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Question:

In a square of side 2 meters, isosceles triangles of equal area are cut from the corners to form a regular octagon. Find the perimeter and area of the regular octagon.

Explanation:

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Let the side of isosceles triangle be x.

∴ The side of octagon x$\sqrt{2}$.

∴ x + x + x$\sqrt{2}$ = 2

∴ 2x + x$\sqrt{2}$ = 2

∴ x = $\frac{2}{2+\sqrt{2}}$

∴ Side of octagon = x$\sqrt{2}$ = $\frac{2\sqrt{2}}{2+\sqrt{2}}$ = $\frac{2}{1+\sqrt{2}}$

∴ Perimeter of octagon = 8 × $\frac{2}{1+\sqrt{2}}$ = $\frac{16}{1+\sqrt{2}}$ units

Area of octagon = Area of square – 4 × Area of isosceles triangle

= 2² - 4 × $\frac{1}{2}$ x²

= 4 - 4 × $\frac{1}{2}$ × ${\left(\frac{\sqrt{2}}{1+\sqrt{2}}\right)}^2$

= 4 - $\frac{4}{3+2\sqrt{2}}$ = $\frac{8(1+\sqrt{2})}{3+2\sqrt{2}}$ sq. units

Hence, option (d).