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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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A
B
C
D
E

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  95% (hard)

Question Stats:

29% (03:40) correct 71% (02:57) wrong based on 14 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 228 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 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] 28 Mar 2019, 19:48
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