SOLUTION: A regular pentagon has sides of 10cm. Find the radius the largest circle which can be drawn inside the pentagon. Actually this question is from the chapter of Triginometry: Tang

Algebra ->  Trigonometry-basics -> SOLUTION: A regular pentagon has sides of 10cm. Find the radius the largest circle which can be drawn inside the pentagon. Actually this question is from the chapter of Triginometry: Tang      Log On


   



Question 207273This question is from textbook Longman Mathematics for IGCSE book 1
: A regular pentagon has sides of 10cm.
Find the radius the largest circle which can be drawn inside the pentagon.
Actually this question is from the chapter of Triginometry: Tangent ratios
Please tell me that how could I use tangent in this particular question.
I request you to solve my question as soon as possible. I would be highly glad. Thankyou very much indeed.
This question is from textbook Longman Mathematics for IGCSE book 1

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
A regular pentagon has sides of 10cm.
Find the radius the largest circle which can be drawn inside the pentagon.
Actually this question is from the chapter of Triginometry: Tangent ratios
Please tell me that how could I use tangent in this particular question.
I request you to solve my question as soon as possible. I would be highly glad. Thankyou very much indeed.

Start with a regular pentagon all sides 10.



And we want the radius of this circle,
which is tangent to all five sides:



Draw in all the "spokes", which will
all be radii of the circle:



The angle marked is %2272%B0%22 because %22360%B0%22%2F5=%2272%B0%22

Next we erase the three upper spokes and concentrate
only on the bottom triangle:



Draw the alititude to the base, which
bisects the 72° angle into two 36°
angles, and which also bisects the base
into two segments, each 5cm.  The desired
circle will be tangent to the base at
the midpoint, and the radius of the
circle will be this altitude, marked
as r.



Observing only the right triangle
on the left,



the side opposite the 36° angle is
5, and r is the side adjacent to
the 36° angle, so we use tangent:

tan%28%2236%B0%22%29=5%2Fr 

Multiply both sides by r:

r%2Atan%28%2236%B0%22%29=5

Divide both sides by tan%28%2236%B0%22%29

r=5%2Ftan%28%2236%B0%22%29=5%2F0.726542528=6.881909602cm

Edwin