Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.



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,

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya

a

X 5 < Y 6 < Z 7  and X<Y<Z  

b

X 5 < Y 6 < Z 7  and X>Y>Z  

c

X 5 > Y 6 > Z 7  and X>Y>Z  

d

  X 5 > Y 6 > Z 7  and X<Y<Z   

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

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;
IMG_256We know that an area of a regular pentagon can be divided into 5 equal triangles of the same area.
So, Pentagon (ABCD) area = 5× area ofΔOCD  
Area ofΔOCD(base=sin72 & apothem=1)= 1 2 ×base×height= base×apothem 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,
Area of ΔOCD= 1 2 ×1×sin72  
= 1 2 ×0.951 =0.951  
So, the area of regular Pentagon (ABCD) area = 5× area ofΔOCD  
= 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;
IMG_256Area of regular hexagon (ABCDEF)  = 6×(area ofΔOCD)  
Area ofΔOCD(base=sin60 & apothem=1)= 1 2 ×base×height= base×apothem 2  
So, on putting the values of base and apothem, we get,
Area of ΔOCD= 1 2 ×sin 60 ×1 = 1 2 × 3 2 = 3 4  
Area of regular hexagon (ABCDEF)  = 6×(area ofΔOCD)  
=6× 3 4 = 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;
IMG_256Area of regular heptagon (ABCDEFG) = 7× area ofΔOCB  
Area ofΔOCB(base= sin(2π) 7 & apothem=1)= 1 2 ×base×height= 1 2 ×base×apothem  
So, on putting the values of base and apothem, we get,
Area of ΔOCB= 1 2 ×1×sin 2π 7 Area of ΔOCB= 1 2 ×sin 2π 7  
We know that sin 2π 7 =0.78183  ,
= 1 2 ×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,
2.377 5 > 2.565 6 > 2.736 7 Or,  X 5 > Y 6 > Z 7  
 
Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

  • promsvg

    live classes

  • promsvg

    progress tracking

  • promsvg

    24x7 mentored guidance

  • promsvg

    study plan analysis

download the app

gplay
mentor

Download the App

gplay
whats app icon
personalised 1:1 online tutoring