# Thread: Area of a Equil. Triangle given 1 measure

1. ## Area of a Equil. Triangle given 1 measure

Here is the question: Gyazo - 8d2dd59242fb7fa392dbdc6fa84e79a5.png (question 25)
Here is my work: Gyazo - 9fcbe6021e7df14bd5c03df38dd82d3a.png

I could've used A=.5*b*h (regular area formula for triangle) but since I've been using A=.5*apothem*perimeter, I figured I'd continue using it. My final answer was 32in squared.

Used 30-60-90 special right triangle rules etc. If something isn't clear in my scrap, let me know.

2. ## Re: Area of a Equil. Triangle given 1 measure

Originally Posted by Iceycold
Here is the question: Gyazo - 8d2dd59242fb7fa392dbdc6fa84e79a5.png (question 25)
Here is my work: Gyazo - 9fcbe6021e7df14bd5c03df38dd82d3a.png

I could've used A=.5*b*h (regular area formula for triangle) but since I've been using A=.5*apothem*perimeter, I figured I'd continue using it. My final answer was 32in squared.

Used 30-60-90 special right triangle rules etc. If something isn't clear in my scrap, let me know.
This isn't right.

You've got a 30/60/90 degree triangle with the side opposite the 30 deg angle, $h=4$.

so by the properties of this triangle the hypotenuse is 8, and the other, $b=4\sqrt{3}$.

The area of a right triangle is $\large \frac{b h}{2}$ and you have 6 of these triangles in your original triangle.

See if you can come up with the correct answer now.

3. ## Re: Area of a Equil. Triangle given 1 measure

Hello, Iceycold!

Find the area.

Code:
              *
*|*
* | *
*  |4 *
*   |   *
*    |    * 2x
*     *     *
*      |      *
*       |2      *
*        |     60 *
* * * * * * * * * * *
x

The "center" (centroid) of a triangle
. . divides a median in the ratio 2:1.

We have a 30-60-90 right triangle
. . with legs $x$ and $6$, and hypotenuse $2x.$

Hence: . $x^2 + 6^2 \:=\:(2x)^2 \quad\Rightarrow\quad x^2 + 36 \:=\:4x^2$

. . . $3x^2 \:=\:36 \quad\Rightarrow\quad x^2 \:=\:12 \quad\Rightarrow\quad x \:=\:2\sqrt{3}$

We have a triangle with base $b = 4\sqrt{3}$ and height $h = 6.$

It area is: . $A \;=\;\tfrac{1}{2}(4\sqrt{3})(6) \;=\;12\sqrt{3}\text{ in}^2$

4. ## Re: Area of a Equil. Triangle given 1 measure

Originally Posted by romsek
This isn't right.

You've got a 30/60/90 degree triangle with the side opposite the 30 deg angle, $h=4$.

so by the properties of this triangle the hypotenuse is 8, and the other, $b=4\sqrt{3}$.

The area of a right triangle is $\large \frac{b h}{2}$ and you have 6 of these triangles in your original triangle.

See if you can come up with the correct answer now.
I believe this is what you're talking about: Gyazo - 6bde3478b337984dd9c270f12dc82a3b.png

If the height is 4, wouldn't the short leg (side opposite of 30deg) 4/sqrt3? Then *2 for the hypotenuse?