Question

In the adjoining figure, ABC is an equilateral triangle inscribing a square of maximum possible area. Again in this squares there is an equilateral triangle whose side is same as that of the square. Further the smaller equilateral triangle inscribes a square of maximum possible area. What is the area of the innermost square if each side of the outermost triangle be 0.01 m ?

(a) (873 – 504√3) cm² 

(b) (738 – 504√3) cm²

(c) (873 – 405√2) cm²

(d) None of these

Answer 1

(a) (873 – 504√3) cm².

Since ∠CPO = ∠COP = 60º, therefore, PCO is also an equilateral triangle. 

Let each side of the square MNOP be x cm. 

Then PC = CO = PO = x cm 

Then in ΔPAM,

\(\frac{PM}{PA}\) = sin 60º

⇒ \(\frac{x}{PA}\) = \(\frac{\sqrt3}{2}\) ⇒ PA = \(\frac{2x}{\sqrt3}\)

∴ AC = AP + PC = \(\frac{2x}{\sqrt3}\) + \(x\)

Given \(\frac{2x}{\sqrt3}\) + \(x\) = 0.01 m = 1 cm

⇒ \(x\) =  \(\frac{\sqrt3}{(2+\sqrt3)}\) cm = √3 (2 - √3) cm         .....(i)

(On rationalising the denominator) 

Now, let each side of the square RSXY be y. Then RT = y 

(∵ RTS is an equilateral triangle)

∴ In ΔRYM, \(\frac{RY}{RM}\) = sin 60º

⇒ \(\frac{y}{RM}\) = \(\frac{\sqrt3}{2}\) ⇒ RM = \(\frac{2y}{\sqrt3}\)

∴ MT = MR + RT = \(\frac{2y}{\sqrt3}\) + y = \(\frac{(2+\sqrt3)}{\sqrt3}y\)

Given MT = x, then

x = \(\bigg(\frac{2+\sqrt3}{\sqrt3}\bigg)y\) ⇒ y = \(\frac{\sqrt3x}{2+\sqrt3}\)

But from (i), x = √3(2 - √3)

∴ y = \(\frac{\sqrt3\sqrt3(2-\sqrt3)}{2+\sqrt3}\) = \(\frac{3.(2-\sqrt3)}{2+\sqrt3}\).\(\frac{(2-\sqrt3)}{(2-\sqrt3)}\)

⇒ y = 3(2 - √3)² = 3(7 - 4√3)

∴ Area of the inner most square = y²

= \(\big(3(7-4\sqrt2)\big)^2\)

= 9(49 + 48 – 56√3) = (873 – 504√3) cm².