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14 Dec 2020, 11:01
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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In the figure above, AC = 10 and ABC is a semicircle. AD = p and DC = q What is the ratio of the areas of triangles ABD and ABC?
(1) p = 9q
(2) AB = \(3\sqrt{10}\)
14 Dec 2020, 11:36
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Answer (C)
Area of triangle is 1/2 * b * h.
Condition (1) gives definitive values for p,q. However, with BD being unknown, area of ABD can't be determined.
Condition (2) gives AB values, and when integrated with 'p' from condition (1), BD can be obtained. And hence areas of both triangles can be definitely computed.
Posted from my mobile device
Area of triangle is 1/2 * b * h.
Condition (1) gives definitive values for p,q. However, with BD being unknown, area of ABD can't be determined.
Condition (2) gives AB values, and when integrated with 'p' from condition (1), BD can be obtained. And hence areas of both triangles can be definitely computed.
Posted from my mobile device
14 Dec 2020, 12:02
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Hi the27s,
I think the answer should be A.
The question asks Area of ADB/Area of ABC.
Area of ADB = 1/2 * BD * p
Area of ABC can be split as Area of ADB + Area of DBC
Which is 1/2 * BD * p + 1/2 * BD * q
Simplifying this becomes 1/2 * BD (p +q)
Now the ratio becomes 1/2 * BD * p/(1/2*BD(p+q))
Cancelling 1/2 * BD in the numerator and denominator the ratio becomes
p/(p+q)
Now statement 1 gives p = 9q
The ratio of areas of ADB/ABC = 9q/(9q+q) = 9/10
Statement 2 : insufficient.
Answer: A
Posted from my mobile device
I think the answer should be A.
The question asks Area of ADB/Area of ABC.
Area of ADB = 1/2 * BD * p
Area of ABC can be split as Area of ADB + Area of DBC
Which is 1/2 * BD * p + 1/2 * BD * q
Simplifying this becomes 1/2 * BD (p +q)
Now the ratio becomes 1/2 * BD * p/(1/2*BD(p+q))
Cancelling 1/2 * BD in the numerator and denominator the ratio becomes
p/(p+q)
Now statement 1 gives p = 9q
The ratio of areas of ADB/ABC = 9q/(9q+q) = 9/10
Statement 2 : insufficient.
Answer: A
Posted from my mobile device
