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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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Updated on: 14 Feb 2012, 08:26
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69% (03:06) correct 31% (03:01) wrong based on 303 sessions

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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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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 $$\frac{AREA_{ABC}}{area_{MNP}}=\frac{AC^2}{MP^2}$$ --> $$\frac{2}{1}=\frac{7^2}{MP^2}$$ --> $$MP^2=\frac{7^2}{2}$$ --> $$MP=\frac{7}{\sqrt{2}}=\frac{7\sqrt{2}}{2}$$.

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

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14 Feb 2012, 08:35
Neat little rule Bunuel ! Kudos !
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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) $$7\sqrt{3}/3$$
Attachments

Triangle.GIF [ 6.91 KiB | Viewed 8740 times ]

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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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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) $$\frac{7[square_root]3[}{square_root]/3}$$

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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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MGMAT 1 --> 530
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GMAT ==> 730
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Posts: 56300

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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?

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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21 Nov 2015, 12:33
Bunuel wrote:
Attachment:
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 $$\frac{AREA_{ABC}}{area_{MNP}}=\frac{AC^2}{MP^2}$$ --> $$\frac{2}{1}=\frac{7^2}{MP^2}$$ --> $$MP^2=\frac{7^2}{2}$$ --> $$MP=\frac{7}{\sqrt{2}}=\frac{7\sqrt{2}}{2}$$.

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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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.

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

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17 Jan 2017, 00:56
lauramo wrote:
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.

Thanks

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

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01 Jul 2018, 23:17
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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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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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