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: List the side lengths of the shrinking triangles.
Starting from a side of 16, joining midpoints repeatedly gives sides 16, 8, 4, 2, 1, and so on forever, each one exactly half of the last, because the segment joining the midpoints of two sides of a triangle is always half the third side, a direct result of the midpoint theorem.

Step 2: Compute the area of each triangle one by one using $A = \dfrac{\sqrt{3}}{4}a^2$.
$A_1 = \dfrac{\sqrt{3}}{4}(16^2) = \dfrac{\sqrt{3}}{4}(256) = 64\sqrt{3}$
$A_2 = \dfrac{\sqrt{3}}{4}(8^2) = \dfrac{\sqrt{3}}{4}(64) = 16\sqrt{3}$
$A_3 = \dfrac{\sqrt{3}}{4}(4^2) = \dfrac{\sqrt{3}}{4}(16) = 4\sqrt{3}$
$A_4 = \dfrac{\sqrt{3}}{4}(2^2) = \dfrac{\sqrt{3}}{4}(4) = \sqrt{3}$

Step 3: Notice the pattern between consecutive areas.
Dividing each area by the one before it: $16\sqrt{3}/64\sqrt{3} = 1/4$, and $4\sqrt{3}/16\sqrt{3} = 1/4$, and $\sqrt{3}/4\sqrt{3} = 1/4$ again. So each area is exactly one quarter of the one before it, confirming a geometric series with first term $64\sqrt{3}$ and common ratio $1/4$.

Step 4: Add up the series using the infinite sum formula.
For a geometric series with $|r| < 1$, the sum to infinity is $\dfrac{\text{first term}}{1-r}$:
\[ \text{Sum} = \dfrac{64\sqrt{3}}{1 - 1/4} = \dfrac{64\sqrt{3}}{3/4} \]

Step 5: Simplify the fraction.
Dividing by $3/4$ is the same as multiplying by $4/3$:
\[ \dfrac{64\sqrt{3}}{3/4} = 64\sqrt{3} \times \dfrac{4}{3} = \dfrac{256\sqrt{3}}{3} \]

Step 6: State the final total.
So the combined area of every triangle in this never ending sequence adds up to $\dfrac{256}{3}\sqrt{3}$ sq cm, matching option (B).
\[ \boxed{\dfrac{256\sqrt{3}}{3} \text{ sq cm}} \]