# How to find the area of an isosceles triangle and hexagon?

• March 10th 2010, 03:43 PM
pocketasian
How to find the area of an isosceles triangle and hexagon?
Hello, I would like some help on finding the area of two different shapes.

The first one is the area of an isosceles triangle with two sides of length 2x and one side of length x. Will someone show me the process of how to get this answer? The answer I got was:
(squareroot(15)*x^2)/2

As for my second area question, how would I find the area of a regular hexagon with sides of length x? The answer I got was:
3squareroot(3)*x^2

With both of these, it appears that I must have forgotten to multiply it by 1/2 at some point, but I can't figure out when.

• March 10th 2010, 09:13 PM
Soroban
Hello, pocketasian!

Are you forgetting the "half" in: . $A \:=\:\tfrac{1}{2}bh$ ?

Quote:

Find the area of an isosceles triangle with two sides of length $2x$ and one side of length $x.$
Code:

              A               o             /|\             / | \           /  |  \       2x /  |  \ 2x         /    |h  \         /    |    \       /      |      \     B o-------o-------o C         ½x  M  ½x       : - - - x - - - :

We have isosceles triangle $ABC$ with $AB \,=\,AC\,=\,2x,$
. . and $BC = x.$

$M$ is the midpoint of $BC\!:\;BM \,=\,MC \,=\,\tfrac{1}{2}x$

$h \,=\,AM$ is the altitude (height} of $\Delta ABC.$

In either of the right triangles: . $h^2 + \left(\tfrac{x}{2}\right)^2 \:=\:(2x)^2$

. . $h^2 + \frac{x^2}{4} \:=\:4x^2 \quad\Rightarrow\quad h^2 \:=\:\frac{15x^2}{4} \quad\Rightarrow\quad\boxed{ h \:=\:\frac{\sqrt{15}}{2}\,x}$

The base of the triangle is: . $\boxed{b \:=\:x}$

Therefore: . $\text{Area} \;=\;\tfrac{1}{2}bh \;=\;\frac{1}{2}(x)\left(\frac{\sqrt{15}}{2}\,x\ri ght) \;=\;\frac{\sqrt{15}}{4}\,x^2$

• March 11th 2010, 12:06 AM
earboth
Quote:

Originally Posted by pocketasian
...

As for my second area question, how would I find the area of a regular hexagon with sides of length x? The answer I got was:
3squareroot(3)*x^2

...

1. A regular hexagon consists of 6 equilateral triangles with the side-length x. The area of one triangle is calculated by:

$A_\Delta = \frac12 \cdot x \cdot h$

2. Use Pythagorean theorem:

$x^2=h^2+\left(\frac x2\right)^2~\implies~h=\frac x2 \cdot \sqrt{3}$

3. The area of the hexagon is calculated by:

$A_6=6 \cdot \frac12 \cdot x \cdot \frac x2 \cdot \sqrt{3}$

Simplify!