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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.
How did you get \(\frac{S^2}{s^2}\), as area of triangle is 1/2 * base * height.