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Triangle.doc In the figure above, three squares and a [#permalink]
 16 Jan 2008, 05:17
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 Triangle.doc [25 KiB]
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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 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
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Re: PS: three squares and a triangle [#permalink]
16 Jan 2008, 05:25
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: PS: three squares and a triangle [#permalink]
16 Jan 2008, 05:31
DA = 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: PS: three squares and a triangle [#permalink]
16 Jan 2008, 06:49
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: PS: three squares and a triangle [#permalink]
16 Jan 2008, 07:22
1
This post received
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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)}
} 
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Re: PS: three squares and a triangle [#permalink]
16 Jan 2008, 07:28
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: PS: three squares and a triangle [#permalink]
16 Jan 2008, 11:24
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: PS: three squares and a triangle [#permalink]
16 Jan 2008, 12:14
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: PS: three squares and a triangle [#permalink]
21 Jan 2008, 08:05
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: PS: three squares and a triangle
[#permalink]
21 Jan 2008, 08:05
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