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17 Aug 2021, 08:00
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Difficulty:
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Question Stats:
39% (01:56) correct
61%
(02:22)
wrong
based on 209
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In the image above, if \(AE = \sqrt{17}\), \(BE = \sqrt{5}\), and \(CE = \sqrt{13}\) then what is the area of square ABCD?
A. 2
B. 3
C. 4
D. 5
E. 6
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Attachment:
Untitled3.png
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17 Aug 2021, 09:13
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Asked: In the image above, if \(AE = \sqrt{17}\), \(BE = \sqrt{5}\), and \(CE = \sqrt{13}\) then what is the area of square ABCD?
Attachment:
Screenshot 2021-08-17 at 9.02.16 PM.png [ 40.65 KiB | Viewed 4749 times ]
z^2 + (x+y)^2 = 17
z^2 + y^2 = 5
(z+x)^2 + y^2 = 13
(x+2y)x = x^2 + 2xy = 17-5 = 12
x(x+2z) = x^2 + 2xz= 13-5 = 8
2x(y-z) = 12-8 = 4
Area of the square ABCD = x^2
If x = 2; y = 2; z = 1; All 3 equations are satisfied.
z^2 + (x+y)^2 = 1^2 + (2+2)^2 = 1 + 16 = 17
z^2 + y^2 = 1^2 + 2^2 = 1+ 4 = 5
(z+x)^2 + y^2 = (1+2)^2 + 2^2 = 9 + 4 = 13
IMO C
23 Nov 2021, 03:42
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Official Solution:
In the image above, if \(AE = \sqrt{17}\), \(BE = \sqrt{5}\), and \(CE = \sqrt{13}\) then what is the area of square ABCD?
A. \(2\)
B. \(3\)
C. \(4\)
D. \(5\)
E. \(6\)
Consider the image below:
Apply Pythagorean theorem to the three triangles shown in the image:
\(a^2+b^2=(\sqrt{5})^2\);
\((x+a)^2+b^2=(\sqrt{13})^2\). Expand: \(x^2+2ax+a^2+b^2=13\). Substitute \(a^2+b^2=5\): \(x^2+2ax+5=13\). So, \(a=\frac{8-x^2}{2x}\);
\((x+b)^2+a^2=(\sqrt{17})^2\). Expand: \(x^2+2bx+b^2+a^2=17\). Substitute \(a^2+b^2=5\): \(x^2+2bx+5=17\). So, \(b=\frac{12-x^2}{2x}\).
Substitute \(a=\frac{8-x^2}{2x}\) and \(b=\frac{12-x^2}{2x}\) into \(a^2+b^2=5\)
\((\frac{8-x^2}{2x})^2+ (\frac{12-x^2}{2x})^2=5\);
\(\frac{(8-x^2)^2}{4x^2}+ \frac{(12-x^2)^2}{4x^2}=5\)
At this point it's better to plug-in the options. We are asked to fins the area of the square, which would be \(x^2\), so plug-in options for \(x^2\) and check which gives the right answer.
Option C (4) works: \(\frac{(8-4)^2}{4*4}+ \frac{(12-4)^2}{4*4}=5\)
Answer: C
In the image above, if \(AE = \sqrt{17}\), \(BE = \sqrt{5}\), and \(CE = \sqrt{13}\) then what is the area of square ABCD?
A. \(2\)
B. \(3\)
C. \(4\)
D. \(5\)
E. \(6\)
Consider the image below:
Apply Pythagorean theorem to the three triangles shown in the image:
\(a^2+b^2=(\sqrt{5})^2\);
\((x+a)^2+b^2=(\sqrt{13})^2\). Expand: \(x^2+2ax+a^2+b^2=13\). Substitute \(a^2+b^2=5\): \(x^2+2ax+5=13\). So, \(a=\frac{8-x^2}{2x}\);
\((x+b)^2+a^2=(\sqrt{17})^2\). Expand: \(x^2+2bx+b^2+a^2=17\). Substitute \(a^2+b^2=5\): \(x^2+2bx+5=17\). So, \(b=\frac{12-x^2}{2x}\).
Substitute \(a=\frac{8-x^2}{2x}\) and \(b=\frac{12-x^2}{2x}\) into \(a^2+b^2=5\)
\((\frac{8-x^2}{2x})^2+ (\frac{12-x^2}{2x})^2=5\);
\(\frac{(8-x^2)^2}{4x^2}+ \frac{(12-x^2)^2}{4x^2}=5\)
At this point it's better to plug-in the options. We are asked to fins the area of the square, which would be \(x^2\), so plug-in options for \(x^2\) and check which gives the right answer.
Option C (4) works: \(\frac{(8-4)^2}{4*4}+ \frac{(12-4)^2}{4*4}=5\)
Answer: C
