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In the figure above triangles ABC and MNP are both isosceles

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In the figure above triangles ABC and MNP are both isosceles  [#permalink]

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New post Updated on: 14 Feb 2012, 08:26
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In the figure above, triangles ABC and MNP are both isosceles. AB is parallel to MN, BC is parallel to NP, the length of AC is 7 and the length of BY is 4. If the area of the unshadd region is equal to the area of the shaded region, what is the length of MP?

A. \(2\sqrt{2}\)

B. \(2\sqrt{7}\)

C. \(\frac{2\sqrt{3}}{3}\)

D. \(\frac{7\sqrt{2}}{2}\)

E. \(\frac{7\sqrt{3}}{3}\)

Originally posted by rxs0005 on 14 Feb 2012, 07:59.
Last edited by Bunuel on 14 Feb 2012, 08:26, edited 1 time in total.
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Re: Triangle base  [#permalink]

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New post 14 Feb 2012, 08:14
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In the figure above, triangles ABC and MNP are both isosceles. AB is parallel to MN, BC is parallel to NP, the length of AC is 7 and the length of BY is 4. If the area of the unshadd region is equal to the area of the shaded region, what is the length of MP?

A. \(2\sqrt{2}\)

B. \(2\sqrt{7}\)

C. \(\frac{2\sqrt{3}}{3}\)

D. \(\frac{7\sqrt{2}}{2}\)

E. \(\frac{7\sqrt{3}}{3}\)

Since the area of unshaded region is equal to the are of shaded region, then the area of the big triangle is twice the area of the little triangle (unshaded region): \(\frac{AREA_{ABC}}{area_{MNP}}=\frac{2}{1}\)

Next, triangles ABC and MNP are similar. In two similar triangles, the ratio of their areas is the square of the ratio of their sides: \(\frac{AREA}{area}=\frac{S^2}{s^2}\). Thus \(\frac{AREA_{ABC}}{area_{MNP}}=\frac{AC^2}{MP^2}\) --> \(\frac{2}{1}=\frac{7^2}{MP^2}\) --> \(MP^2=\frac{7^2}{2}\) --> \(MP=\frac{7}{\sqrt{2}}=\frac{7\sqrt{2}}{2}\).

Answer: D.
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Re: In the figure above triangles ABC and MNP are both isosceles  [#permalink]

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New post 14 Feb 2012, 08:35
Neat little rule Bunuel ! Kudos !
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Triangle  [#permalink]

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New post Updated on: 31 Mar 2012, 16:37
In the …figure below, triangles ABC and MNP are both isoceles. AB is parallel to MN, BC is parallel to NP, the length of AC is 7 and the length of BY is 4. If the area of the unshaded region is equal to the area of the shaded region, what is the length of MP?

(A) \(2\sqrt{2}\)
(B) \(2\sqrt{7}\)
(C) \(2\sqrt{3}/3\)
(D) \(7\sqrt{2}/2\)
(E) \(7\sqrt{3}/3\)
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Originally posted by enigma123 on 31 Mar 2012, 16:34.
Last edited by enigma123 on 31 Mar 2012, 16:37, edited 1 time in total.
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Re: Triangle  [#permalink]

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New post 31 Mar 2012, 16:37
enigma123 wrote:
In the …figure below, triangles ABC and MNP are both isoceles. AB is parallel to MN, BC is parallel to NP, the length of AC is 7 and the length of BY is 4. If the area of the unshaded region is equal to the area of the shaded region, what is the length of MP?

(A) \(2\sqrt{2}\)
(B) \(\sqrt{2[square_root]7}[/square_root]\)
(C) \(\frac{2[square_root]3[}{square_root]/3}\)
(D) \(\frac{7[square_root]2[}{square_root]/2}\)
(E) \(\frac{7[square_root]3[}{square_root]/3}\)


Merging similar topics. Please ask if anything remains unclear.
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Re: Triangle base  [#permalink]

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New post 31 Mar 2012, 17:02
Since the area of unshaded region is equal to the area of shaded region, then the area of the big triangle is twice the area of the little triangle (unshaded region): \(\frac{AREA_{ABC}}{area_{MNP}}=\frac{2}{1}\)

How did you get the above?
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Re: Triangle base  [#permalink]

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New post 31 Mar 2012, 17:07
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enigma123 wrote:
Since the area of unshaded region is equal to the area of shaded region, then the area of the big triangle is twice the area of the little triangle (unshaded region): \(\frac{AREA_{ABC}}{area_{MNP}}=\frac{2}{1}\)

How did you get the above?


Unshaded+Shaded=ABC --> since Unshaded=Shaded then Unshaded+Unshaded=ABC --> 2*Unshaded=ABC --> ABC/Unshaded=2/1 --> ABC/MNP=2/1.

Hope it's clear.
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Re: In the figure above triangles ABC and MNP are both isosceles  [#permalink]

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New post 21 Nov 2015, 12:33
Bunuel wrote:
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The attachment Triangle.PNG is no longer available
In the figure above, triangles ABC and MNP are both isosceles. AB is parallel to MN, BC is parallel to NP, the length of AC is 7 and the length of BY is 4. If the area of the unshadd region is equal to the area of the shaded region, what is the length of MP?

A. \(2\sqrt{2}\)

B. \(2\sqrt{7}\)

C. \(\frac{2\sqrt{3}}{3}\)

D. \(\frac{7\sqrt{2}}{2}\)

E. \(\frac{7\sqrt{3}}{3}\)

Since the area of unshaded region is equal to the are of shaded region, then the area of the big triangle is twice the area of the little triangle (unshaded region): \(\frac{AREA_{ABC}}{area_{MNP}}=\frac{2}{1}\)

Next, triangles ABC and MNP are similar. In two similar triangles, the ratio of their areas is the square of the ratio of their sides: \(\frac{AREA}{area}=\frac{S^2}{s^2}\). Thus \(\frac{AREA_{ABC}}{area_{MNP}}=\frac{AC^2}{MP^2}\) --> \(\frac{2}{1}=\frac{7^2}{MP^2}\) --> \(MP^2=\frac{7^2}{2}\) --> \(MP=\frac{7}{\sqrt{2}}=\frac{7\sqrt{2}}{2}\).

Answer: D.



How did you get \(\frac{S^2}{s^2}\), as area of triangle is 1/2 * base * height.
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Re: In the figure above triangles ABC and MNP are both isosceles  [#permalink]

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New post 16 Jan 2017, 13:55
Good morning everyone,

How did you get to this MP=72‾√=72‾√2MP=72=722 ? I guess my question is more how do you put the root from the denominator up to the numerator.

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Re: In the figure above triangles ABC and MNP are both isosceles  [#permalink]

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New post 17 Jan 2017, 00:56
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Re: In the figure above triangles ABC and MNP are both isosceles  [#permalink]

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New post 01 Jul 2018, 23:17
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Re: In the figure above triangles ABC and MNP are both isosceles   [#permalink] 01 Jul 2018, 23:17
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