In the adjoining figure, ABC is an equilateral triangle inscribing a square of maximum possible area. Again in this squares -Maths 9th

In the adjoining figure, ABC is an equilateral triangle inscribing a square of maximum possible area. Again in this squares -Maths 9th
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1 Answer

Answer :

(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,\(rac{PM}{PA}\) = sin 60º⇒ \(rac{x}{PA}\) = \(rac{\sqrt3}{2}\) ⇒ PA = \(rac{2x}{\sqrt3}\)∴ AC = AP + PC = \(rac{2x}{\sqrt3}\) + \(x\)Given \(rac{2x}{\sqrt3}\) + \(x\) = 0.01 m = 1 cm⇒ \(x\) =  \(rac{\sqrt3}{(2+\sqrt3)}\) 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, \(rac{RY}{RM}\) = sin 60º⇒ \(rac{y}{RM}\) = \(rac{\sqrt3}{2}\) ⇒ RM = \(rac{2y}{\sqrt3}\)∴ MT = MR + RT = \(rac{2y}{\sqrt3}\) + y = \(rac{(2+\sqrt3)}{\sqrt3}y\)Given MT = x, thenx = \(\bigg(rac{2+\sqrt3}{\sqrt3}\bigg)y\) ⇒ y = \(rac{\sqrt3x}{2+\sqrt3}\)But from (i), x = √3(2 - √3)∴ y = \(rac{\sqrt3\sqrt3(2-\sqrt3)}{2+\sqrt3}\) = \(rac{3.(2-\sqrt3)}{2+\sqrt3}\).\(rac{(2-\sqrt3)}{(2-\sqrt3)}\)⇒ 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.
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Last Answer : Under root 3/4 side square. Got it

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