geom help~ finding areas and altitudes given only specific

[cheetahbelly13]

1. i need help figuring out how to find the area for this problem:
(heres the question)

Find the area of an equilateral triangle with alt. (8square roots of 3)

I also need help with this one:


The legs of a right triangle are 12, 35. Find the altitude to the hypotenuse.

 

[galactus]

For equilateral triangles, let h=altitude and s=length of side.

\(\displaystyle \H\\h=\frac{\sqrt{3}}{2}s\)

Area=\(\displaystyle \L\\\frac{\sqrt{3}}{4}s^{2}\)

 

[cheetahbelly13]

what if your not given the length of the sides.
how do you find the area of this equilateral triangle when your only given the altitude?

would you use the 30-60-90 triangle to get it.

i was given the altitude which was 8sqroot3. and i found by using 30-60-90 triangle form that the area is 32sqroot3. is this right?

 

[tkhunny]

Re: geom help~ finding areas and altitudes given only specif

The legs of a right triangle are 12, 35. Find the altitude to the hypotenuse.

There are quite a few ways to do this one. What's the easiest?

This one is Mostly Geometry
Area = ½*12*35
Hypotenuse = sqrt(12^2 + 35^2)
Area = ½*Hypotenuse*(Altitude to Hypotenuse)

 

[galactus]

Plug in your value for h and solve for s, then use s in the area equation.

 

[soroban]

Re: geom help~ finding areas and altitudes given only specif

Hello, cheetahbelly13!

1. Find the area of an equilateral triangle with altitude \(\displaystyle \,8\sqrt{3}\)

Did you make a sketch?

Let \(\displaystyle x\) = side of the equilateral triangle.

              *
             /:\
            / : \
           /  :  \
        x /   :  _\
         /    :8√3 \
        /     :     \
       /      :      \
      *-------+-------*
         x/2

You already know the height of the triangle.

Now use Pythagorus to find \(\displaystyle x\), the base.

2. The legs of a right triangle are 12, 35.
Find the altitude to the hypotenuse.

You know, a sketch would really help . . .

            C
            *
           /: \
          / :   \
      12 /  :     \ 35
        /   :a      \
       /    :         \
      * - - + - - - - - *
      A     D           B
               37

In right triangle \(\displaystyle ABC:\;AC\,=\,12,\;BC\,=\,35\)
\(\displaystyle \;\;\)Using Pythagorus: \(\displaystyle AB\,=\,37\)

In \(\displaystyle \Delta ADC:\;\sin A\,=\,\frac{a}{12}\)

In \(\displaystyle \Delta ABC:\;\sin A\,=\,\frac{35}{37}\)

Hence: \(\displaystyle \L\,\frac{a}{12}\,=\,\frac{35}{37}\;\;\Rightarrow\;\;a\,=\,\frac{420}{37}\)