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in Perimeter and Area of Plane Figures by (23.5k points)
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The hypotenuse of an isosceles right-angled triangle is q. If we describe equilateral triangles (outwards) on all its three sides, then the total area of the re-entrant hexagon thus obtained is.

(a) q2 (√3 + 2) 

(b) q2 \(\frac{2\sqrt3+1}{4}\)

(c) \(\frac{q^2(4\sqrt3-1)}{4}\)

(d) \(\frac{q^2(5\sqrt3-2)}{4}\)

1 Answer

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Best answer

(b)  \(\frac{q^2}{4}\) (2√3 + 1).

AC = q, ∠ABC = 90º 

⇒ q = \(\sqrt{AB^2+BC^2}\)

⇒ q = \(\sqrt{2x^2}\)

⇒ q2 = 2x2\(x\) = \(\frac{q}{\sqrt2}\)

∴ Area of the re-entrant hexagon = Sum of areas of (ΔABC + ΔADC + ΔBFC + ΔAEB) 

\(\frac{1}{2}\)\(\frac{q}{\sqrt2}\) x \(\frac{q}{\sqrt2}\) + \(\frac{\sqrt3}{4}\)q2\(\frac{\sqrt3}{4}\)\(\big(\frac{q}{\sqrt2}\big)^2\) + \(\frac{\sqrt3}{4}\)\(\big(\frac{q}{\sqrt2}\big)^2\)

\(\frac{q^2}{4}\) + \(\frac{\sqrt3}{4}\)q\(\frac{\sqrt3}{8}\)q\(\frac{\sqrt3q^2}{8}\) = \(\frac{q^2}{4}\) (2√3 + 1).

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