Given an equilateral triangle $\mathrm{T} 1$ with side $24 \mathrm{~cm}$, a second triangle $\mathrm{T} 2$ is formed by joining the midpoints of the sides of $\mathrm{T} 1$. Then a third triangle $\mathrm{T} 3$ is formed by joining the midpoints of the sides of $\mathrm{T} 2$. If this process of forming triangles is continued, the sum of the areas, in sq cm , of infinitely many such triangles $\mathrm{T} 1, \mathrm{~T} 2, \mathrm{~T} 3, \ldots$ will be

When we join the midpoints of an equilateral triangle's sides, we use the midpoint theorem. The line joining two midpoints of a triangle is parallel to the third side and exactly half its length.

If triangle T1 has side 24 cm:

Triangle T2 (formed by joining midpoints of T1) has side = 24/2 = 12 cm

Triangle T3 (formed by joining midpoints of T2) has side = 12/2 = 6 cm

Triangle T4 has side = 6/2 = 3 cm

And so on...

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For any equilateral triangle with side length 'a', the area is: Area = $\dfrac{\sqrt{3}}{4} \times a^2$

An equilateral triangle can be split into two right triangles. Using basic trigonometry, the height is $\dfrac{\sqrt{3}}{2} \times$ side, so area = $\dfrac{1}{2} \times$ base $\times$ height = $\dfrac{\sqrt{3}}{4} \times a^2$.

Now let's calculate each area:

T1: Area = $\dfrac{\sqrt{3}}{4} \times 24^2$

= $\dfrac{\sqrt{3}}{4} \times 576$

= $144\sqrt{3}$ sq cm

T2: Area = $\dfrac{\sqrt{3}}{4} \times 12^2$

= $\dfrac{\sqrt{3}}{4} \times 144$

= $36\sqrt{3}$ sq cm

T3: Area = $\dfrac{\sqrt{3}}{4} \times 6^2$

= $\dfrac{\sqrt{3}}{4} \times 36$

= $9\sqrt{3}$ sq cm

T4: Area = $\dfrac{\sqrt{3}}{4} \times 3^2$

= $\dfrac{\sqrt{3}}{4} \times 9$

= $2.25\sqrt{3}$ sq cm

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Notice the pattern in areas: $144\sqrt{3}, 36\sqrt{3}, 9\sqrt{3}, 2.25\sqrt{3}, ...$

Each area is $\tfrac{1}{4}$ of the previous area! This happens because:

Side length gets halved each time

Area depends on side²

So area ratio = $\left(\dfrac{1}{2}\right)^2 = \dfrac{1}{4}$

This forms a geometric series with:

First term (a) = $144\sqrt{3}$

Common ratio (r) = $\dfrac{1}{4}$

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For an infinite geometric series with first term 'a' and common ratio 'r' (where |r| < 1), the sum is:

Sum = $\dfrac{a}{1-r}$

As we add more and more terms, they get smaller and smaller, approaching zero. The sum approaches a finite limit.

Applying this formula:

a = $144\sqrt{3}$

r = $\dfrac{1}{4}$

Sum = $144\sqrt{3} \div \left(1 - \dfrac{1}{4}\right)$

= $144\sqrt{3} \div \dfrac{3}{4}$

= $144\sqrt{3} \times \dfrac{4}{3}$

= $192\sqrt{3}$ sq cm

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The sum of areas of all infinitely many triangles = $192\sqrt{3}$ sq cm

When areas follow a geometric pattern with ratio less than 1, their infinite sum has a beautiful, finite value!