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11 Jun 2022, 09:45
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Rectangle ABCD has AB =8 and BC = 6. Point M is the midpoint of diagonal AC, and E is on AB with ME perpendicular to AC. What is the area of triangle AME?
(A) 65/8
(B) 25/3
(C) 9
(D) 75/8
(E) 85/8
Are You Up For the Challenge: 700 Level Questions
(A) 65/8
(B) 25/3
(C) 9
(D) 75/8
(E) 85/8
Are You Up For the Challenge: 700 Level Questions
PyjamaScientist
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12 Jun 2022, 03:47
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MissBong
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12 Jun 2022, 11:39
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\(AC^2\) =\(AB^2\) +\( BC^2\)
AC= 10
If we consider \(\angle\) BAC = \(\alpha \),
then tan \(\alpha\) = \(\frac{6}{8}\)
We can also write tan \(\alpha\) = \(\frac{ME}{5}\) [ 5, because M is mid point of AC]
Therefore,\(\frac{ME}{5}\)=\(\frac{6}{8}\)
ME= \(\frac{30}{8}\)
Area=\( \frac{1}{2}\)* Base * Height =\(\frac{1}{2}\)* AM* ME = \(\frac{1}{2}\)*5*\(\frac{30}{8}\)
Area =\(\frac{75}{8} \) (D)
AC= 10
If we consider \(\angle\) BAC = \(\alpha \),
then tan \(\alpha\) = \(\frac{6}{8}\)
We can also write tan \(\alpha\) = \(\frac{ME}{5}\) [ 5, because M is mid point of AC]
Therefore,\(\frac{ME}{5}\)=\(\frac{6}{8}\)
ME= \(\frac{30}{8}\)
Area=\( \frac{1}{2}\)* Base * Height =\(\frac{1}{2}\)* AM* ME = \(\frac{1}{2}\)*5*\(\frac{30}{8}\)
Area =\(\frac{75}{8} \) (D)
