Areas of circles and triangles

Perimeter of a triangle:

The perimeter of a triangle is the sum of its 3 sides.

In the following figure we have:

$P=a+b+c$ is the $perimeter$ of the triangle.

We have used the semi-perimeter of the triangle when we deomnstrated the Heron formula.

It is simply $s=\frac{a+b+c}{2}$

The student learns to calculate perimeters to be able to tackle problems involving costs of fences for example.

Example:

A location has a form of a triangle. The three sides have measurements of 250ft, 185ft and 215 ft. A fence costing 7 dollars per linear foot has to be constructed around the location, leaving an entry of 5 ft. The company hired for the construction charges 6 dollars for each linear foot. What will be the total cost of the fence?

Solution:

The perimeter of the triangle:

$p=250+185+215=650$

However, the entry of 5 ft will not get the fence.

We get:

$fence=650-5=645 feet$

Total cost per foot:

$Cost=7+6=13$

Total cost for constructing the fence:

$Total=13 \times 645=8385$ dollars.

Area of a triangle:

The area of a triangle can be calculated using methods depending on information we have:

For a scalene triangle:

-Taking any altitude (as height, h) and its opposite side (as base, b).

The Area is:

$A=\frac{b\times h}{2}$

We can also write:

$A=\frac{1}{2}bh$

For an isosceles triangle, given the three sides: a,a and b

We can calculate $h$

$h^2=a^2-\left(\frac{b}{2}\right)^{2}$

Finally:

$h=\frac{1}{2}\sqrt{4a^2-b^2}$

And the area of the triangle:

$A=\frac{1}{4}b\sqrt{4a^2-b^2}$

If we use the Heron formula, we find the same formula

$A=\sqrt{s(s-a)(s-b)(s-c}$

In our situation $b=a$ and $c=b$

That means $s=\frac{1}{2}(2a+b)$

$s-a=\frac{1}{2}(2a+b-2a)=\frac{1}{2}(b)$

$s-b=s-a=\frac{1}{2}(2a+b-2a)=\frac{1}{2}(b)$

$s-c=s-b=\frac{1}{2}(2a+b-2b)=\frac{1}{2}(2a-b)$

We get the following area:

$A=\sqrt{(\frac{2a+b}{2})(\frac{b}{2})(\frac{b}{2})(\frac{2a-b}{2})}$

$A=\sqrt{(\frac{b^2}{4})(\frac{4a^2-b^2}{4}})$

Finally:

$A=\frac{1}{4}b\sqrt{4a^2-b^2}$

Special isosceles and right triangle:

In this situation we have :

$h=\frac{1}{2}b$

Without going far, we can notice that each of the right triangle legs can be the base (a) or the height (a):

$A=\frac{1}{2}a\times a$

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

Using b:

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

This can be found from the previous formula as well.

Equilateral Triangle

We have the nomrmal formula:

$A=\frac{1}{2}bh$

$h^2=a^2-\frac{a^2}{4}$

$h^2=\frac{1}{4}(4a^2-a^2)$

$h^2=\frac{1}{4}(3a^2)$

$h=\frac{\sqrt{3}a}{2}$

Now for the area we have $b=a$

$A=\frac{1}{2}(a)(\frac{\sqrt{3}a}{2})$

finally:

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

Area Triangle using angles:

We can see that:

$Area=\frac{1}{2}ch$

$\sin(A)=\frac{h}{b}$

So: $h=b\sin(A)$

The Area:

$Area=\frac{1}{2}bc\sin(A)$

We can find all three relationships:

$Area=\frac{1}{2}bc\sin(A)$

$Area=\frac{1}{2}ac\sin(B)$

$Area=\frac{1}{2}ab\sin(C)$

Circumference of a circle

The Circumference $C$ of a circle is as follows:

$C=2\pi r$

where $r$ is the radius of the circle.

Since the diameter $D=2r$ , we can write:

$C=D\times\pi$

Areas and circumferences of circular figures can be calculated without replacing $\pi$ with its value. That means in terms of $\pi$ .

Length of an arc:

The length of an arc of a circle is related to the central angle $\alpha$ that intercepts it.

If $\alpha$ is in $radians$ , we have:

$m\stackrel\frown{AB}=\alpha r$

For an angle in $degrees$ , we use the proportion to the circumference:

The length of $m\stackrel\frown{AB}=\frac{\alpha^{\circ}}{360}2\pi r$

This is the same as:

$m\stackrel\frown{AB}=\frac{\alpha^{\circ}}{180}\pi r$

Area of a Circle of radius r:

If $r$ is the radius of a circle, the $Area$ is:

$Area=\pi r^{2}$

This formula is widely used in geometry and other fields.

Area of a Sector AOB of a circle:

If $r$ is the radius of a circle, $\alpha$ the central angle (or intercepted arc) in $radians$ , the $Area$ of $Sector \; AOB$ is:

$Area\; AOB=\frac{\alpha}{2\pi}\pi r^2$

We simplify and get:

$Area\; AOB=\frac{\alpha}{2}r^2$

If $\alpha$ the central angle (or intercepted arc) is in $degrees$ :

$Area\; AOB=\frac{\alpha^{\circ}}{360}\pi r^2$

This is a simple conversion of $\alpha$ radians above. $\alpha (rad)=\frac{\alpha^{\circ}\pi}{180}$

Area of a segment of a circle

This can be approached 2 ways:

-Area of the sector minus area of isosceles triangle $AOB$

-By deriving the formula.

The central angle’s bisector bisects the chord AB at D.

We get

$\sin\frac{\alpha}{2}=\frac{AD}{OA}$

But $OA=r$

$AD=r\sin\frac{\alpha}{2}$

OD is the altitude:

$OD=h$

$h^2=r^2-(AD)^2$

$h^2=r^2-r^2\sin^{2}\frac{\alpha}{2}$

$h=\sqrt{r^2-r^2\sin^{2}\frac{\alpha}{2}}$

$h=r\sqrt{1-\sin^{2}\frac{\alpha}{2}}$

But we know that:

$1-\sin^{2}\frac{\alpha}{2}=\cos^{2}\frac{\alpha}{2}$

That returns:

$h=r\cos\frac{\alpha}{2}$

The Area of $\triangle AOB$ is:

$Area\;\triangle AOB=AD \cdot h$

We plug in:

$Area\;\triangle AOB=r\sin\frac{\alpha}{2}\cdot r\cos\frac{\alpha}{2}$

Finally:

$Area\;\triangle AOB=\frac{1}{2}r^2\sin\alpha$

For sector AOB:

$Area\;Sector\; AOB=\frac{\alpha}{2}r^2$

We take the difference to get the area of the segment:

$Area\; Segment=\frac{1}{2}\alpha r^2-\frac{1}{2}r^2\sin\alpha$

$Area\; Segment=\frac{1}{2}r^2 \left(\alpha-\sin\alpha \right)$

Post Views: 269

Areas of circles and triangles

Areas of circles and triangles

Perimeter of a triangle:

Area of a triangle:

Special isosceles and right triangle:

Equilateral Triangle

Area Triangle using angles:

Circumference of a circle

Length of an arc:

Area of a Circle of radius r:

Area of a Sector AOB of a circle:

Area of a segment of a circle

Be the first to comment

Leave a Reply Cancel reply