# Geometry - word problem

• Jan 13th 2006, 08:40 AM
The Preacher
Geometry - word problem
Hey, thanks for taking a look at this.

A parallelogram with sides of 6 and 10 has an area of 30(sqrt)2. Find the measure of each angle of the parallelogram.

Small angle _____

Large angle _____

I apologize that my first posting here had to be urgent, my school day is over soon, and I'd like some work to show for it, but I'm stuck. I appreciate any assistance.

So far I've got is this:

360 - x2 = y2

X being either the small or large angle and y taking the place of the opposite.

That's all I have so far, I'd like a hint in the right direction to finding one of the angles. The normal forum I post on for this sort of stuff is not working right now. Thanks again.

God bless y'all,
-The Preacher
• Jan 13th 2006, 09:03 AM
TD!
If we take the largest side (10) to be the base and we call the heigth h, then the area is given by 10h. But we know the area, so we can find h.

By drawing the parallellogram and this height, you can form a right triangle (at the left or right side of the parallellogram).
This right triangle has one side h (which we know now) and hypotenuse 6. With the pythagorean theorem, you can even find the last side.

Now you have this triangle and you know all sides, use some trigonometry to fnd the angles. For example, check this topic.
• Jan 13th 2006, 09:11 AM
The Preacher
Quote:

Originally Posted by TD!
If we take the largest side (10) to be the base and we call the heigth h, then the area is given by 10h. But we know the area, so we can find h.

By drawing the parallellogram and this height, you can form a right triangle (at the left or right side of the parallellogram).
This right triangle has one side h (which we know now) and hypotenuse 6. With the pythagorean theorem, you can even find the last side.

I think that I wouldn't have any problem with this if I knew how to get the height from the area. Radicals confuse me. Do you think you could help me out?
• Jan 13th 2006, 09:14 AM
TD!
Of course. For a parallellogram, the area is given by height x base.
By choice, we take the long side as base (which is 10 here) and call the height h, we then have:

$\displaystyle 10h = 30\sqrt 2 \Leftrightarrow h = \frac{{30\sqrt 2 }} {{10}} = 3\sqrt 2$
• Jan 13th 2006, 09:21 AM
The Preacher
Quote:

Originally Posted by TD!
Of course. For a parallellogram, the area is given by height x base.
By choice, we take the long side as base (which is 10 here) and call the height h, we then have:

$\displaystyle 10h = 30\sqrt 2 \Leftrightarrow h = \frac{{30\sqrt 2 }} {{10}} = 3\sqrt 2$

Okay... so $\displaystyle 3\sqrt 2$ is the height. Uh... like I said, radicals confuse me. Is there a way to change that into an integer? I'm sorry to require so much assistance in order to understand. I appreciate your time, TD!. =]

God bless y'all,
-The Preacher
• Jan 13th 2006, 09:24 AM
TD!
Quote:

Originally Posted by The Preacher
Okay... so $\displaystyle 3\sqrt 2$ is the height. Uh... like I said, radicals confuse me. Is there a way to change that into an integer? I'm sorry to require so much assistance in order to understand. I appreciate your time, TD!. =]

Well, the square root of two is irrational. That means that if you'd want to write it in decimals, it will have infinitely many decimals, just like pi. You can of course just truncate the expansion, then you're left with an approximation. The square root of two is more or less equal to 1.414.
I usually just leave the radicals and work with them, since that's exact. If you're uncomfortable with that, you could use an approximation.
• Jan 13th 2006, 09:26 AM
The Preacher
Yeah, I found that out before (the square root of two having too many decimals), that's why I was asking. Dang, I'm having so much trouble with this problem.

Wait... I need to find the sine ratio. I could take the $\displaystyle 3\sqrt 2$ and the hypotenuse (6) and find the opposite angle.

Soo... that'd be $\displaystyle $\frac{{3\sqrt 2 }}{6}$$.

Hold on... I am bad at math, but I'm going to try to figure out the angle from this. I can do that, right?
• Jan 13th 2006, 09:29 AM
TD!
Check the topic I linked in my first post, there are the 3 formula's you could possibly use.
• Jan 13th 2006, 09:31 AM
The Preacher
Alright, thanks. Sorry for ignoring it. :o

EDIT:

Okay... I looked at it. So I need to divide $\displaystyle 3\sqrt 2$ by 6. How do I divide radicals? :confused:

Sorry to seem so helpless... I really do stink at math.
• Jan 13th 2006, 09:33 AM
TD!
No problem. Try it and if it doesn't work out, ask for help :)
• Jan 13th 2006, 09:34 AM
The Preacher
Lol, I edited my message. Sorry for the bump.
• Jan 13th 2006, 09:34 AM
ThePerfectHacker
There is a theorem from trigonometry which states the area of a parallelogram with sides $\displaystyle a,b$ and angle $\displaystyle \theta$ is $\displaystyle A=ab\sin \theta$.
Now you said its sides are 6 and 10. And area of $\displaystyle 30\sqrt{2}$ Thus, by the formula $\displaystyle 60\sin \theta=30\sqrt{2}$ divide by 60, thus,
$\displaystyle \sin \theta=\frac{\sqrt{2}}{2}$. Now that happens when $\displaystyle \theta=45,135$.
Q.E.D.
• Jan 13th 2006, 09:40 AM
The Preacher
:eek:

I think I just got really confused. Couldn't I just have found the angle by dividing $\displaystyle 3\sqrt 2$ by 6? Maybe not. Maybe I've got all my sine stuff out of whack. Sorry.

=]

-The Preacher

EDIT:

Nevermind, I just went back some in my textbook, reviewed, and was able to find out how to solve this with TD!'s help. Thanks for helping me out, guys. I appreciate it.

God bless y'all,
-The Preacher
• Jan 13th 2006, 10:01 AM
TD!
Quote:

Originally Posted by The Preacher
:eek:

I think I just got really confused. Couldn't I just have found the angle by dividing $\displaystyle 3\sqrt 2$ by 6? Maybe not. Maybe I've got all my sine stuff out of whack. Sorry.

Yes, that would have given you the sine of that angle.

$\displaystyle \sin \left( \alpha \right) = \frac{{3\sqrt 2 }} {6} \Leftrightarrow \sin \left( \alpha \right) = \frac{{\sqrt 2 }} {2} \Leftrightarrow \alpha = 45^\circ \vee \alpha = 135^\circ$
• Jan 13th 2006, 12:40 PM
ticbol
Quote:

Originally Posted by The Preacher
Hey, thanks for taking a look at this.

A parallelogram with sides of 6 and 10 has an area of 30(sqrt)2. Find the measure of each angle of the parallelogram.

Small angle _____

Large angle _____

I apologize that my first posting here had to be urgent, my school day is over soon, and I'd like some work to show for it, but I'm stuck. I appreciate any assistance.

So far I've got is this:

360 - x2 = y2

X being either the small or large angle and y taking the place of the opposite.

That's all I have so far, I'd like a hint in the right direction to finding one of the angles. The normal forum I post on for this sort of stuff is not working right now. Thanks again.

God bless y'all,
-The Preacher

I do not belong to the group that gives "answers" or help in forms of hints/guides/hanging solutions/the likes.
I belong to the group that gives complete/not partial/detailed answers.

Let us go first to your 360 -x2 = y2.
Your intention is correct, but you should know how to present it to us so that there'd be less confusion.
x2 or y2 should have been x*2 or y*2. Or, the usual 2x or 2y.
"x2" is mostly read as x, sub 2.

Let x = small angle, and y = large angle.

If you don't have the figure of the said parallelogram, draw it on paper. [Always work, or play, with a figure. Make that a habit.]
Let us say the parallelogram is ABCD, where:
--- angle A = angle C = angle x
--- angle B = angle D = angle y
--- sides BC and AD are horizontal

Area of parallelogram = base times altitude, this altitude being perpendicular to the base.
There is no perpendicular lines shown on the figure yet, so draw a line from B that is perpendicular to AD, and call is new line or altitude, h.
Then,
(AB)*h = 30sqrt(2)
10*h = 30sqrt(2)
Divide both sides by 1o,
h = 3sqrt(2) ----------------***

In the right triangle formed by AB, h and portion of AD,
sin(angle A) = h /6
sin(x) = [3sqrt(2)] /6
sin(x) = sqrt(2) /2

Either get the decimal equivalent of that and then find the x by using a calculator, or, if you know from memory that sin(45 degrees) is [sqrt(2)]/2 or 1/[sqrt(2)], then
x = arcsin[sqrt(2) /2] = 45deg --------answer.

Then, using your
360 -2x = 2y,
360 -2*45 = 2y
270 = 2y