The sides of one regular hexagon is larger than that of the other regular hexagon by 1 cm . If the product of their
areas is 243, then find the sides of both the regular hexagons .

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Solution

Let the side of the first regular hexagon be x cm. Then, the side of second regular hexagon will be (x + 1) cm.

We know that a hexagon is a combination of six equal equilateral triangles.
Thus, the area of a regular hexagon is, A = 6 × Area of one equilateral triangle.
$W e know that area of an equilateral triangle is \frac{\sqrt{3}}{4} {(Side)}^{2} . Therefore, the area of regular hexagon is A = 6 [\frac{\sqrt{3}}{4} {(Side)}^{2}] = \frac{3 \sqrt{3}}{2} {(S i d e)}^{2} And, the area of the first regular hexagon with side x cm is A_{2} = \frac{3 \sqrt{3}}{2} {(x)}^{2} Now, the area of thesecond regu lar hexagon with side (x + 1) cm is A_{1} = \frac{3 \sqrt{3}}{2} {(x + 1)}^{2} Thus, by the given condition, we get : \frac{3 \sqrt{3}}{2} {(x + 1)}^{2} \times \frac{3 \sqrt{3}}{2} {(x)}^{2} = 243 \Rightarrow \frac{27}{4} x^{2} {(x + 1)}^{2} = 243 \Rightarrow x^{2} {(x + 1)}^{2} = \frac{243 \times 4}{27} = 36 \Rightarrow x (x + 1) = 6 \Rightarrow x^{2} + x = 6 \Rightarrow x^{2} + x - 6 = 0 On splititng the middle term x as 3 x - 2 x, we get : x^{2} + 3 x - 2 x - 6 = 0 \Rightarrow (x + 3) (x - 2) = 0 \Rightarrow x + 3 = 0 o r x - 2 = 0 \Rightarrow x = - 3 o r x = 2 Since x is the side of the regular hexagon, which cannot be negative, therefore, x = 2 cm Thus, the side of the regular hexagon is 2 cm and the side of the second regular hexagon is 3 cm$

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