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In the figure above, the area ΔABC is 6. If BC is 1/3 the length of AB

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In the figure above, the area ΔABC is 6. If BC is 1/3 the length of AB  [#permalink]

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New post 25 Dec 2018, 08:40
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A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

73% (01:11) correct 27% (01:12) wrong based on 50 sessions

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Re: In the figure above, the area ΔABC is 6. If BC is 1/3 the length of AB  [#permalink]

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New post 25 Dec 2018, 09:19
Bunuel wrote:
Image
In the figure above, the area ΔABC is 6. If BC is 1/3 the length of AB, then AC =


A. \(\sqrt{2}\)

B. 2

C. 4

D. 6

E. \(2\sqrt{10}\)




Attachment:
2018-12-25_1928.png

x^2=36

x=6
36+4= hypotenuse

sqrt 40 = 2 sqrt 10

IMO E
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Re: In the figure above, the area ΔABC is 6. If BC is 1/3 the length of AB  [#permalink]

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New post 27 Dec 2018, 10:39
Could you expllain me why 36+4 = hypotenuse ?
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Re: In the figure above, the area ΔABC is 6. If BC is 1/3 the length of AB  [#permalink]

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New post 27 Dec 2018, 10:50
2
shobhitraj wrote:
Could you expllain me why 36+4 = hypotenuse ?


Here are my 2 cents on this.

Area of a triangle ABC = 1/2 base * height (1)

Given BC = 1/3 * AB => AB = 3 BC

So if you follow (1),

1/2 * AB * BC = 6
BC* BC = 4
BC = 2

Giving AB as 6

Now since from the figure < B = 90 we can say that AB and BC are the legs of the triangle.
From Hypotenuse Theorem => AB^2 + BC^2 = AC^2
AC^2 = 36 + 4
AC = \(\sqrt{40}\)
AC = 2\(\sqrt{10}\)
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Re: In the figure above, the area ΔABC is 6. If BC is 1/3 the length of AB  [#permalink]

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New post 28 Dec 2018, 01:15
1
Quote:
Could you expllain me why 36+4 = hypotenuse ?


shobhitraj

given :0.5*AB*BC=6
and BC=1/3 * AB
let AB=x

so
1/2 * x^2*1/3= 6
x= 6
and AB = 4

so AC^2= 6^2+4^2 = sqrt 40 = 2sqrt 10
option E

Hope this helps
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Re: In the figure above, the area ΔABC is 6. If BC is 1/3 the length of AB   [#permalink] 28 Dec 2018, 01:15
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