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Area of right triangle when base splits are known

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Area of right triangle when base splits are known

by saxenashobhit » Thu Jul 07, 2011 5:21 am
https://www.scribd.com/doc/58999877/GMAT ... et-6-Quant

Question 9 - It has image of right triangle

In the diagram above, is a right angle, and is perpendicular to . If has a length of 25 and has a length of 4, what is the area of ?

125

145

240

290

It cannot be determined
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by Frankenstein » Thu Jul 07, 2011 6:38 am
Hi,
Let PQ = a, QR = b, QS = c,
QS^2 = a^2 - 25^2
QS^2 = b^2 - 4^2
So, a^2 - b^2 = 25^2 - 4^2 = 609
and a^2+b^2 = (25+4)^2 = 841
So, a^2 = 725 and b^2 = 116
Area of triangle is (1/2)a.b = (1/2)sqrt(725*116) = (1/2)sqrt(29*25*29*4) = 29*5 = 145

Hence, B
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by winniethepooh » Thu Jul 07, 2011 10:02 am
Hey Frnkenstine, I presume the triangle not to b a 30-60-90 or 45-45-90 triangle
so hyp square = 25 + 4 = 29 square
Also , the lengths of the legs should be in the ratio 3:4
So, 29/5 = 5.8 * 7 = 40.6
In the ratio of 3:4 = 17.4 and 23.2
So these should be the length and base of the triangle. Therefore, area = 1/2 (17.4 * 23.2)= comes 201.84= can you help?
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by Frankenstein » Thu Jul 07, 2011 10:10 am
winniethepooh wrote:Hey Frnkenstine, I presume the triangle not to b a 30-60-90 or 45-45-90 triangle
so hyp square = 25 + 4 = 29 square
Also , the lengths of the legs should be in the ratio 3:4
So, 29/5 = 5.8 * 7 = 40.6
In the ratio of 3:4 = 17.4 and 23.2
So these should be the length and base of the triangle. Therefore, area = 1/2 (17.4 * 23.2)= comes 201.84= can you help?
That's a big mistake. There can be infinitely many Pythagorean triplets.
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by winniethepooh » Thu Jul 07, 2011 10:16 am
but the sum of their squares does give 841!
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by Frankenstein » Thu Jul 07, 2011 10:25 am
winniethepooh wrote:but the sum of their squares does give 841!
Why wouldn't they give? You have calculated them by taking them in the ratio of 3:4:5
If you take them in the ratio of 1:1:sqrt(2), you will get another triplet.
In this ways you can get infinitely many.
For example: if the hypotenuse o a right triangle is 10, the other 2 sides can be any number of pairs,
they can be (6,8), (5sqrt2, 5sqrt2)...so on.
But check if you are getting the same value for QS^2 by calculating (PQ^2 - 25^2) and (QR^2 - 4^2) separately. You won't get.
Moreover, your values of 17.4, 23.2 cannot be PQ because in triangle PQS, PQ is hypotenus and one of the other sides is 25. How can the hypotenuse be smaller than either legs.
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by winniethepooh » Thu Jul 07, 2011 10:32 am
Righto!!
Thanks man.
How did you get 725 and 116?
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by Frankenstein » Thu Jul 07, 2011 10:39 am
winniethepooh wrote:Righto!!
Thanks man.
How did you get 725 and 116?
QS^2 = a^2 - 25^2
QS^2 = b^2 - 4^2
So, a^2 - b^2 = 25^2 - 4^2 = 609 ---equation (1)
and a^2+b^2 = (25+4)^2 = 841 --- equation (2)
So, a^2 = 725 and b^2 = 116
Solving equations (1) and (2)
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