geometry : acute triangle

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geometry : acute triangle

by khilolik » Mon Aug 30, 2010 1:38 pm
Guys, please help me,


I am brushing my math skills these days with Kaplan Gmat math book - and have a trouble to find the way to solve the following.

We have a n acute triangle with sides 1 and 3, need to find hypotenuse (third side). I applied Pythagorean theorem- it does not match my answer. I do not why and the book did not provided explanation to this question. only the answer 8 square root. Please heeeeeeeeeeeeeeeeeeeeeelpppppppp :([/spoiler][/list]

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by Gurpinder » Mon Aug 30, 2010 1:58 pm
khilolik wrote:Guys, please help me,


I am brushing my math skills these days with Kaplan Gmat math book - and have a trouble to find the way to solve the following.

We have a n acute triangle with sides 1 and 3, need to find hypotenuse (third side). I applied Pythagorean theorem- it does not match my answer. I do not why and the book did not provided explanation to this question. only the answer 8 square root. Please heeeeeeeeeeeeeeeeeeeeeelpppppppp :([/spoiler][/list]
Can you post the exact question please.
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by khilolik » Mon Aug 30, 2010 2:09 pm
please open the attachment Thanks so much!!!!!!![/img]
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by Gurpinder » Mon Aug 30, 2010 2:59 pm
You cannot use Pythagorean theorem directly - this is an acute triangle.

Do it like this (attached image).

Image

Now calculate the missing side for triangle A first by using Pythagorean theorem.


.5^2+b^2=3^2
b^2=8.75

Now solve for triangle B.

.5^2+8.75=c^2
.25+8.75=c^2
9=c^2
c=3.

Therefore the missing side y = 3.
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by Stuart@KaplanGMAT » Mon Aug 30, 2010 3:14 pm
Gurpinder wrote:You cannot use Pythagorean theorem directly - this is an acute triangle.

Do it like this (attached image).

Image

Now calculate the missing side for triangle A first by using Pythagorean theorem.


.5^2+b^2=3^2
b^2=8.75

Now solve for triangle B.

.5^2+8.75=c^2
.25+8.75=c^2
9=c^2
c=3.

Therefore the missing side y = 3.
Hi,

the only reason you got y=3 is because you actually assumed that y=3 earlier in your math.

By having the height bisect the "1" side, you're assuming that it's an isosceles triangle; of course when you assume it's an isosceles triangle you'll get y=3.

Unfortunately I don't have a copy of the workbook handy - I'd need to see the exact question to give you the proper solution.
Image

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by Gurpinder » Mon Aug 30, 2010 3:21 pm
Oh,

hmmm.....Thanks.

Can you please post your explanation when you get a chance?

You can't use the sine law right?

Thanks,
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by khilolik » Tue Aug 31, 2010 3:14 am
thanks guys for helping, I think i have to explain the question in details. The answer is 8 with radical sign. The picture has acute triangle with sides - x=1, z=3, y=?? -where y is hypotenuse. Sorry can't provide the question from the book- i need to scan it. but it has exactly the same image of a triangle (see the link) I right side is equal to 1, base is equal to 3 hypotenuse y ?



https://www.regentsprep.org/regents/math ... /stri2.gif

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by khilolik » Tue Aug 31, 2010 3:21 am
1^2 + x^2 = 3^2
1 + x^2 = 9
x^2 = 8
x = root8 - this is what I received from Stuarsky


looks like you need to consider y as one of the legs, not hypotenuse. Thanks guys for trying to help me!