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

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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?

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

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?

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?

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): \(\frac{AREA_{ABC}}{area_{MNP}}=\frac{2}{1}\)

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.

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

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26 Sep 2013, 10:56

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

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03 Oct 2014, 22:21

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

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20 Nov 2015, 19:03

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

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21 Nov 2015, 11:33

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.

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

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11 Jan 2017, 05:44

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