Question

An equilateral triangle is formed by joining the midpoints of the sides of a given equilateral triangle. A third equilateral triangle is formed inside the second equilateral triangle in the same way, and so on. If this process continues indefinitely, then the sum of the areas of all such triangles, when the side of the first triangle is 16 cm, is:

• A. \(256\sqrt{3}\) sq cm
• B. \(\dfrac{256}{3}\sqrt{3}\) sq cm
• C. \(\dfrac{64}{3}\sqrt{3}\) sq cm
• D. \(64\sqrt{3}\) sq cm

Hint:

Each new triangle has half the side and one quarter the area of the one before it, so sum the resulting infinite geometric series with first term 64 root 3 and ratio 1/4.

Answer 1

The Correct Option is B

Step 1: Recall the area formula for an equilateral triangle.
For an equilateral triangle of side a, the area is:
\[ \text{Area} = \dfrac{\sqrt{3}}{4}a^2 \]
We use this for every triangle in the sequence.

Step 2: Find the area of the first triangle.
The first triangle has side 16 cm.
\[ A_1 = \dfrac{\sqrt{3}}{4}(16)^2 = \dfrac{\sqrt{3}}{4}(256) = 64\sqrt{3} \text{ sq cm} \]

Step 3: Find how the side length changes at each step.
Joining the midpoints of an equilateral triangle's sides creates a new equilateral triangle whose side is exactly half the original side, a basic result of the midpoint theorem.
So the sides go 16, 8, 4, 2, and so on, each one half of the one before, forming a geometric sequence with common ratio \(\dfrac{1}{2}\).

Step 4: Find how the area changes at each step.
Since area depends on the square of the side, halving the side means the area becomes \(\left(\dfrac{1}{2}\right)^2 = \dfrac{1}{4}\) of the previous area.
So the areas \(A_1, A_2, A_3, \ldots\) form a geometric sequence too, with first term \(A_1 = 64\sqrt{3}\) and common ratio \(r = \dfrac{1}{4}\).

Step 5: Sum the infinite geometric series of areas.
For an infinite geometric series with first term a and ratio r (where \(|r|<1\)), the sum is \(\dfrac{a}{1-r}\). Here:
\[ \text{Sum} = \dfrac{64\sqrt{3}}{1 - \tfrac{1}{4}} = \dfrac{64\sqrt{3}}{\tfrac{3}{4}} = 64\sqrt{3} \times \dfrac{4}{3} = \dfrac{256\sqrt{3}}{3} \]

Final Answer:
The sum of the areas of all the triangles formed this way, without end, is \(\dfrac{256}{3}\sqrt{3}\) sq cm. \[ \boxed{\dfrac{256}{3}\sqrt{3} \text{ sq cm}} \]