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In the figure above triangles ABC and MNP are both isosceles [#permalink]
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14 Feb 2012, 07:59
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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. \(\frac{7\sqrt{3}}{3}\)
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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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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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14 Feb 2012, 08:35
Neat little rule Bunuel ! Kudos !
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Triangle [#permalink]
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31 Mar 2012, 16:34
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, 16:37, edited 1 time in total.



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Re: Triangle [#permalink]
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31 Mar 2012, 16:37



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Re: Triangle base [#permalink]
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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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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 \(\frac{AREA_{ABC}}{area_{MNP}}=\frac{AC^2}{MP^2}\) > \(\frac{2}{1}=\frac{7^2}{MP^2}\) > \(MP^2=\frac{7^2}{2}\) > \(MP=\frac{7}{\sqrt{2}}=\frac{7\sqrt{2}}{2}\). 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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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.
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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




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