Ratio between areas

Geometry Level 4

Consider a pentagon and an hexagon, both with side L L . Let A A be the area of the pentagon, and B B the area of the hexagon. If the ratio A : B A:B can be expressed as a + b c d \frac{\sqrt{a+b \sqrt{c}}}{d} , then find a + b + c + d a+b+c+d .


The answer is 128.

Alan Enrique Ontiveros Salazar by Alan Enrique Ontiveros S… , 22

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.

2 solutions

Shriram Lokhande
Jul 17, 2014

we will use the following formula for area of an regular polygon. A = 1 4 n s 2 cot ⁡ 18 0 ∘ n A = \frac{1}{4}ns^2\cot\frac{180^{\circ}}{n} where n is number of sides and s the length of a side . a r e a o f p e n t a g o n a r e a o f h e x a g o n = 1 4 5 L 2 cot ⁡ 36 1 4 6 L 2 cot ⁡ 30 \frac{area of pentagon}{area of hexagon}=\frac{\frac{1}{4}5L^2\cot36}{\frac{1}{4}6L^2\cot30} ⇒ 5 cot ⁡ 36 6 cot ⁡ 30 \Rightarrow \frac{5\cot36}{6\cot30} ⇒ 5 1 + 2 5 6 3 \Rightarrow \frac{5\sqrt{1+\frac{2}{\sqrt{5}}}}{6\sqrt{3}} ⇒ 25 + 10 5 6 3 } \Rightarrow \frac{\sqrt{25+10\sqrt{5}}}{6\sqrt{3}}\} ⇒ 75 + 50 5 18 \Rightarrow \frac{\sqrt{75+50\sqrt{5}}}{18} hence we get our answer as a + b + c + d = 128 a+b+c+d=\boxed{128}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

how have you got c o t 36 cot36 = 1 + 2 5 1+\frac{2}{\sqrt5} . pls explain

Chirayu Bhardwaj - 5 years, 3 months ago
Saad Akbar Khan
Feb 28, 2014

Hint: multiply and divide the resultant of A/B by sqrt(3)

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...