100 Follower Special

Geometry Level 5

A B C D E F G ABCDEFG is a regular heptagon with centre O O . Let there be an incircle in △ O D E \triangle ODE with centre P P , and let the point of tangency of the incircle and the line E D ED be Q Q .

Given that the side length of the heptagon is 5, find the area of P B Q G PBQG correct to 2 decimal places.

Try the 50 follower special .


The answer is 7.08.

View wiki
Michael Fuller by Michael Fuller , 22, United Kingdom

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Ahmad Saad
Feb 11, 2016

1 Helpful 0 Interesting 0 Brilliant 0 Confused

R = 5 2 ∗ S i n π 7 = 5.7619......... ( 1 ) A r e a Δ O E D = R 2 ∗ S i n 2 ∗ π 7 ∗ 1 2 = 12.9783. P e r i m e t e r = 2 ∗ R + 5 = 16.5238. ∴ r = A r e a 1 2 ∗ p e r i m e t e r = 12.9783 ∗ 2 16.5238 = 1.5708... ( 2 ) L = ⊥ d i s t a n c e o f B f r o m P Q = R ∗ S i n 2 ∗ π 7 = 4.5048.......... ( 3 ) R e q u i r e d a r e a = Δ s P B Q + Q G P = 2 ∗ 1 2 ∗ r ∗ L = 7.0762 N o t e : − Δ s P B Q = Q G P R=\dfrac 5 {2*Sin\frac \pi 7}=5.7619.........(1)\\ Area ~\Delta OED=R^2*Sin\frac{2*\pi} 7*\frac 1 2=12.9783.\\ Perimeter=2*R+5=16.5238. \\ \therefore ~r=\dfrac{Area}{\frac 1 2 *perimeter}=\dfrac{12.9783*2}{16.5238}=\color{#3D99F6}{1.5708...(2)}\\ L=\bot ~ distance ~ of ~ B ~ from ~~PQ = R*Sin\frac{2*\pi} 7=\color{#3D99F6}{4.5048.}.........(3)\\ Required ~ area ~=\Delta s ~PBQ+QGP=2*\frac 1 2 *r*L=\Large \color{#D61F06}{7.0762}\\ Note:- ~ \Delta s ~PBQ=QGP

Niranjan Khanderia - 5 years, 4 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...