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Triangle ABC has a right angle at B. Point D is the foot of the altitu

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Triangle ABC has a right angle at B. Point D is the foot of the altitu  [#permalink]

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New post 28 Mar 2019, 05:55
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Question Stats:

43% (03:12) correct 57% (03:10) wrong based on 37 sessions

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Triangle ABC has a right angle at B. Point D is the foot of the altitude from B, AD = 3, and DC = 4. What is the area of triangle ABC ?


(A) \(4\sqrt{3}\)

(B) \(7\sqrt{3}\)

(C) 21

(D) \(14\sqrt{3}\)

(E) 42

Attachment:
4bea2faf81f4ba6dca40fdb9ce720c9ed82b168c.png
4bea2faf81f4ba6dca40fdb9ce720c9ed82b168c.png [ 5.59 KiB | Viewed 619 times ]

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Re: Triangle ABC has a right angle at B. Point D is the foot of the altitu  [#permalink]

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New post 16 Oct 2019, 10:56
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TheNightKing wrote:
chetan2u Can you help with this one?

I believe Ypsychotic is saying CD/BD=BD/AD. From this we can find BD and with that we can find area.

Can you share your approach?



Hi,

Yes similar triangle property can be used to find BF and as you too have mentioned that BD =\(\sqrt{AD*CD}=\sqrt{12}\) and I too will do it same way.

But even the choices give away the answer and we should be able to exploit this too.
The area is \(\frac{1}{2}*7*BD\).
So our answer should normally be multiple of 7.
A is out.
Now take C. If (1/2)*7*BD=21...BD=6
But is it possible .. NO as in triangle aBDC, BC will become >7, that is a side of a right angled triangle will become bigger than the hypotenuse AC or 3+4.
So answer is less than 21 for sure. Only B left.
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Re: Triangle ABC has a right angle at B. Point D is the foot of the altitu  [#permalink]

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New post 28 Mar 2019, 19:48
Bunuel wrote:
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Triangle ABC has a right angle at B. Point D is the foot of the altitude from B, AD = 3, and DC = 4. What is the area of triangle ABC ?


(A) \(4\sqrt{3}\)

(B) \(7\sqrt{3}\)

(C) 21

(D) \(14\sqrt{3}\)

(E) 42

Attachment:
4bea2faf81f4ba6dca40fdb9ce720c9ed82b168c.png


Angle ACD=Angle BAD, Because if Angle c=x, angle a= 90.-x, in Triangle ABD, Angle ABD=180-(90+90-x)
=x, Similarly, Angle BCD= Angle ABD,

Using Similarlity,

Let AD=x,

\(\frac{4}{x}=\frac{x}{3}\)

\(x=2\sqrt{3}\)

\(Area=1/2*2\sqrt{3}*(4+3)=7\sqrt{3}\)

IMO, Option B
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Re: Triangle ABC has a right angle at B. Point D is the foot of the altitu  [#permalink]

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New post 16 Oct 2019, 09:45
chetan2u Can you help with this one?

I believe Ypsychotic is saying CD/BD=BD/AD. From this we can find BD and with that we can find area.

Can you share your approach?
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Re: Triangle ABC has a right angle at B. Point D is the foot of the altitu   [#permalink] 16 Oct 2019, 09:45
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