Consider a circle inscribed in a square. A rectangle with a base of 2 cm and a height of 1 cm is inserted in the space between one of the vertices of the square and the circumferen... Consider a circle inscribed in a square. A rectangle with a base of 2 cm and a height of 1 cm is inserted in the space between one of the vertices of the square and the circumference in such a way that one vertex of the rectangle coincides with that of the square and the opposite vertex lies on the circumference. Calculate the radius of the circle.
Understand the Problem
The problem describes a circle inscribed in a square. A rectangle with base 2 cm and height 1 cm is placed in the corner of the square such that one vertex is on the corner of the square, and the opposite vertex lies on the circumference of the circle. The goal is to calculate the radius of the circle.
Answer
A) 5 cm
Answer for screen readers
A) 5 cm
Steps to Solve
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Visualize the problem Imagine the square with an inscribed circle. The rectangle sits in the corner, touching both sides of the square and with its opposite vertex on the circle.
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Set up a coordinate system Place the corner of the square at the origin (0,0) of a coordinate system.
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Define the circle's center and equation The circle's center will be at (r, r), where r is the radius. The equation of the circle is $(x-r)^2 + (y-r)^2 = r^2$.
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Determine the coordinates of the rectangle's vertex on the circumference The vertex of the rectangle on the circumference is at the point (2, 1).
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Substitute the coordinates into the circle's equation Substitute x=2 and y=1 into the equation of the circle: $(2-r)^2 + (1-r)^2 = r^2$.
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Solve for r Expand the equation: $4 - 4r + r^2 + 1 - 2r + r^2 = r^2$ Simplify: $r^2 - 6r + 5 = 0$ Factor: $(r-5)(r-1) = 0$ So, $r = 5$ or $r = 1$.
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Consider the possible solutions If $r = 1$, the circle would be too small to contain the rectangle with a base of 2. So, $r = 1$ is an extraneous solution. Therefore, the radius is $r = 5$.
A) 5 cm
More Information
The radius of the circle is 5 cm. We found this by setting up a coordinate system, expressing the circle's equation, substituting the coordinates of the rectangle's vertex that lies on the circumference, and solving the resulting quadratic equation. We discarded the extraneous solution $r=1$ because it was geometrically impossible.
Tips
A common mistake is to incorrectly set up the equation of the circle or to incorrectly substitute the coordinates of the point. Another mistake is to forget to check for extraneous solutions. When solving for $r$, both roots must be checked in the context of the problem. Failing to do so might lead you to include an extraneous solution and choose the wrong answer.
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