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In the figure above, three squares and a triangle have areas
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Updated on: 23 Aug 2014, 06:23
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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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Originally posted by chica on 16 Jan 2008, 04:17.
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
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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
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16 Jan 2008, 04:31
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: In the figure above, three squares and a triangle have areas
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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
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16 Jan 2008, 06:22
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: In the figure above, three squares and a triangle have areas
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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
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16 Jan 2008, 10:24
A=144, a=12 B=81, b=9 C=225, c=15
Area of a triangle(X) = sqrt(S(Sa)(Sb)(Sc)) S= (a+b+c)/2 = (12+9+15)/2 = 18
X = sqrt(18(1812)(189)(1815)) = 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
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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
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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
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23 Feb 2018, 19:38
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