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In Episode 3 of our GMAT Ninja Critical Reasoning series, we tackle Discrepancy, Paradox, and Explain an Oddity questions. You know the feeling: the passage gives you two facts that seem completely contradictory....- May 08
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23 Aug 2022, 22:52
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In the figure, triangle ABC has a right angle at B and has a perimeter of 390. Line segment BD is perpendicular to AC. BD has a length of 60. AB > BC. All the sides of all the three triangles are integers. The ratio of the area of triangle ABD to the area of triangle BDC is closest to which integer?
Screenshot 2022-08-24 112113.png [ 11.3 KiB | Viewed 3846 times ]
A. 2
B. 3
C. 4
D. 5
E. 6
Attachment:
Screenshot 2022-08-24 112113.png [ 11.3 KiB | Viewed 3846 times ]
A. 2
B. 3
C. 4
D. 5
E. 6
29 Aug 2022, 01:49
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Few Pythagorean Triplets are
3,4,5 with sum of sides = 12
5,12,13 with sum of sides = 30
8,15,17 with sum of sides = 40
7,24,25 with sum of sides = 56
9,40,41 with sum of sides = 90
In the question, the perimeter is given as 390 (= 13 x 30), therefore its a triplet of 5,12,13
hence sides of triangle ABC are 65,156,169
Now we are required to find ratio of area of triangle \((\frac{ABD}{BCD}) \)= \(\frac{AD}{CD}\) (as BD is height of both triangles)
We know that AD + CD = 169
from Pythagorean Theorem in Triangle BCD,
\(BC^2\) = \(BD^2\) + \(CD^2\)
\(65^2\) = \(60^2 + CD^2\) (Given: BC < AC)
solving CD = 25
therefore, AD = 169 - 25 = 144
therefore, \(\frac{AD}{CD} \) = \(\frac{144}{25}\) = 6 (approx)
Answer : E
3,4,5 with sum of sides = 12
5,12,13 with sum of sides = 30
8,15,17 with sum of sides = 40
7,24,25 with sum of sides = 56
9,40,41 with sum of sides = 90
In the question, the perimeter is given as 390 (= 13 x 30), therefore its a triplet of 5,12,13
hence sides of triangle ABC are 65,156,169
Now we are required to find ratio of area of triangle \((\frac{ABD}{BCD}) \)= \(\frac{AD}{CD}\) (as BD is height of both triangles)
We know that AD + CD = 169
from Pythagorean Theorem in Triangle BCD,
\(BC^2\) = \(BD^2\) + \(CD^2\)
\(65^2\) = \(60^2 + CD^2\) (Given: BC < AC)
solving CD = 25
therefore, AD = 169 - 25 = 144
therefore, \(\frac{AD}{CD} \) = \(\frac{144}{25}\) = 6 (approx)
Answer : E
General Discussion
28 Aug 2022, 01:28
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Area of triangle ABD: \(\frac{BD*AD}{2}\)
Area of triangle BDC: \(\frac{DC*BD}{2}\)
The ratio of the area of triangle ABD to the area of triangle BDC: \(\frac{\frac{BD*AD}{2}}{\frac{DC*BD}{2}}\)
\(=\frac{BD*AD}{2}*\frac{2}{DC*BD}\)
\(= \frac{AD}{DC}\)
So really all we need to find is the ratio of AD to DC.
AD & DC: As \( \triangle ABD\) and \( \triangle BDC\) are similar we can write: \(\frac{AD}{BD} = \frac{BD}{DC}\)
\(\frac{AD}{DC} = \frac{(BD)^2}{(DC)^2}\)
Plugging in BD = 60: \(\frac{AD}{DC} = \frac{(60)^2}{(DC)^2}\)
So we need to only solve for DC and plug it back into the above equation.
DC: Given the information we are told in the question stem, one can deduce that \( \triangle ABC\) is a Pythagorean triplet of (5, 12, 13) which is 13 times bigger. \(\frac{390}{(5+12+13)} = 13\). Which means that \(BC= (5*13) = 65\), \(AB = (12*13) = 156\) & \(AC = (13*13) = 169\)
As \( \triangle ABC\) and \( \triangle BDC\) are similar we can write:
\(\frac{BC}{AB} = \frac{DC}{BD}\)
\(\frac{65}{156} = \frac{DC}{60}\)
\(DC = 25\)
Plugging this back into \(\frac{AD}{DC} = \frac{(60)^2}{(DC)^2}\):
\(\frac{AD}{DC} = \frac{(60)^2}{(25)^2}\)
\(\frac{AD}{DC} = \frac{(60)^2}{(DC)^2}\)
\(\frac{AD}{DC}≈ 6\)
Area of triangle BDC: \(\frac{DC*BD}{2}\)
The ratio of the area of triangle ABD to the area of triangle BDC: \(\frac{\frac{BD*AD}{2}}{\frac{DC*BD}{2}}\)
\(=\frac{BD*AD}{2}*\frac{2}{DC*BD}\)
\(= \frac{AD}{DC}\)
So really all we need to find is the ratio of AD to DC.
AD & DC: As \( \triangle ABD\) and \( \triangle BDC\) are similar we can write: \(\frac{AD}{BD} = \frac{BD}{DC}\)
\(\frac{AD}{DC} = \frac{(BD)^2}{(DC)^2}\)
Plugging in BD = 60: \(\frac{AD}{DC} = \frac{(60)^2}{(DC)^2}\)
So we need to only solve for DC and plug it back into the above equation.
DC: Given the information we are told in the question stem, one can deduce that \( \triangle ABC\) is a Pythagorean triplet of (5, 12, 13) which is 13 times bigger. \(\frac{390}{(5+12+13)} = 13\). Which means that \(BC= (5*13) = 65\), \(AB = (12*13) = 156\) & \(AC = (13*13) = 169\)
As \( \triangle ABC\) and \( \triangle BDC\) are similar we can write:
\(\frac{BC}{AB} = \frac{DC}{BD}\)
\(\frac{65}{156} = \frac{DC}{60}\)
\(DC = 25\)
Plugging this back into \(\frac{AD}{DC} = \frac{(60)^2}{(DC)^2}\):
\(\frac{AD}{DC} = \frac{(60)^2}{(25)^2}\)
\(\frac{AD}{DC} = \frac{(60)^2}{(DC)^2}\)
\(\frac{AD}{DC}≈ 6\)
