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In triangle XYZ above, the length of XZ is 6/7 of the length of altitu

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In triangle XYZ above, the length of XZ is 6/7 of the length of altitu [#permalink]

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New post 01 Sep 2017, 00:00
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In triangle XYZ above, the length of XZ is 6/7 of the length of altitude h. What is the area of ∆ XYZ in terms of h?

(A) h^2/3
(B) 3h^2/7
(C) 3h/7
(D) 6h^2/7
(E) 12h^2/7


[Reveal] Spoiler:
Attachment:
2017-09-01_1157.png
2017-09-01_1157.png [ 4.64 KiB | Viewed 895 times ]
[Reveal] Spoiler: OA

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Re: In triangle XYZ above, the length of XZ is 6/7 of the length of altitu [#permalink]

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New post 01 Sep 2017, 00:56
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If the length of XZ is \(\frac{6}{7}\) the length of the altitude 'h', the base of the triangle(XZ) is \(\frac{6z}{7}\)

We know that area is \(\frac{1}{2}\)*Base*Height

Therefore, Area = \(\frac{1}{2}\)*\(\frac{6z}{7}\)*z = \(\frac{3z^2}{7}\)(Option B)
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Re: In triangle XYZ above, the length of XZ is 6/7 of the length of altitu [#permalink]

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New post 02 Sep 2017, 04:26
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Area = \(\frac{1}{2}\) \(* h * XZ\)

= \(\frac{1}{2} * h * \frac{6h}{7}\)

= \(3h^{2}/7\)

Answer: B

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Re: In triangle XYZ above, the length of XZ is 6/7 of the length of altitu [#permalink]

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New post 02 Sep 2017, 06:37
Bunuel wrote:
Image
In triangle XYZ above, the length of XZ is 6/7 of the length of altitude h. What is the area of ∆ XYZ in terms of h?

(A) h^2/3
(B) 3h^2/7
(C) 3h/7
(D) 6h^2/7
(E) 12h^2/7


[Reveal] Spoiler:
Attachment:
2017-09-01_1157.png


Area of Triangle =1/2 base* height

Base XZ= 6/7 h
Height = h
so Area= 1/2* 6/7* h* h
=3/7 h^2

So ans: B

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Re: In triangle XYZ above, the length of XZ is 6/7 of the length of altitu [#permalink]

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New post 07 Sep 2017, 06:13
Bunuel wrote:
Image
In triangle XYZ above, the length of XZ is 6/7 of the length of altitude h. What is the area of ∆ XYZ in terms of h?

(A) h^2/3
(B) 3h^2/7
(C) 3h/7
(D) 6h^2/7
(E) 12h^2/7


[Reveal] Spoiler:
Attachment:
2017-09-01_1157.png


We can create the following equation:

XZ = base = (6/7)h

So, the area of the triangle is [(6/7)h x h]/2 = (6h^2/7)/2 = 6h^2/14 = 3h^2/7.

Answer: B
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Re: In triangle XYZ above, the length of XZ is 6/7 of the length of altitu [#permalink]

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New post 08 Sep 2017, 03:59
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The correct answer is Option B.

Bunuel : please change the OA after review.

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Re: In triangle XYZ above, the length of XZ is 6/7 of the length of altitu [#permalink]

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New post 08 Sep 2017, 04:08

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Re: In triangle XYZ above, the length of XZ is 6/7 of the length of altitu [#permalink]

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New post 08 Sep 2017, 05:02
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Re: In triangle XYZ above, the length of XZ is 6/7 of the length of altitu   [#permalink] 08 Sep 2017, 05:02
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