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11 Feb 2020, 10:25
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Difficulty:
85%
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Question Stats:
29% (02:27) correct
71%
(03:38)
wrong
based on 14
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A square whose diagonal measures 30√2 units is symmetrical about the x-axis and the y-axis. Two circles, each centred at the point (-1,0), are to be drawn inside but not touching the square. If the radius of each circle is an integer, in how many ways can the circles be drawn such that the point (5,0) lies between the two circles?
A. 27
B. 32
C. 35
D. 40
E. 45
A. 27
B. 32
C. 35
D. 40
E. 45
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12 Feb 2020, 02:06
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Solution:
- • Diagonal of square = \(30√2\) units.
• Let the sides of the square be x units
- o \(x^2+x^2 = 900*2\)
o \(2x^2 = 900*2\)
o \(x^2 = 900\)
o \(x = 30,\) here we discarded the case of \(-30\) because the length can never be negative.
- • It means the length of the side of the square above the x-axis and below the x-axis are equal.
- • It means the length of the side of the square right of the y-axis and length left of the y-axis is equal.
- • The radius of one circle must be less than \(5-(-1)= 6\) and the other must be greater than \(6\) and less than \(14\). Then the point \((5,0)\) will be between the two circles.
• The number of ways in which the radius of the circle can be less than \(6 = 5\) (when the radius is \(1, 2, 3, 4, and 5\))
• The number of ways in which the radius of the circle can be between \(6\) and \(14 = 7\) (when the radius is \(7, 8, 9, 10 , 11 , 12, and 13\))
• Total number of required ways = \(5*7=35\)
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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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