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C
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In the figure, ABCD is a square, and BC is tangent to the circle with radius 3. If PC = 2, then what is the area of square ABCD ?
(A) 9
(B) 13
(C) 16
(D) 18
(E) 25
Attachment:
#GREpracticequestion In the figure, ABCD is a square (1).jpg [ 32.05 KiB | Viewed 3694 times ]
09 Sep 2022, 02:45
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\(OP = 3\), since \(OP\) is the radius
\(OB= 3\), \(OB\) is the radius of the circle
\(OC = OP+PC = 3+2=5\)
Angle OBC is right angle, since the angle between radius and tangent to a circle at the point of contact is \(90\).
So triangle OBC is a right angled triangle.
Therefore
\(OC^2= OB^2+BC^2\)
\(5^2= 3^2+BC^2\)
\(BC^2=25-9\)
\(BC^2=16\)
\(BC=4\)
Since \(ABCD\) is a square , area of a square is = \(Side*Side\)
\(Area = 4*4 = 16\), since BC is the side of the square.
Correct answer is Option C
\(OB= 3\), \(OB\) is the radius of the circle
\(OC = OP+PC = 3+2=5\)
Angle OBC is right angle, since the angle between radius and tangent to a circle at the point of contact is \(90\).
So triangle OBC is a right angled triangle.
Therefore
\(OC^2= OB^2+BC^2\)
\(5^2= 3^2+BC^2\)
\(BC^2=25-9\)
\(BC^2=16\)
\(BC=4\)
Since \(ABCD\) is a square , area of a square is = \(Side*Side\)
\(Area = 4*4 = 16\), since BC is the side of the square.
Correct answer is Option C
Attachments
practicequestion In the figure, ABCD is a square (1).jpg [ 20.6 KiB | Viewed 1830 times ]
