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In the figure above, three squares and a triangle have areas

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In the figure above, three squares and a triangle have areas [#permalink] New post 16 Jan 2008, 04:17
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Question Stats:

81% (02:32) correct 19% (01:21) wrong based on 41 sessions
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In the figure above, three squares and a triangle have areas of A, B, C, and X as shown. If A = 144, B = 81, and C = 225, then X =

(A) 150
(B) 144
(C) 80
(D) 54
(E) 36

[Reveal] Spoiler:
I solved the problem via quadratic equation:
12^2 - (15 -x)^2 = 9^2 - x^2
Is there a better way to solve the problem? Or the trick here is to quickly solve the quadratic equation?

Thank you
[Reveal] Spoiler: OA

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Last edited by Bunuel on 23 Aug 2014, 06:23, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: In the figure above, three squares and a triangle have areas [#permalink] New post 16 Jan 2008, 04:25
chica wrote:
Attachment:
Triangle.doc

In the figure above, three squares and a triangle have areas of A, B, C, and X as shown. If A = 144, B=81, and C=225, then X =
(A) 150
(B) 144
(C) 80
(D) 54
(E) 36

I solved the problem via quadratic equation:
12^2 - (15 -x)^2 = 9^2 - x^2
Is there a better way to solve the problem? Or the trick here is to quickly solve the quadratic equation?

Thank you

I am not sure what is being asked but I think we should find the area of triangle, since triangle has sides 9,12,15 it is right triangle and area is height multiplied by half base * 1/2
S=12(height, because side=15 is hypotenuse)*9(base)=54
D
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Re: In the figure above, three squares and a triangle have areas [#permalink] New post 16 Jan 2008, 04:31
Expert's post
D

A = 144=12², B=81=9², and C=225=15²
A²+B²=C² ==> ABC is a right triangle.
S=12*9/2=54

Your figure looks like these ones: http://en.wikipedia.org/wiki/Pythagorean_theorem
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Re: In the figure above, three squares and a triangle have areas [#permalink] New post 16 Jan 2008, 05:49
From the figure each side of the triangle is a side of three different squares.
Now given the area of square you get the side.
Formula Side * Side =Area of square.
So 15, 12, 9 are three ides of triangle.

Notice that the numbers fit pythogram theorem.
a^2+b^2 =c^2 . So its is right angle triangle. Area of =b*h/2

Base =12, Height =9. Largest side is hypotenuse

Area =54

BTW: How did you get quadratic equation from the figure?
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Re: In the figure above, three squares and a triangle have areas [#permalink] New post 16 Jan 2008, 06:22
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Expert's post
I just try a new feature with mimeTeX......

\picture(400) {
(80,150) {\line(150,0)}
(230,150) {\line(-20,50)}
(210,200) {\line(-130,-50)}
(80,150) {\line(-55,130)}
(25,280) {\line(130,55)}
(155,335) {\line(55,-135)}
(210,200) {\line(50,20)}
(260,220) {\line(20,-50)}
(280,170) {\line(-50,-20)}
(230,150) {\line(0,-150)}
(230,0) {\line(-150,0)}
(80,0) {\line(0,150)}
}


Code:
\picture(400) {
(80,150) {\line(150,0)}
(230,150) {\line(-20,50)}
(210,200) {\line(-130,-50)}
(80,150) {\line(-55,130)}
(25,280) {\line(130,55)}
(155,335) {\line(55,-135)}
(210,200) {\line(50,20)}
(260,220) {\line(20,-50)}
(280,170) {\line(-50,-20)}
(230,150) {\line(0,-150)}
(230,0) {\line(-150,0)}
(80,0) {\line(0,150)}
}

:help2
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Re: In the figure above, three squares and a triangle have areas [#permalink] New post 16 Jan 2008, 06:28
Wow, walker, you are taking mimeTeX farther than I'd ever be able to :) Thanks for popularizing this tool, +1.
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Re: In the figure above, three squares and a triangle have areas [#permalink] New post 16 Jan 2008, 10:24
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A=144, a=12
B=81, b=9
C=225, c=15

Area of a triangle(X) = sqrt(S(S-a)(S-b)(S-c))
S= (a+b+c)/2 = (12+9+15)/2 = 18

X = sqrt(18(18-12)(18-9)(18-15))
= sqrt( 18 * 6 * 9 * 3)
= sqrt( (2*9) * (2*3) * 9 * 3)
= 54
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Re: In the figure above, three squares and a triangle have areas [#permalink] New post 16 Jan 2008, 11:14
chica wrote:
Attachment:
Triangle.doc

In the figure above, three squares and a triangle have areas of A, B, C, and X as shown. If A = 144, B=81, and C=225, then X =
(A) 150
(B) 144
(C) 80
(D) 54
(E) 36

I solved the problem via quadratic equation:
12^2 - (15 -x)^2 = 9^2 - x^2
Is there a better way to solve the problem? Or the trick here is to quickly solve the quadratic equation?

Thank you


Im guessing your looking for the area of the triangle.

12*9/2 = 54 D.
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Re: In the figure above, three squares and a triangle have areas [#permalink] New post 21 Jan 2008, 07:05
Travel09 wrote:
From the figure each side of the triangle is a side of three different squares.
Now given the area of square you get the side.
Formula Side * Side =Area of square.
So 15, 12, 9 are three ides of triangle.

Notice that the numbers fit pythogram theorem.
a^2+b^2 =c^2 . So its is right angle triangle. Area of =b*h/2

Base =12, Height =9. Largest side is hypotenuse

Area =54

BTW: How did you get quadratic equation from the figure?


I got trapped by the picture even though it was not drawn to the scale.. and did not notice that the triangle - was actually right triangle :( . So, I draw another height.. and solved the problem that way. The equation aimed to find the new height. This is how I got the quadratic equation.
It worked, unfortunately, not for GMAT when you are pressed on time..

Thanks for helping me realize my careless on this one :)
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Re: In the figure above, three squares and a triangle have areas [#permalink] New post 23 Aug 2014, 05:13
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Re: In the figure above, three squares and a triangle have areas   [#permalink] 23 Aug 2014, 05:13
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