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In the figure above, AC = 10 and ABC is a semicircle. AD = p and DC =

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In the figure above, AC = 10 and ABC is a semicircle. AD = p and DC = [#permalink] New post   14 Dec 2020, 11:01
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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 = \(3\sqrt{10}\)
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D
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Re: In the figure above, AC = 10 and ABC is a semicircle. AD = p and DC = [#permalink] New post   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

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Re: In the figure above, AC = 10 and ABC is a semicircle. AD = p and DC = [#permalink] New post   14 Dec 2020, 11:36
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.

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Re: In the figure above, AC = 10 and ABC is a semicircle. AD = p and DC = [#permalink] New post   16 Feb 2021, 21:17
Tough problem. Does anyone have the answer? It says D.
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Re: In the figure above, AC = 10 and ABC is a semicircle. AD = p and DC = [#permalink] New post   16 Feb 2021, 21:19
mcmoorthy wrote:
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

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Nice. Why isn't B sufficient?
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Re: In the figure above, AC = 10 and ABC is a semicircle. AD = p and DC = [#permalink] New post   16 Feb 2021, 21:43
CEdward wrote:
Tough problem. Does anyone have the answer? It says D.

Answer is indeed D
B is also sufficient.
triangle ABD and BDC are similar.
AC 10, AB 3√10 so BC √10

By similar traingle we can calculate the height if both traingles in terms of P and Q.
thus we can get the desired area ratio

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In the figure above, AC = 10 and ABC is a semicircle. AD = p and DC = [#permalink] New post   17 Feb 2021, 10:58
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mcmoorthy wrote:
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


B is also sufficient. Since triangle ABD and ABC are similar, therefore the ratio of areas of these triangles is equal to the ratio of squares of the sides of these triangles.

Hence ratio of areas of ABD and ABC= (AB^2/AC^2)
Since AB=3*sqrt10 and AC=10
Therefore ratio of areas of ABD and ABC = 9/10

Therefore answer is D.
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Re: In the figure above, AC = 10 and ABC is a semicircle. AD = p and DC = [#permalink] New post   20 Jan 2024, 15:07
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Re: In the figure above, AC = 10 and ABC is a semicircle. AD = p and DC = [#permalink]
20 Jan 2024, 15:07
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