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16 Sep 2014, 00:37
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In triangle ABC, points M, N, and O are the midpoints of sides AB, BC, and AC, respectively. What is the area of triangle MON?
(1) The area of ABC is \(\frac{\sqrt{3}}{4}\)
(2) ABC is an equilateral triangle with a height of \(\frac{\sqrt{3}}{2}\)
(1) The area of ABC is \(\frac{\sqrt{3}}{4}\)
(2) ABC is an equilateral triangle with a height of \(\frac{\sqrt{3}}{2}\)
16 Sep 2014, 00:37
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Official Solution:
In triangle ABC, points M, N, and O are the midpoints of sides AB, BC, and AC, respectively. What is the area of triangle MON?
Look at the diagram below:
MN, NO, and OM are midsegments of triangle ABC (a midsegment is a line segment joining the midpoints of two sides of a triangle). An important property of a midsegment is: the midsegment is always half the length of the third side. So, \(MN=\frac{AC}{2}\), \(NO=\frac{AB}{2}\), and \(OM=\frac{BC}{2}\).
Furthermore, since each side of triangle MNO is half of the side of triangle ABC, these triangles are similar (the ratio of all the sides is the same). An important property of similar triangles is: if two similar triangles have sides in the ratio \(\frac{x}{y}\), their areas are in the ratio \(\frac{x^2}{y^2}\).
Since the sides of the two similar triangles, MNO and ABC, are in the ratio 1:2, their areas are in the ratio 1:4. Therefore, (the area of MNO) = (the area of ABC)/4.
So, to find the area of MNO, we need to determine the area of ABC.
(1) The area of ABC is \(\frac{\sqrt{3}}{4}\).
Sufficient.
(2) ABC is an equilateral triangle with a height of \(\frac{\sqrt{3}}{2}\).
We can determine the area of an equilateral triangle with the given height. Sufficient.
Answer: D
In triangle ABC, points M, N, and O are the midpoints of sides AB, BC, and AC, respectively. What is the area of triangle MON?
Look at the diagram below:
MN, NO, and OM are midsegments of triangle ABC (a midsegment is a line segment joining the midpoints of two sides of a triangle). An important property of a midsegment is: the midsegment is always half the length of the third side. So, \(MN=\frac{AC}{2}\), \(NO=\frac{AB}{2}\), and \(OM=\frac{BC}{2}\).
Furthermore, since each side of triangle MNO is half of the side of triangle ABC, these triangles are similar (the ratio of all the sides is the same). An important property of similar triangles is: if two similar triangles have sides in the ratio \(\frac{x}{y}\), their areas are in the ratio \(\frac{x^2}{y^2}\).
Since the sides of the two similar triangles, MNO and ABC, are in the ratio 1:2, their areas are in the ratio 1:4. Therefore, (the area of MNO) = (the area of ABC)/4.
So, to find the area of MNO, we need to determine the area of ABC.
(1) The area of ABC is \(\frac{\sqrt{3}}{4}\).
Sufficient.
(2) ABC is an equilateral triangle with a height of \(\frac{\sqrt{3}}{2}\).
We can determine the area of an equilateral triangle with the given height. Sufficient.
Answer: D
General Discussion
19 Aug 2016, 03:54
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I think this is a high-quality question and I agree with explanation.
