word problem

thetomps

Junior Member
Joined
Sep 26, 2005
Messages
59
Find the length, in inches, of the larger side of a right triangle with an area of 180 square inches if its hypotenuse is 41 inches long.
 
Draw the right triangle, with the short leg vertical, the long leg horizontal, and the hypotenuse slanty diagonal.

The short leg then is the height, and the long leg is the base.

The area "A" for a triangle with base "b" and height "h" is given by:

. . . . .A = (1/2)bh

You are given a value for A. Plug this in, and solve for one of the remaining variables.

For any right triangle, you have:

. . . . .a<sup>2</sup> + b<sup>2</sup> = c<sup>2</sup>

...or, in this case:

. . . . .h<sup>2</sup> + b<sup>2</sup> = 41<sup>2</sup>

You solved for one of the two variables in terms of the other. Substitute that into the above equation, and solve.

If you get stuck, please reply showing how far you have gotten. Thank you.

Eliz.
 
To check the answer to any "solving" problem, plug it back into the original exercise. So:

. . . . .a<sup>2</sup> + 12<sup>2</sup> = 41<sup>2</sup>
. . . . .a<sup>2</sup> + 144 = 1681
. . . . .a<sup>2</sup> = 1537
. . . . .a = 39.20459....

. . . . .A = (1/2)bh
. . . . .A = (1/2)(39.20459...)(12)
. . . . .A = 235.22754....

So "12" does not look like being a valid solution.

Please reply showing your steps, so we can help you find any errors. Thank you.

Eliz.
 
So you're saying now that you've gotten the legs as being a = 35.8 (approximately) and b = 20. Then:

. . . . .A = (1/2)(35.8)(20) = 358

This is also not equal to the given area of 180 in<sup>2</sup>.

Please reply showing your steps. Thank you.

Eliz.
 
Try following the step-by-step instructions, provided above. If you get stuck, please reply showing your work.

Thank you.

Eliz.
 
thetomps said:
a^2 + 35^2 = 41^2
a^2 + 1225 = 1681
a^2 = 456
I'm sorry, but I don't follow...?

Where does the original exercise say that one of the legs has length 35 inches, and that the area is 456 in<sup>2</sup>? I'm not seeing that...?

Eliz.
 
josh123 said:
then help them instead of making them go thru this!
Please review the thread. You will find step-by-step instructions provided in the first reply.

...or did you mean "do it for him"? Sorry, but it's another tutor that does that.

Eliz.
 
josh is right you aren't being too helpful, you should try other methods when trying to tutor.

oh and sacrasium and nastyness doesn't help when someone is struggling and just comes here for help, NOT ANSWERS OR CHEATING.
 
thetomps said:
josh is right you aren't being too helpful, you should try other methods when trying to tutor. oh and sacrasium and nastyness doesn't help when someone is struggling and just comes here for help, NOT ANSWERS OR CHEATING.
I'm sorry that you view step-by-step instructions as being "sarcasm" and "nastiness", but if you're not wanting the answer, I'm not sure what else you have in mind for "help".

Please reply with clarification. Thank you.

Eliz.
 
Leave Stapel alone! If you have a problem with her, you have a problem with me! She is one of the best, if not the best, tutor on any math site on the internet! Don't feel bad, Stapel, these two have a history of bothering the tutors.
 
HAPPY JOSH IS A NEW MEMBER AND I DON'T HAVE HISTORY OF CRAP! YOU ARE JUST A TERRIBLE TUTOR. THAT IS YOUR FING HISTORY!
 
Hey Stomps, settle down and LISTEN!

You were told the area of a right triangle is (the product of the 2 legs) / 2

Legs are a and b: so area = ab / 2; so ab/2 = 180; ab = 360; b = 360/a

Since a^2 + b^2 = c^2 (c is the hypotenuse);
then a^2 + (360/a)^2 = 41^2

Now solve for a.
If you can't do that, are we supposed to perform miracles for you?
 
Denis said:
...If you can't do that, are we supposed to perform miracles for you?
I think possibly some of the students have become spoiled, and are now expecting all the tutors to provide fully-worked solutions to everything they post.

Following instructions is, after all, harder than just copying the answers into one's homework, not matter how simple and straight-forward those instructions might have been.

Eliz.
 
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