Problem of the Week

Problem 28 Solution

Congratulations to Patryk Odzimek from Spalding Academy, Markus Kuan from Northstowe Secondary College, Anna Poliakoff from Hitchin Girls’ School, and Matthew Whitten from The Cavendish School for successfully solving the problem!

a) We can divide the regular polygon into 6 congruent triangles, as shown above. Then, apply the triangle area formula $\frac{1}{2}ab\sin{C}$ , using the two side lengths A and the angle between them:

Area of each congruent triangle:

$\frac{1}{2}(A)(A)(\sin{\frac{360}{6}})$

$=\frac{1}{2}(A^2)(\sin{60})$

$=\frac{\sqrt{3}}{4}A^2$

Total area of the 6 congruent triangles:

$6(\frac{\sqrt{3}}{4}A^2)$

$=\frac{3\sqrt{3}}{2}A^2$

b) Similarly, divide the regular polygon into 12 congruent triangles:

Area of each congruent triangle:

$\frac{1}{2}(A)(A)(\sin{\frac{360}{12}})$

$=\frac{1}{2}(A^2)(\sin{30})$

$=\frac{1}{4}A^2$

Total area of the 12 congruent triangles:

$12(\frac{1}{4}A^2)$

$=3A^2$

c) This is similar to the previous expressions, but let n be the total number of sides in the regular polygon:

Area of each congruent triangle:

$\frac{1}{2}(A)(A)(\sin{\frac{360}{n}})$

$=\frac{1}{2}(A^2)(\sin{\frac{360}{n}})$

Total area of the n congruent triangles:

$n(\frac{1}{2}(A^2)(\sin{\frac{360}{n}}))$

$=\frac{n}{2}(A^2)(\sin{\frac{360}{n}})$

Many of you used radians to solve this part, so for clarity and reference, the equivalent expression in radians is:

$=\frac{n}{2}(A^2)(\sin{\frac{2\pi}{n}})$

Note: Radians are not part of the GCSE Maths / Further Maths curriculum. Do not worry if you don’t understand this.

d) As the number of sides increases to infinity, the shape approaches a circle, where A is the radius of the circle.

Area of the shape therefore approaches $\pi(A^2)$ , and therefore the coefficient of $A^2$ approaches $\pi$ .

Many of you used a small angle approximation using radians, so a solution of this will be made for clarity and reference:

Note: Small angle approximations can only be done in radians, neither of which are part of the GCSE Maths / Further Maths curriculum. Do not worry if you don’t understand this.

As the total number of sides increases, the congruent triangles approach a very small internal angle.

For small angles, the sine of a small angle approaches the angle value itself (in radians):

$\lim_{n\to\infty}\sin(\frac{2\pi}{n})=\frac{2\pi}{n}$

Therefore, the expression for the area of a regular polygon with n sides as n approaches infinity is:

$\lim_{n\to\infty}\frac{n}{2}(A^2)(\sin{\frac{2\pi}{n}})=\frac{n}{2}(A^2)({\frac{2\pi}{n}})$

$=\pi(A^2)$

The coeffcient of $A^2$ therefore approaches $\pi$ .

Problem of the Week

Problem 28 Solution

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