of equilateral triangle is sqrt(3)/4* side^2
Area of equilateral triangle with side 24 (T1) = sqrt(3)/4 * 24^2 = 144sqrt(3)
Every triangle formed by joining the midpoints of previous triangle will have 1/4 area of that triangle
Area
Sum of areas = T1+T2+.....
= 144* sqrt(3) * (1+ 1/4 + 1/4^2+....infinite terms)
This a geometric progression with a= 144 sqrt(3)
And r=1/4
Sum of infinite geometric progression is a/(1-r)
= 144 sqrt(3) / (1-1/4)
= 144 sqrt(3) /(3/4)
= 192 sqrt(3)
IMO B
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