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CAT 2025 Lesson : Quadrilaterals - Area of Regular Polygons

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2.4 Area of Regular Polygons

From the centre of any regular
nn-sided polygon, if we draw lines to all the vertices of the regular polygon, we get nn congruent triangles. The area of o]ne of these triangles multiplied by nn is the area of the regular polygon. Therefore, for an nn-sided polygon, where a is the length of each side and hh is the length of the altitude drawn from the centre to any side, the area of the polygon =n×12×a×h=12nah = n \times \dfrac {1}{2} \times a \times h = \dfrac{1}{2}nah

Area of a regular hexagon is
332a2\dfrac{3 \sqrt{3}}{2}a^2 and area of a regular octagon is 2(1+2)a2 2(1 + \sqrt{2} )a^2 . Knowing this will help solve direction questions on these two figures.

Example 7

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 =(82)×180o8=135o = \dfrac {(8 - 2) \times 180^{ \mathrm{o}}}{8} = 135^{\mathrm{o}}

ABEF is a rectangle in a regular octagon.

\angle HAF == \angle GFA =45o= 45^{\mathrm{o}} .

\triangle APH and \triangle FQG are 45o45o90o45^{\mathrm{o}} - 45^{\mathrm{o}} - 90^{\mathrm{o}} triangles.

As the hypotenuse AH
== 1, AP == FQ =12= \dfrac {1}{\sqrt{2}}

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

AF
=12+1+12=1+22=1+2 = \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+2)=1+2= 1 (1 + \sqrt{2}) = 1 + \sqrt{2}

Answer:
1+21 + \sqrt{2} cm


Example 8

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

Solution

Interior angle of Hexagon =(62)×180o6=120o =\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
=1202=60o = \dfrac{120}{2} = 60^{\mathrm{o}}


In right-angled \triangle BFE, sin 60o60^{\mathrm{o}} = BFBE\dfrac{BF}{BE} = 32\dfrac {\sqrt{3}}{2}

Answer:
3:2\sqrt{3} : 2


Example 9

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 =120o = 120^{\mathrm{o}}

As OA is the angle bisector,
\angle OAP =60o= 60^{\mathrm{o}}

\triangle OPA is a 30o60o90o30^{\mathrm{o}} - 60^{\mathrm{o}} - 90^{\mathrm{o}} triangle, where sides are in the ratio 1 : 3\sqrt{3} : 2

∴ OP
= = AP ×3=3a2 \times \sqrt3 = \dfrac {\sqrt{3}a}{2}

Similarly, R is the mid-point of PQ and
\angleOPR = 60o60^{\mathrm{o}}

\triangleORP is also a 30o60o90o30^{\mathrm{o}} - 60^{\mathrm{o}} - 90^{\mathrm{o}} triangle, where OP is the hypotenuse.

∴ PR
=OP2 = \dfrac{\mathrm{OP}}{2} =3a4 = \dfrac {\sqrt{3}a}{4} and PQ = = 2PR =3a2 = \dfrac{\sqrt{3}a}{2}

Ratio of areas of 2 similar hexagons = Ratio of squares of their sides =
(3a2)2\left (\dfrac {\sqrt{3}a}{2}\right)^2 : a2a^2 =34= \dfrac{3}{4} : 1 = 3 : 4

Answer:
3:4 3 : 4


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