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23 Nov 2021, 03:41
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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. \(6\)
23 Nov 2021, 03:41
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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
