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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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Bunuel
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In the figure above, the area ΔABC is 6. If BC is 1/3 the length of AB [#permalink] New post   25 Dec 2018, 08:40
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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. \(2\sqrt{10}\)


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Archit3110
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Re: In the figure above, the area ΔABC is 6. If BC is 1/3 the length of AB [#permalink] New post   25 Dec 2018, 09:19
Bunuel wrote:

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. \(2\sqrt{10}\)




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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] 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] New post   27 Dec 2018, 10:50
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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] New post   28 Dec 2018, 01:15
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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] New post   19 Oct 2020, 08:53
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Re: In the figure above, the area ΔABC is 6. If BC is 1/3 the length of AB [#permalink]
19 Oct 2020, 08:53
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