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

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Math Expert
Joined: 02 Sep 2009
Posts: 44294
In triangle XYZ above, the length of XZ is 6/7 of the length of altitu [#permalink]

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01 Sep 2017, 01:00
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Difficulty:

35% (medium)

Question Stats:

71% (00:46) correct 29% (01:09) wrong based on 78 sessions

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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 [ 4.64 KiB | Viewed 1157 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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01 Sep 2017, 01:56

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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02 Sep 2017, 05:26

Area = $$\frac{1}{2}$$ $$* h * XZ$$

= $$\frac{1}{2} * h * \frac{6h}{7}$$

= $$3h^{2}/7$$

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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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02 Sep 2017, 07:37
Bunuel wrote:

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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07 Sep 2017, 07:13
Bunuel wrote:

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.

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Joined: 05 Oct 2014
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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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08 Sep 2017, 04:59
1
KUDOS
The correct answer is Option B.

Bunuel : please change the OA after review.
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Joined: 02 Sep 2009
Posts: 44294
Re: In triangle XYZ above, the length of XZ is 6/7 of the length of altitu [#permalink]

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08 Sep 2017, 05:08
chipsy wrote:
The correct answer is Option B.

Bunuel : please change the OA after review.

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Edited. Thank you.
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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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08 Sep 2017, 06:02

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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, 06:02
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