Let X, Y, Z be respectively the areas of a regular pentagon, regular hexagon and regular heptagon which are inscribed in a circle of radius 1. Then,

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X 5 < Y 6 < Z 7 a n d X < Y < Z

X 5 < Y 6 < Z 7 a n d X > Y > Z

X 5 > Y 6 > Z 7 a n d X > Y > Z

X 5 > Y 6 > Z 7 a n d X < Y < Z

answer is D.

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Detailed Solution

Given,
X, Y, Z are the faces of a regular pentagon, a regular hexagon, and a regular heptagon, and we have to find the correct relationship between these three regular polygons.
Regular polygons can be divided into triangles. The content of regular polygons can be found by multiplying the number of sides and the area of one triangle contained in it.
Let us first consider the case of a regular pentagon. The question says that a regular pentagon is inscribed in a circle of radius 1. So we can draw the figure with all the details as shown below;

We know that an area of a regular pentagon can be divided into 5 equal triangles of the same area.
So, Pentagon (ABCD) area

= 5 × a r e a o f Δ O C D

A r e a o f Δ O C D (b a s e = s i n 72 ∘ & a p o t h e m = 1) = 12 × b a s e × h e i g h t = b a s e × a p o t h e m 2

The radius of the incircle is the apothem of the regular polygon and here the height of the triangle is the apothem of the regular polygon.
So if we put in the values of the base and the apothem, we get,

A r e a o f Δ O C D = 12 × 1 × s i n 72 ∘

= 12 × 0.951 = 0.951

So, the area of regular Pentagon (ABCD) area

= 5 × a r e a o f Δ O C D

= 5 × 0.951 2 = 2.377

Similarly, we find the area of a regular hexagon (ABCDEF). The question says that a regular hexagon is inscribed in a circle of radius 1. So we can draw the figure with all the details as shown below;

Area of regular hexagon (ABCDEF)

= 6 × (a r e a o f Δ O C D)

A r e a o f Δ O C D (b a s e = s i n 60 ∘ & a p o t h e m = 1) = 12 × b a s e × h e i g h t = b a s e × a p o t h e m 2

So, on putting the values of base and apothem, we get,

A r e a o f Δ O C D = 12 × sin 60 ∘ × 1 = 12 × 32 = 34

Area of regular hexagon (ABCDEF)

= 6 × (a r e a o f Δ O C D)

= 6 × 34 = 10.26 4 = 2.565

Similarly, we find the area of a regular heptagon (ABCDEFG). The question says that a regular heptagon is inscribed in a circle of radius 1. So we can draw the figure with all the details as shown below;

Area of regular heptagon (ABCDEFG)

= 7 × a r e a o f Δ O C B

A r e a o f Δ O C B (b a s e = s i n (2 π) 7 & a p o t h e m = 1) = 12 × b a s e × h e i g h t = 12 × b a s e × a p o t h e m

So, on putting the values of base and apothem, we get,

A r e a o f Δ O C B = 12 × 1 × s i n 2 π 7 A r e a o f Δ O C B = 12 × s i n 2 π 7

We know that

sin 2 π 7 = 0.78183

= 12 × 0.78183 = 0.390915

So, the area of regular heptagon (ABCDEFG)

= 7 × (0.390915) = 2.736

So, the area of regular heptagon (ABCDEFG) will be 2.736
So, we have X = 2.377, Y = 2.565 and Z = 2.736.
From this value we have,
2.377 < 2.565 < 2.736
X < Y < Z
Also,