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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] New post 14 Feb 2012, 06:59
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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. \(\frac{7\sqrt{3}}{3}\)
[Reveal] Spoiler: OA

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Last edited by Bunuel on 14 Feb 2012, 07:26, edited 1 time in total.
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Re: Triangle base [#permalink] New post 14 Feb 2012, 07:14
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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] New post 14 Feb 2012, 07:35
Neat little rule Bunuel ! Kudos !
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Triangle [#permalink] New post 31 Mar 2012, 15:34
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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Last edited by enigma123 on 31 Mar 2012, 15:37, edited 1 time in total.
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Re: Triangle [#permalink] New post 31 Mar 2012, 15:37
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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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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: Triangle base [#permalink] New post 31 Mar 2012, 16:02
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] New post 31 Mar 2012, 16:07
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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.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis ; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) ; 12. Tricky questions from previous years.

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: In the figure above triangles ABC and MNP are both isosceles [#permalink] New post 26 Sep 2013, 10:56
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Re: In the figure above triangles ABC and MNP are both isosceles [#permalink] New post 03 Oct 2014, 22:21
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Re: In the figure above triangles ABC and MNP are both isosceles   [#permalink] 03 Oct 2014, 22:21
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In the figure above triangles ABC and MNP are both isosceles

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