Find the area

[The Preacher]

Hey, it's good to know there's places like this around for help if I need it.

Here's my problem:

Find the area of parallelogram ABCD given angle A = 60 degrees, and given the following measures.

Thanks to ImageShack for Free Image Hosting

AB = 10 in.; AD = 6 in. A = (blank) times the square root of (blank) sq. in.

I'm supposed to fill in the blanks, but I'm not too great with those square-root-typey problems. If you could show me how to do this, I might be able to understand it and figure it out myself next time.

Thanks a lot,

-The Preacher

 

[galactus]

The area of a parallelogram is base * height. To find the height (h), use AD times the sine of your angle.

 

[The Preacher]

The area of a parallelogram is base * height.

I know how to find the area of a parallelogram, they want me to fill in those blanks on (blank) times the square root of (blank).

I have no idea how I'm supposed to find what square root to use, and what number to multiply it by. The info I provided is all the stuff they gave me.

To find the height (h), use AD times the sine of your angle.

I forgot what sine is, and the sad part is that was probably only one unit ago. Man, I stink at geometry.

Thanks, galactus.

 

[galactus]

When you use the sine you will wind up with an answer in terms of what you want.

\(\displaystyle sin(60)=\frac{sqrt{3}}{2}\). Now multipy by base and height.

\(\displaystyle (10)(6)sin(60)=30\sqrt{3}\)

A="30 times the square root of 3"

 

[The Preacher]

When you use the sine you will wind up with an answer in terms of what you want.

\(\displaystyle sin(60)=\frac{sqrt{3}}{2}\). Now multipy by base and height.

I feel like I'm testing your patience. Sorry.

Thanks for helping me out.

-The Preacher

 

[happy]

When you use the sine you will wind up with an answer in terms of what you want.

\(\displaystyle sin(60)=\frac{sqrt{3}}{2}\). Now multipy by base and height.

I feel like I'm testing your patience. Sorry.

Thanks for helping me out.

-The Preacher

Don't give up so fast. Do you really think you can learn this in one day? Do as many problems as you can. :wink:

 

[The Preacher]

When you use the sine you will wind up with an answer in terms of what you want.

\(\displaystyle sin(60)=\frac{sqrt{3}}{2}\). Now multipy by base and height.

I feel like I'm testing your patience. Sorry.

Thanks for helping me out.

-The Preacher

Don't give up so easily.

Nope, not giving up. =] He gave me what I wanted, and I wanted to apologize for testing his patience. Lol. The sad thing is I still don't know what to do with that stuff. I may have to start geometry over, just so I can remember all the rules and stuff. That'd really stink, though.

God bless y'all.

 

[galactus]

You're not testing my patience at all. I think you were just misunderstanding the problem. They wanted the answer in terms of "something times the square root of something". They didn't necessarily mean use square roots to solve.

 

[The Preacher]

You're not testing my patience at all. I think you were just misunderstanding the problem. They wanted the answer in terms of "something times the square root of something". They didn't necessarily mean use square roots to solve.

Well, that's the problem. Whenever it wants something times the square root of something, I start to get frazzled. It's confusing to me, I definitely don't think mathematically.

For example, I when I look at \(\displaystyle \frac{sqrt{3}}{2}\), it just looks like a funny line, then a three over a two. I understand it means "square root of three over two", but I don't really know what square root of three over two is. I'm not sure if I'm doing a good job of communicating what's confusing me, but oh well.

Again, thanks for all your help.

 

[galactus]

If you need something in decimal form it would be .866025403784....

Do you know what a square root is?. It's a number, which when squared (multiplied times itself) equals 3.....in this case. It complements the square.

\(\displaystyle (sqrt{3})^{2}=(3^{\frac{1}{2}})^{2}=3\)

The square root is the second root, cube root is the third root, and so forth.

If it makes it easier, think of the square root as something to the one-half power.

Or is it just the radical sign\(\displaystyle \sqrt{}\) that spooks you?.

 

[The Preacher]

If you need something in decimal form it would be .866025403784....

Do you know what a square root is?. It's a number, which when squared (multiplied times itself) equals 3.....in this case. It complements the square.

\(\displaystyle (sqrt{3})^{2}=(3^{\frac{1}{2}})^{2}=3\)

The square root is the second root, cube root is the third root, and so forth.

If it makes it easier, think of the square root as something to the one-half power.

Or is it just the radical sign\(\displaystyle \sqrt{}\) that spooks you?.

Haha, thanks, I know what a square root is. It's seeing a number with a radical sign positioned over another number that spooks me. I usually don't know what to do with it.

God bless y'all.

Jovially (I located the explanation of sine in my "textbook"),
-The Preacher

 

[The Preacher]

Well, with the much appreciated help from the posters in this 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'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.

-The Preacher

 

[happy]

Hey Preach,

In order to help you better, you may find it beneficial to post new problems in new threads. :wink:

 

[The Preacher]

Hey Preach,

In order to help you better, you may find it beneficial to post new problems in new threads. :wink:

Ah, thanks happy. I'll get right on that.