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What is the area of triangle MNO in the figure above?
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08 Dec 2022, 10:10
08 Dec 2022, 10:10
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82% (02:07) correct
17% (02:56) wrong based on 23 sessions
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What is the area of triangle MNO in the figure above?
\(3\)
\(6\)
\(4 \sqrt{2} + 2 \sqrt{5} \)
\(2 \sqrt{10} \)
\(6 \sqrt{10}\)
What is the area of triangle MNO in the figure above?
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11 Dec 2022, 05:00
11 Dec 2022, 05:00
1
Expert Reply
OE
The slope of MO is 1, so it makes a 45 degree angle with the positive x-axis. Similarly, the slope of NO is −1, so it makes another 45 degree angle with the positive x-axis. The sum of the degree measures of these angles is 90, so MNO is a right triangle. Therefore, MO and NO are the base and height of triangle MNO. To find the area of the triangle, you need to find the length of MO and NO. Drop a perpendicular from point M to the y-axis, to form an isosceles right triangle whose hypotenuse is MO. Each leg of this triangle has length 1 so, \(MO = \sqrt{2}\) . Similarly, dropping a perpendicular line from the y-axis to point N creates another isosceles right triangle, whose legs have length 3, and whose hypotenuse is NO. Therefore, \(NO = 3 \sqrt{2}\) . So the area of triangle MNO is\( \frac{1}{2 } * b*h = \frac{1}{2} (\sqrt{2})(3 \sqrt{2} )=3\) , and the answer is choice (A).
The slope of MO is 1, so it makes a 45 degree angle with the positive x-axis. Similarly, the slope of NO is −1, so it makes another 45 degree angle with the positive x-axis. The sum of the degree measures of these angles is 90, so MNO is a right triangle. Therefore, MO and NO are the base and height of triangle MNO. To find the area of the triangle, you need to find the length of MO and NO. Drop a perpendicular from point M to the y-axis, to form an isosceles right triangle whose hypotenuse is MO. Each leg of this triangle has length 1 so, \(MO = \sqrt{2}\) . Similarly, dropping a perpendicular line from the y-axis to point N creates another isosceles right triangle, whose legs have length 3, and whose hypotenuse is NO. Therefore, \(NO = 3 \sqrt{2}\) . So the area of triangle MNO is\( \frac{1}{2 } * b*h = \frac{1}{2} (\sqrt{2})(3 \sqrt{2} )=3\) , and the answer is choice (A).
Re: What is the area of triangle MNO in the figure above?
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30 Jul 2024, 21:26
30 Jul 2024, 21:26
Carcass wrote:
OE
The slope of MO is 1, so it makes a 45 degree angle with the positive x-axis. Similarly, the slope of NO is −1, so it makes another 45 degree angle with the positive x-axis. The sum of the degree measures of these angles is 90, so MNO is a right triangle. Therefore, MO and NO are the base and height of triangle MNO. To find the area of the triangle, you need to find the length of MO and NO. Drop a perpendicular from point M to the y-axis, to form an isosceles right triangle whose hypotenuse is MO. Each leg of this triangle has length 1 so, \(MO = \sqrt{2}\) . Similarly, dropping a perpendicular line from the y-axis to point N creates another isosceles right triangle, whose legs have length 3, and whose hypotenuse is NO. Therefore, \(NO = 3 \sqrt{2}\) . So the area of triangle MNO is\( \frac{1}{2 } * b*h = \frac{1}{2} (\sqrt{2})(3 \sqrt{2} )\) , and the answer is choice (A).
The slope of MO is 1, so it makes a 45 degree angle with the positive x-axis. Similarly, the slope of NO is −1, so it makes another 45 degree angle with the positive x-axis. The sum of the degree measures of these angles is 90, so MNO is a right triangle. Therefore, MO and NO are the base and height of triangle MNO. To find the area of the triangle, you need to find the length of MO and NO. Drop a perpendicular from point M to the y-axis, to form an isosceles right triangle whose hypotenuse is MO. Each leg of this triangle has length 1 so, \(MO = \sqrt{2}\) . Similarly, dropping a perpendicular line from the y-axis to point N creates another isosceles right triangle, whose legs have length 3, and whose hypotenuse is NO. Therefore, \(NO = 3 \sqrt{2}\) . So the area of triangle MNO is\( \frac{1}{2 } * b*h = \frac{1}{2} (\sqrt{2})(3 \sqrt{2} )\) , and the answer is choice (A).
I didnt Understood..Kindly please explain
