Category Archives: Finding an area

Area of Regular Polygons

Finding the area of a regular polygon when you know the side length

Find the area of an n-sided regular polygon if you know the side length, l.

An octagon for a visual reference

Find the h of the triangle in terms of l and theta.

tan(\theta)=\frac{\frac{l}{2}}{h}

h=\frac{l}{2tan(\theta)}

Remember the area of a triangle is A=\frac{1}{2}bh

Hence, A=\frac{1}{2} l \times \frac{l}{2tan(\theta)}=\frac{l^2}{4tan(\theta)}

And \theta=\frac{360}{2n}=\frac{180}{n}

Therefore A=\frac{l^2}{4tan(\frac{180}{n})}

There are n triangles in an n-sided polygon

(1)   \begin{equation*}A=\frac{nl^2}{4tan(\frac{180}{n})}\end{equation*}

Find the area of a hexagon with side length 10cm.
A=\frac{6\times10^2}{4tan(\frac{180}{6})}
A=\frac{600}{4(\frac{1}{\sqrt{3}})}
A=150\sqrt{3} cm^2

Finding the area of a polygon if you know the inradius or the apothem

The apothem and the inradius are the same. It is the radius of the incircle.

Find the area of the triangle in terms of a and theta.

tan(\theta)=\frac{\frac{l}{2}}{a}

l=2atan(\theta)

A=\frac{1}{2}bh

A=\frac{1}{2}2atan(\theta)a=a^2tan(\theta)

And \theta=\frac{180}{n}

Hence for an n-sided polygon

(2)   \begin{equation*}A=na^2tan(\frac{180}{n})\end{equation*}

Find the area of a regular pentagon with apothem 4.5cm
A=5\times 4.5^2tan(\frac{180}{5})
A=73.56cm^2

Finding the area of a regular polygon given the circumradius

The circumradius is the radius of the circumscribed circle (R in the diagram above)

Remember the area of triangle formula

A=\frac{1}{2}absin(\theta)

A=\frac{1}{2}R^2sin(\theta)

\theta=\frac{360}{n}

Hence, A=\frac{1}{2}R^2sin(\frac{360}{n})

Hence, for an n-sided polygon

(3)   \begin{equation*}A=\frac{nR^2sin(\frac{360}{n})}{2}\end{equation*}

Find the area of a regular octagon inscribed in a circle of radius 10cm.
A=\frac{8\times 10^2sin(45)}{2}
A=200\sqrt{2}cm^2

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Filed under Area, Area of Triangles (Sine), Finding an area, Non-Right Trigonometry, Regular Polygons, Right Trigonometry, Year 11 Mathematical Methods

Intersecting Circles

Two circles of radius L and 2L intersect as shown. What is the area of the shaded region?

From Professor Povey’s Perplexing Problems

My plan is to find the sum of the area of the two segments (see below).

Construct triangles

The diagonal of the square (the pink line above) has length

=\sqrt{(2L)^2+(2L)^2}=\sqrt{8L^2}=2\sqrt{2}L

From the pink triangle in the above diagram, I am going to find the angles using the cosine rule.

cos\theta=\frac{L^2+(2\sqrt{2}L)^2-(2L)^2}{2(L)(2\sqrt{2}L)}

cos\theta=\frac{5L^2}{4\sqrt{2}L^2}

cos\theta=\frac{5}{4\sqrt{2}}

\theta=0.487

cos\alpha=\frac{(2L)^2+(2\sqrt{2}L)^2-L^2}{(2\sqrt{2}L)(2L)}

cos\alpha=\frac{11L^2}{8\sqrt{2}L^2}

cos\alpha=\frac{11}{8\sqrt{2}}

\alpha=0.236

The green quadrilateral is a kite, which means the diagonals are perpendicular.

This means the segment angles are 2\theta and 2\alpha (because the triangles are isosceles and the diagonal is perpendicular to the base of the triangles).

Area of green segment

A=\frac{1}{2}r^2(\theta-sin\theta)

A=\frac{1}{2}L^2(0.9734-sin(0.9734))=0.0733L^2

Area of yellow segment

A=\frac{1}{2}4L^2(0.4721-sin(0.4721))=0.0035

Total Area =0.0733L^2+0.0035L^2=0.108L^2

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Filed under Finding an area, Geometry, Professor Povey