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23 May 2019, 12:12
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C
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Difficulty:
55%
(hard)
Question Stats:
64% (02:17) correct
36%
(02:14)
wrong
based on 103
sessions
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A triangle ABC has area 32 sq units and its side BC, of length 8 units, lies on the line x = 4. Then the shortest possible distance between A and the point (0,0) is
A. 2
B. \(2\sqrt{2}\)
C. 4
D. \(4\sqrt{2}\)
E. 6
A. 2
B. \(2\sqrt{2}\)
C. 4
D. \(4\sqrt{2}\)
E. 6
23 May 2019, 12:25
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base = 8
Area = 1/2 * b * h = 1/2 * 8 * h = 32
h = 8
So point A can be on x=-4 or x=12
For shortest distance from origin the point has to be on x=-4 and it has to be on the x-axis
So A is (-4,0) and the distance is 4
Ans C
Posted from my mobile device
Area = 1/2 * b * h = 1/2 * 8 * h = 32
h = 8
So point A can be on x=-4 or x=12
For shortest distance from origin the point has to be on x=-4 and it has to be on the x-axis
So A is (-4,0) and the distance is 4
Ans C
Posted from my mobile device
03 Jul 2021, 04:58
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The 2 possible distances between Point A and Point (0,0) are:
A) |12-0| = 12 units & B) |-4-0| = 4 units
Shortest is 4 units (C)
A) |12-0| = 12 units & B) |-4-0| = 4 units
Shortest is 4 units (C)
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