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Circles Inscribed in Squares: 7 Hard Geometry Problems & Solutions

Updated on March 25, 2015
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TR Smith is a product designer and former teacher who uses math in her work every day.

Geometry problems involving circles inscribed in squares and rectangles are common on standardized tests such as the GMAT, GED, and SAT. Typical problems ask you to solve for the radius or area of a figure, or find the ratio of measurements between various figures. Most problems can be solved by applying properties of circles and and the Pythagorean Theorem.

For these problems, the most important geometric property of circles is that if two circles are tangent to one another, then the line joining their centers also passes through the circles' point of tangency. Another property to remember is that any line segment that connects the center of a circle to a point on the circumference has a length equal to the radius.

Try your hand at these 7 classic problems plus 2 bonus problems.


Problem 1

A circle with a radius of 1 is inscribed within a 2-by-2 square. A smaller circle is inscribed in the corner such that it is tangent to two adjacent sides of the square and the larger circle. What is the radius of this smaller circle?

Solution: Call the radius of the smaller circle r. If you connect the centers of the two circles with a straight line, it will pass through the point of tangency and this line segment will have a length of 1 + r. This line makes a 45-degree angle with the corner of the square. As shown in the figure on the right, you can add two more lines to form an isosceles right triangle whose hypotenuse is 1 + r, and whose two legs are both 1 - r.

Since it is an isosceles right triangle, the hypotenuse must be sqrt(2) times longer than each leg. Thus we have the equation

1 + r = sqrt(2)(1 - r)
r + sqrt(2)r = sqrt(2) - 1
r(1 + sqrt(2)) = sqrt(2) - 1
r = [sqrt(2) -1]/[sqrt(2) + 1]
r = 3 - 2*sqrt(2)
r ≈ 0.171573


Problem 2

Two circles each with a radius of 1 and tangent to each other are inscribed in a 2-by-4 rectangle. A smaller circle is inscribed in the space between the circles and the long edge of the rectangle, such that it is tangent to both circles and the edge of the rectangle. What is the radius of this smaller circle?

Solution: A right triangle can be formed that connects the center of the smaller circle, the center of one of the larger circles, and the tangent point between the two larger circles. Call the radius of the smaller circle r. Then the legs of this triangle have lengths of 1 and 1 - r, and the hypotenuse has a length of 1 + r. This gives us the following Pythagorean equation to solve for r:

1^2 + (1-r)^2 = (1+r)^2
1 + 1 - 2r + r^2 = 1 + 2r + r^2
2 - 2r = 1 + 2r
1 = 4r
r = 1/4.


Problem 3

A circle is inscribed within a quarter-circle sector. It is tangent to the arc of the sector and the two perpendicular radii of the sector. What is the ratio of the area of the circle to the area of the sector?

Solution: Since this is a ratio problem, it does not matter what the radius of each circle is, only their sizes relative to one another. Therefore we can simplify the problem by setting the radius of the sector equal to 1. If we draw a line from the vertex of the sector to the point on the arc where it is tangent to the circle, this line segment has a length of 1 because it is a radius.

If we call the radius of the smaller circle r, we can see that the length of this line segment is also equal to r + sqrt(2)r. Therefore we can solve for r:

r + sqrt(2)r = 1
[1 + sqrt(2)]r = 1
r = 1/[1 + sqrt(2)]
r = sqrt(2) - 1

The area of the quarter circle is π/4. The area of the smaller inscribed circle is π(3 - 2*sqrt(2)). Thus, the ratio of the circle to the sector is

[ π(3 - 2*sqrt(2)) ] / [ π/4 ]
= 12 - 8*sqrt(2)
≈ 0.68629


Problem 4

Four circles, each with a radius of 1, are arranged so that they are tangent to two others and their centers form the corners of a square. A smaller circle is inscribed in the space bounded by the four circles, and this smaller circle is tangent to each of the other four. What is its radius?

Solution: The square formed by the four circles has an edge length of 2. The diagonal of this square is 2*sqrt(2). If the radius of the smaller circle is r, the diagonal of the square is also equal to 1 + 2r + 1. Thus we can solve for r:

1 + 2r + 1 = 2*sqrt(2)
2 + 2r = 2*sqrt(2)
1 + r = sqrt(2)
r = sqrt(2) - 1
r ≈ 0.41421


Problem 5

A circle is inscribed within a square. A smaller square is inscribed within this circle. What is the ratio of the area of the larger square to that of the smaller square?

Solution: This is a classic math brain teaser question that requires no sophisticated geometry or algebra knowledge. Spin the circle so that the corners of the inscribed square lie on the centers of the edges of the larger square. Now you can see that the smaller square occupies exactly half the space of the larger square. Therefore the area ratio is 2:1.


Problem 6

A square is inscribed in a circle. Within this square is inscribed a smaller circle. What is the ratio of the area of the larger circle to that of the smaller circle?

Solution: Inscribe another smaller square within the smaller circle. Call the are of the large square Q, the area of the small square q, the area of the large circle M, and the area of the small circle m. We know that M/Q = m/q by properties of scale. And from the previous problem we know that Q/q = 2. Therefore

M/Q = m/q
Mq = mQ
M/m = Q/q
M/m = 2

So the ratio of the area of the large circle to the area of the small circle is also 2:1.


Problem 7

The figure above shows a square with a side length of 1. Along the edges are four half-circles with each a radius of 0.5. Their intersections inside the square create eight regions: four like the one shown in pink, and four like the one shown in yellow. What are the areas of the pink and yellow regions?

Solution: Call the area of the pink region P and the area of the yellow region Y. We know that 4P + 4Y equals the area of the square. We also know that 2Y + P equals the area of one half-circle within the square. This information gives us a set of two equations in two unknowns.

4P + 4Y = 1
2Y + P = π/8

If we multiply the second equation by 2 and subtract it from the first we get

2P = 1 - π/4
P = 1/2 - π/8
P ≈ 0.1073

Solving back for Y gives us

Y = π/8 - 1/4
Y ≈ 0.1427


BONUS Problem #1

Two circles of equal size are inscribed in a square such that the circles are tangent to each other and to two sides of the square. The centers of the circle lie on the square's diagonal. If the radius of each circle is 1, what is the side length of the square?

Solution: As shown in the figure at left, if you connect the centers of the squares with a line segment, this line segment can be made into the hypotenuse of an isosceles right triangle. Because this line segment passes through the point of tangency between the two circles, its length is 2. The other two sides of the triangle have a length of 2/sqrt(2), which is equivalent to sqrt(2).

Now draw a line segment that connects the center of a circle with the point on the square's edge where it is tangent. Do this for both circles for all four points of tangency on the square's boundary. You can now see that the side length of the square is equal to

1 + sqrt(2) + 1

= 2 + sqrt(2).


BONUS Problem #2

Three circles have a radius of 6 and are inscribed in a rectangle as shown in the figure above. What is the area and perimeter of the rectangle?

Solution: It's easy to see that the width of the rectangle is four times the radius of the circles, 4*6 = 24. The height of the rectangle is a little harder to work out, but it can be figured in the same way as in Bonus Problem #1 by making a triangle with the radii and centers of the circles, as in the image on the right.

The triangle formed by the centers of the circles is equilateral with a side length of 12, and its height is 6*sqrt(3). Therefore, the height of the bounding rectangle is 6 + 6*sqrt(3) + 6 = 12 + 6*sqrt(3).

The area of the rectangle is 24*(12 + 6*sqrt(3)) = 288 + 144*sqrt(3). The perimeter of the rectangle is 24 + 12 + 6*sqrt(3) + 24 + 12 + 6*sqrt(3) = 72 + 12*sqrt(3).


Some Related Geometry Facts and Formulas

  • If a square and a circle have equal area, then the diameter of the circle is 2/sqrt(π) times the side length of the square. Equivalently, the side length of the square is sqrt(π)/2 times the diameter of the circle.
  • If a square and circle have equal perimeter, then the diameter of the circle is 4/π times the side length of the square. Equivalently, the side length of the square is π/4 times the diameter of the circle.
  • In terms of area, the largest quadrilateral that can be inscribed in a circle is a square.


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    • profile image

      xcvbxcvbxcvb 3 years ago

      this is VERY helpful, i was wondering if you could add another section to explain how to solve the following similar problem:

      2 equal sized circles are inscribed in a square such that the circles are tangent to each other and to 2 sides of the square. the centers of the circle lie on the diagonal of the square. if the radius of the circles is 1, what is the size of the square?

      sorry for lack of a picture but I hope that was enough info so that you can visualize it. thank you!!

    • calculus-geometry profile image
      Author

      TR Smith 3 years ago from Eastern Europe

      Not a problem at all. I actually had planned to include that one, but decided to stop at seven.

    • profile image

      kuroko 2 years ago

      can you give me the solution which 3 circles with radius 6 are inscribe in a rectangle? find the perimeter of the rectangle .

    • calculus-geometry profile image
      Author

      TR Smith 2 years ago from Eastern Europe

      @kuroko, I added it as a second bonus problem.

    • Blackspaniel1 profile image

      Blackspaniel1 2 years ago

      As a mathematician I could enjoy such puzzles, but can see them really challenging some. Yet, what better way to get confidence is there than for a person to solve two or three themselves. Nice problems

    • profile image

      Ben 2 years ago

      Voted up, very challenging and helpful for geometry study. If I may ask a related math problem, if a quadrilateral is inscribed in a circle, what is the maximum area and maximum perimeter it can have? Thank you.

    • calculus-geometry profile image
      Author

      TR Smith 2 years ago from Eastern Europe

      The maximum area and maximum perimeter for an inscribed quadrilateral are both achieved by a square. This means for a circle of radius R the max area is 2R^2 and the max perimeter is 4*sqrt(2)R.

    • profile image

      Equu 12 months ago

      A square has two circular arcs drawn inside of it. The centers of the arcs are opposite vertices of the square. If these two arcs touch at a single point without overlapping, what is the their combined area?

    • calculus-geometry profile image
      Author

      TR Smith 12 months ago from Eastern Europe

      Hi Equu, just to clarify, I think you mean to say these are quarter circle arcs and that they are wholly contained inside the square. Otherwise you can construct an arc at one corner that nearly touches the opposite corner and then fills up nearly the whole square, in which case the limit of the combined area would just be the area of the square.

      Anyway, if you suppose that the side length of the square is 1 and the diagonal length of the square is sqrt(2), then then the sum of the radii of the two arcs is sqrt(2) and they touch somewhere along the diagonal. If one of the radii is R, then the other radius is sqrt(2) - R. The combined area of the two quarter circular arcs is

      (pi/4)[ R^2 + (sqrt(2) - R)^2 ]

      Also, R cannot be greater than 1 and it cannot be less than sqrt(2) - 1.

      I suppose there is more to this problem than just finding an expression for the area. If this is a calculus problem, then it's likely you need to find the value of R that either minimizes or maximizes the total area. I'll let you work that out.

    • jonnycomelately profile image

      Alan 11 months ago from Tasmania

      TR, thank you for this fascinating hub.

      I have never been great at maths, but just getting a bit interested, because:

      I am trying to help a neighbour's son (13) improve his maths. He is certainly not dumb or unintelligent, but willing to open his mind. I am too!

      My focus so far is talking about ratio, because this is inherently part of so many facets of life: angles, triangles, fuel mix, road speed, levers, gears, etc. Mathisfun is proving a valuable Website.

      I will come back to more of your hubs as time goes by.

    • profile image

      soumyabrata chaudhuri 3 months ago

      in a circle of radius 4 cm ,10 small circles of the same radius are inscribed along the circumference,such that each circle is in contact with the other and in touch with the circumference.find the radii of the small circles?

    • Onno Kramer profile image

      Onno Kramer 2 months ago from Weesp

      Considering BONUS Problem #2, what is the radius of an inner circle which fits exactly between the other three bigger circles?

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