CAT 2025 Lesson : Quadrilaterals - Area of Regular Polygons

ABCEFGH is a regular octagon with each side measuring 1 cm. What is the area of ABEF?

Solution

Let P and Q be the point of intersection of perpendiculars from H and G to AF respectively.

Interior angle of Octagon $= \dfrac {(8 - 2) \times 180^{ \mathrm{o}}}{8} = 135^{\mathrm{o}}$ ABEF is a rectangle in a regular octagon. ∴ $\angle$ HAF $=$ $\angle$ GFA $= 45^{\mathrm{o}}$. $\triangle$ APH and $\triangle$ FQG are $45^{\mathrm{o}} - 45^{\mathrm{o}} - 90^{\mathrm{o}}$ triangles. As the hypotenuse AH $=$ 1, AP $=$ FQ $= \dfrac {1}{\sqrt{2}}$

As GHPQ is a rectangle, GH $=$ PQ $=$ 1

AF $= \dfrac {1}{\sqrt{2}} + 1 + \dfrac {1}{\sqrt{2}} = 1 + \dfrac {2}{\sqrt{2}} = 1 + \sqrt{2}$

Area of rectangle ABEF $=$ AB $\times$ AF $= 1 (1 + \sqrt{2}) = 1 + \sqrt{2}$

Answer: $1 + \sqrt{2}$ cm

All sides of the hexagon ABCDEF are equal. What is the ratio of BF : BE?

Solution

Interior angle of Hexagon $=\dfrac{(6 - 2 ) \times 180^{ \mathrm{o}}}{6} = 120^{\mathrm{o}}$ As BC and FE are opposite sides, BF $\bot$ FE. As B and E are opposite vertices, BE passes through the centre of the hexagon O. The line joining the vertex to the centre is an angle bisector. ∴ ∠BEF $= \dfrac{120}{2} = 60^{\mathrm{o}}$

In right-angled $\triangle$ BFE, sin $60^{\mathrm{o}}$ = $\dfrac{BF}{BE}$ = $\dfrac {\sqrt{3}}{2}$

Answer: $\sqrt{3} : 2$

When the mid-points of each of the sides of a regular hexagon are joined, we get another regular hexagon. What is the ratio of areas of the smaller hexagon to that of the larger one?

Solution

In hexagon ABCDEF, O is the centre and a is the length of each side. P and Q are the mid-points of AB and BC respectively. ∴ OP $\bot$ AB.

Interior angle of hexagon $= 120^{\mathrm{o}}$ As OA is the angle bisector, $\angle$ OAP $= 60^{\mathrm{o}}$ $\triangle$ OPA is a $30^{\mathrm{o}} - 60^{\mathrm{o}} - 90^{\mathrm{o}}$ triangle, where sides are in the ratio 1 : $\sqrt{3}$ : 2 ∴ OP $=$ AP $\times \sqrt3 = \dfrac {\sqrt{3}a}{2}$ Similarly, R is the mid-point of PQ and $\angle$OPR = $60^{\mathrm{o}}$ $\triangle$ORP is also a $30^{\mathrm{o}} - 60^{\mathrm{o}} - 90^{\mathrm{o}}$ triangle, where OP is the hypotenuse.

∴ PR $= \dfrac{\mathrm{OP}}{2}$ $= \dfrac {\sqrt{3}a}{4}$ and PQ $=$ 2PR $= \dfrac{\sqrt{3}a}{2}$

Ratio of areas of 2 similar hexagons = Ratio of squares of their sides =$\left (\dfrac {\sqrt{3}a}{2}\right)^2$ : $a^2$ $= \dfrac{3}{4}$ : 1 = 3 : 4

Answer: $3 : 4$