Another parallelogram problem

The Preacher

Junior Member
Well, with the much appreciated help from the posters in my other "Find the area" thread, and soroban, I was able to solve that problem, and one other.

Now I've got this one. If AD wasn't a radical number thingy, I might be able to do it, but radicals confuse me. Here's what I've got so far:

AXD is a 30-60-90 right triangle, and angle XAD = 30 degrees (thanks, soroban).

The side opposite the 30 degree angle is half the hypotenuse. Now, I'm having trouble with what exactly half of $$\displaystyle $4\sqrt 3$$$ is.

After I find what $$\displaystyle $4\sqrt 3$$$ is, how do I put it into the Pythagorean equation? I want to use the Pythagorean Theorem to find AX, and that will be my height, but I'm used to using regular numbers in those, and these radical numbers confuse me.

Thanks for helping me out, I hope I can learn this stuff so I don't have to ask so much. Let me know if I start to wear out my welcome with requests.

-The Preacher

Denis

Senior Member
Preacher Kid said: Here's what I've got so far:
AXD is a 30-60-90 right triangle, and angle XAD = 30 degrees (thanks, soroban).
The side opposite the 30 degree angle is half the hypotenuse. Now, I'm having trouble with what exactly half of $$\displaystyle $4\sqrt 3$$$ is.
After I find what $$\displaystyle $4\sqrt 3$$$ is, how do I put it into the Pythagorean equation? I want to use the Pythagorean Theorem to find AX, and that will be my height, but I'm used to using regular numbers in those, and these radical numbers confuse me.
Half of 4sqrt(3) = 2sqrt(3); same as 4x / 2 = 2x: YES, it's that easy :wink:

So height XA = sqrt[(4sqrt(3))^2 - (2sqrt(3))^2] : compliments of Big Pete Pythagoras

Remember that (as example) (4sqrt(3))^2 = 16 * 3 = 48 : rule: (asqrt(b))^2 = a^2 * b

Complete the XA calculation: you'll get a nice integer !

I'm sure you can finish it, right?

The Preacher

Junior Member
Denis said:
Half of 4sqrt(3) = 2sqrt(3); same as 4x / 2 = 2x: YES, it's that easy :wink: :shock: You mean I don't have to change the number behind the radical!? Now that I think about it, it makes sense. Multiplying the square root of three by two would be half of what it would be if I multiplied it by four. Sweet.

Now... is it supposed to look like this?:

Height XA =$$\displaystyle \ \sqrt {4\sqrt 3 ^2 - 2\sqrt {3^2 } } \$$

That would mean that height XA = $$\displaystyle \ \sqrt {2\sqrt 3 ^2 } \$$, right?

Now... please forgive me, but I don't know where to go from here. Math confuses me so much. I'm sorry.

God bless y'all,
-The Preacher

Gene

Senior Member
Squaring a square root gives whatever is under the radical whether it is (sqrt(x))^2 or
sqrt(x²) the result is x.
Your example gives sqrt(6).

The Preacher

Junior Member
Gene said:
Squaring a square root gives whatever is under the radical whether it is (sqrt(x))^2 or
sqrt(x²) the result is x.
Your example gives sqrt(6).
Ah, thanks for that, Gene. Whenever I look at these problems, I tend to see the symbols, and that's it, and that is how I get confused. I don't think in numbers, so most times when I see these symbols, it doesn't register too well. Thanks to that last bit of help, I was able to solve the problem.

I'll try to keep what you said in mind. Thanks again.

God bless you all,
-The Preacher