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18 Mar 2020, 02:46
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In the figure above, ABCD is a rectangle, and each of AP and CQ is perpendicular to BD. If DP = PQ = QB, what is the ratio of AB to AD?
(A) \(\sqrt{2}\) to 1
(B) \(\sqrt{3}\) to 1
(C) \(\sqrt{3}\) to \(\sqrt{2}\)
(D) 2 to 1
(E) 2 to \(\sqrt{3}\)
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18 Mar 2020, 05:01
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AB=a, and AD=b
DP = PQ = QB = k
\(3k = \sqrt{a^2+b^2}\)
Triangle APD is a right angle triangle. Hence,
\(AD^2 - AP^2 = k^2\) (pythagoras theorem)........(1)
Also, APB is a right angle triangle. Hence,
\(AB^2 - AP^2 = 4k^2\).......(2)
Subtract (1) from (2)
\(a^2 - b^2 = 3k^2\)
\(a^2 - b^2 = \frac{(a^2+b^2)}{3}\)
\(\frac{2a^2}{3} = \frac{4b^2}{3}\)
\(\frac{a}{b} = \sqrt{2}\)
DP = PQ = QB = k
\(3k = \sqrt{a^2+b^2}\)
Triangle APD is a right angle triangle. Hence,
\(AD^2 - AP^2 = k^2\) (pythagoras theorem)........(1)
Also, APB is a right angle triangle. Hence,
\(AB^2 - AP^2 = 4k^2\).......(2)
Subtract (1) from (2)
\(a^2 - b^2 = 3k^2\)
\(a^2 - b^2 = \frac{(a^2+b^2)}{3}\)
\(\frac{2a^2}{3} = \frac{4b^2}{3}\)
\(\frac{a}{b} = \sqrt{2}\)
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18 Mar 2020, 05:21
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A perpendicular drawn to the hypotenuse of a right triangle, divides the right triangle into two similar right triangles each of which is also similar to the original triangle
By this property, \(ABD\), \(ADP\) and \(ABP\) are similar triangles
\(\frac{AB}{AD}=\frac{AP}{DP}=\frac{BP}{AP}\)
\(AP^2=BP*DP=2DP^2\) (Since \(BP=2DP\))
\(AP=\sqrt{2}DP\)
So \(\frac{AB}{AD}=\frac{\sqrt{2}DP}{DP} = \frac{\sqrt{2}}{1}\)
Answer is (A)
By this property, \(ABD\), \(ADP\) and \(ABP\) are similar triangles
\(\frac{AB}{AD}=\frac{AP}{DP}=\frac{BP}{AP}\)
\(AP^2=BP*DP=2DP^2\) (Since \(BP=2DP\))
\(AP=\sqrt{2}DP\)
So \(\frac{AB}{AD}=\frac{\sqrt{2}DP}{DP} = \frac{\sqrt{2}}{1}\)
Answer is (A)
