Pentagon, length, area question. Is there a quicker solution than mine?

[MathsFormula]

Hello

The area of a regular pentagon is 600 cm²

Calculate the length of one side. The area of a regular pentagonis 600 cm². Calculate the length of one side of the pentagon.


My solution gives the correct answer but I would like toknow if there is a quicker method with High School level trigonometry:


The angles formed by triangles (in the centre) that splitthe pentagon into 5 equal parts is 360/5 = 72 degrees.

Let the sides of the triangles from the centre of thepentagon (apex of the triangles) be X

Area of triangle formula is 0.5*a*b*SinC

Area of triangle formula = 0.5 * X * X * Sin72

There are 5 triangles that form the 600cm2 pentagon so one trianglehas an area of 120cm²

Therefore 120 = 0.5*X²*Sin72

X²=252.35
X = 15.88 cm

X= 15.88

The angle of the base of the triangles is (180-72)/ 2 = 54degrees

Again using the formula Areaof triangle formula is 0.5*a*b*SinC

Let the length of one side of the pentagon = Y


Area of triangle = 120 = 0.5*15.88*Y*Sin54
Y = 18.68cm

 

[ksdhart]

To be honest, I wouldn't worry too much about having the fastest solution. Getting the right answer is far more important. However, in this case, I do know of a method that might be quicker, if your concern is running out of time on a test or something. You may not have learned these formulas in your class, but the general forms of area of a regular polygon can be immensely helpful to know. They are:

$\displaystyle A=\frac{1}{4}ns^2\cdot cot\left(\frac{\pi}{n}\right)$ and $\displaystyle A=nr^2\cdot tan\left(\frac{\pi}{n}\right)$

Where n is the number of sides, s is the side length, and r is the radius. Since you were asked to find the side length, I'd use the first formula and solve for s.

$\displaystyle 600=\frac{5}{\:4}s^2 \cdot \:cot\left(\frac{\pi}{5}\right)$

$\displaystyle s^2=480 \cdot tan\left(\frac{\pi }{5}\right)$

$\displaystyle s=\sqrt{480 \cdot tan\left(\frac{\pi }{5}\right)}$

$\displaystyle s \approx 18.674$

 

[MathsFormula]

To be honest, I wouldn't worry too much about having the fastest solution. Getting the right answer is far more important. However, in this case, I do know of a method that might be quicker, if your concern is running out of time

YES I was looking for a faster more slick solution to the question. You provided it but I think you're right when you say that know the formulae.
Is that COT that you've written? I thought perhaps you meant COS. I've never seen COT in a text book which means it's far beyond my needs

Thanks for your advice.

 

[Deleted member 4993]

YES I was looking for a faster more slick solution to the question. You provided it but I think you're right when you say that know the formulae.
Is that COT that you've written? I thought perhaps you meant COS. I've never seen COT in a text book which means it's far beyond my needs

Thanks for your advice.

COT → Cotangent → 1/(Tan)