## Answers

** 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 because 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: Multiply both sides by r: Divide both sides by Edwin **.

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. <pre><font size = 4 color = "indigo"><b> Start with a regular pentagon all sides 10. {{{drawing(300,350,-1.5,1.5,-1.5,1.5, line(-.588,-.809,.588,-.809), line(.951,.309,.588,-.809), line(.951,.309,0,1), line(-.951,.309,0,1), line(-.951,.309,-.588,-.809), locate(-.1,-.809,10) )}}} And we want the radius of this circle, which is tangent to all five sides: {{{drawing(300,300,-1.5,1.5,-1.5,1.5, line(-.588,-.809,.588,-.809), line(.951,.309,.588,-.809), line(.951,.309,0,1), line(-.951,.309,0,1), line(-.951,.309,-.588,-.809), locate(-.1,-.809,10), circle(0,0,.8) )}}} Draw in all the "spokes", which will all be radii of the circle: {{{drawing(300,350,-1.5,1.5,-1.5,1.5, line(-.588,-.809,.588,-.809), line(.951,.309,.588,-.809), line(.951,.309,0,1), line(-.951,.309,0,1), line(-.951,.309,-.588,-.809), locate(-.1,-.809,10), line(-.588,-.809,0,0), line(.951,.309,0,0), line(.588,-.809,0,0), line(0,0,0,1), line(-.951,.309,0,0), line(0,0,-.588,-.809), locate(-.1,-.2,"72°") )}}} The angle marked is {{{"72°"}}} because {{{"360°"/5="72°"}}} Next we erase the three upper spokes and concentrate only on the bottom triangle: {{{drawing(300,350,-1.5,1.5,-1.5,1.5, line(-.588,-.809,.588,-.809), line(.951,.309,.588,-.809), line(.951,.309,0,1), line(-.951,.309,0,1), line(-.951,.309,-.588,-.809), locate(-.1,-.809,10), line(-.588,-.809,0,0), line(.588,-.809,0,0), line(0,0,-.588,-.809), locate(-.1,-.2,"72°") )}}} 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. {{{drawing(300,350,-1.5,1.5,-1.5,1.5, line(-.588,-.809,.588,-.809), line(.951,.309,.588,-.809), line(.951,.309,0,1), line(-.951,.309,0,1), line(-.951,.309,-.588,-.809), locate(-.25,-.809,5), line(-.588,-.809,0,0), line(.588,-.809,0,0), rectangle(0,-.809,0-.1,-.809+.1), line(0,0,-.588,-.809), locate(-.25,-.36,"36°"), locate(.05,-.36,r), line(0,0,0,-.809) )}}} Observing only the right triangle on the left, {{{drawing(300,350,-1.5,1.5,-1.5,1.5, line(-.588,-.809,.588,-.809), line(.951,.309,.588,-.809), line(.951,.309,0,1), line(-.951,.309,0,1), line(-.951,.309,-.588,-.809), locate(-.25,-.809,5), line(-.588,-.809,0,0), rectangle(0,-.809,0-.1,-.809+.1), line(0,0,-.588,-.809), locate(-.25,-.36,"36°"), locate(.05,-.36,r), line(0,0,0,-.809) )}}} the side opposite the 36° angle is 5, and r is the side adjacent to the 36° angle, so we use tangent: {{{tan("36°")=5/r}}} Multiply both sides by r: {{{r*tan("36°")=5}}} Divide both sides by {{{tan("36°")}}} {{{r=5/tan("36°")=5/0.726542528=6.881909602cm}}} Edwin</pre>.