JEE Mains · Maths · STD 11 - 8. sequence and series
Let \(\mathrm{ABC}\) be an equilateral triangle. A new triangle is formed by joining the middle points of all sides of the triangle \(\mathrm{ABC}\) and the same process is repeated infinitely many times. If \(\mathrm{P}\) is the sum of perimeters and \(Q\) is be the sum of areas of all the triangles formed in this process, then :
- A \(\mathrm{P}^2=36 \sqrt{3} \mathrm{Q}\)
- B \(\mathrm{P}^2=6 \sqrt{3} \mathrm{Q}\)
- C \(P=36 \sqrt{3} Q^2\)
- D \(\mathrm{P}^2=72 \sqrt{3} \mathrm{Q}\)
Answer & Solution
Correct Answer
(A) \(\mathrm{P}^2=36 \sqrt{3} \mathrm{Q}\)
Step-by-step Solution
Detailed explanation
Area of first \(\Delta=\frac{\sqrt{3} \mathrm{a}^2}{4}\) Area of second \(\Delta=\frac{\sqrt{3} a^2}{4} \frac{a^2}{4}=\frac{\sqrt{3} a^2}{16}\) Area of third \(\Delta=\frac{\sqrt{3} \mathrm{a}^2}{64}\) sum of area…
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