Circles and Geometric Shapes | EOT

Question 18

Two parallel chords of lengths $10\text{ cm}$ and $24\text{ cm}$ are on the same side of the centre of a circle. The distance between the chords is $7\text{ cm}$ . Find the radius of the circle.

Solution

Understand the Question

When identical circles of radius $r$ are arranged in a regular grid inside a rectangle, each circle fits perfectly inside a square bounding box of side length $2r$ .
The area of each circle is $\pi r^2$ , and the area of its bounding square is $(2r)^2 = 4r^2$ .
Therefore, the ratio of the area of the circles to the area of the rectangle is always constant at $\dfrac{\pi r^2}{4r^2} = \dfrac{\pi}{4}$ , regardless of the number of circles or their radius.

Step 1 · Make a Conjecture

Let each circle have radius $r$ , so its area is $\pi r^2$ and diameter is $2r$ .

Let $N$ circles be arranged in $m$ rows and $n$ columns, so $N = m \times n$ . Diagram 1

$\text{Width} = n \times 2r, \quad \text{Height} = m \times 2r$

\begin{aligned} \text{Area of rectangle} &= (n \times 2r) \times (m \times 2r) \\[0.6em] &= n \times m \times 4r^2 \\[0.6em] &= N \times 4r^2 \end{aligned}

\begin{aligned} \text{Ratio} &= \dfrac{\text{Total area of circles}}{\text{Area of rectangle}} \\[0.6em] &= \dfrac{N \times \pi r^2}{N \times 4r^2} \\[0.6em] &= \dfrac{\pi}{4} \end{aligned}

Conjecture: The area occupied by the circles is always $\dfrac{\pi}{4}$ of the area of the rectangle.

Step 2 · Test for 10 Circles

Arrange $10$ circles in a $2 \times 5$ grid ( $m = 2, n = 5$ ) with radius $r = 1\text{ unit}$ :

$\text{Width} = 5 \times (2 \times 1) = 10\text{ units}$
$\text{Height} = 2 \times (2 \times 1) = 4\text{ units}$
$\text{Area of rectangle} = 10 \times 4 = 40\text{ sq units}$
$\text{Total area of 10 circles} = 10 \times \pi (1)^2 = 10\pi\text{ sq units}$

Checking conjecture: $\dfrac{\pi}{4} \times 40 = 10\pi$

The conjecture holds for $10$ circles.

Step 3 · Test for 20 Circles

Arrange $20$ circles in a $4 \times 5$ grid ( $m = 4, n = 5$ ) with radius $r = 1\text{ unit}$ :

$\text{Width} = 5 \times (2 \times 1) = 10\text{ units}$
$\text{Height} = 4 \times (2 \times 1) = 8\text{ units}$
$\text{Area of rectangle} = 10 \times 8 = 80\text{ sq units}$
$\text{Total area of 20 circles} = 20 \times \pi (1)^2 = 20\pi\text{ sq units}$

Checking conjecture: $\dfrac{\pi}{4} \times 80 = 20\pi$

The conjecture holds for $20$ circles.

Step 4 · Test for 50 Circles

Arrange $50$ circles in a $5 \times 10$ grid ( $m = 5, n = 10$ ) with radius $r = 1\text{ unit}$ :

$\text{Width} = 10 \times (2 \times 1) = 20\text{ units}$
$\text{Height} = 5 \times (2 \times 1) = 10\text{ units}$
$\text{Area of rectangle} = 20 \times 10 = 200\text{ sq units}$
$\text{Total area of 50 circles} = 50 \times \pi (1)^2 = 50\pi\text{ sq units}$

Checking conjecture: $\dfrac{\pi}{4} \times 200 = 50\pi$

The conjecture holds for $50$ circles.

Step 5 · Prove the Conjecture

For any grid of $N$ identical circles of radius $r$ in $m$ rows and $n$ columns ( $N = m \times n$ ):

$\text{Width of rectangle} = n \times (2r)$ $\text{Height of rectangle} = m \times (2r)$

\begin{aligned} \text{Area of rectangle} &= (n \times 2r) \times (m \times 2r) \\[0.6em] &= n \times m \times 4r^2 \\[0.6em] &= N \times 4r^2 \end{aligned}

\begin{aligned} \dfrac{\text{Total area of circles}}{\text{Area of rectangle}} &= \dfrac{N \pi r^2}{N \times 4r^2} \\[0.6em] &= \dfrac{\pi}{4} \end{aligned}

Thus, the area occupied by the circles is always $\dfrac{\pi}{4}$ of the rectangle's area.

Answer

The area occupied by the circles is always $\dfrac{\pi}{4}$ times the area of the rectangle.

Common Mistakes

Using Radius Instead of Diameter: Taking the side lengths of the rectangle as $n \times r$ and $m \times r$ instead of using the diameter $2r$ .
Assuming Packing Fraction Depends on Circle Size or Count: Thinking the ratio $\dfrac{\pi}{4} \approx 78.54\%$ changes when more circles are added; the ratio remains identical for any rectangular grid of tangent circles.

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Q10

A cyclic quadrilateral has sides $5$ , $5$ , $12$ , $12$ units. Find its area.

Q11

Consider a cyclic quadrilateral. Without drawing its circumcircle, how can we find out whether the centre of the circumcircle lies inside the quadrilateral or outside? What is the best way of finding out?

Q12

When two chords intersect, each of them is divided into two line segments. Show that if the intersecting chords are of equal length, then the line segments of one chord are equal to the corresponding line segments of the other chord.

Q13

Draw a circle in which a chord of $6\text{ cm}$ length stands at a distance of $3\text{ cm}$ from the centre.

(Hint: Is it a circumcircle of a suitable triangle?)

Q14

Show that rectangle is the only parallelogram that can be inscribed in a circle.

Q15

Show that if a rectangle is inscribed in a circle, then the point of intersection of its diagonals must lie at the centre of the circle.

Q16

Consider all chords of a circle of a fixed length. What is the shape formed by the midpoints of all these chords?

Q17

In a circle with centre $O$ , chords $AB$ and $AC$ are congruent. Explain why this statement is true: "The centre of the circle lies on the angle bisector of $\angle BAC$ ".

Q18

Two parallel chords of lengths $10\text{ cm}$ and $24\text{ cm}$ are on the same side of the centre of a circle. The distance between the chords is $7\text{ cm}$ . Find the radius of the circle.

Q19

A regular hexagon is inscribed in a circle of radius $r$ . Find the length of the sides of the hexagon and the distance of each side from the centre of the circle.

Q20

A quadrilateral $MNOP$ is inscribed in a circle. If $MN$ is a diameter, what can you say about $\angle MOP$ and $\angle MNP$ ? Explain your reasoning.

Q21

Let $ABCD$ be a cyclic quadrilateral. Explain why the exterior angle at any vertex is equal to the interior opposite angle (e.g., $\angle \text{CDE} = \angle \text{ABC}$ , where $E$ is a point on the extension of side $CD$ ).

Q22

"There is no chord of a circle that is longer than its diameter." How do you justify this statement?

Q23

Let $A$ be any point within a given circle with centre $O$ . Show that the shortest chord of the circle that passes through point $A$ is the one that is perpendicular to $OA$ .

Q24

How would you use the following figure to justify the statement that the angle in a semicircle is $90^\circ$ ?

Q25

In a circle, two chords $CC'$ and $DD'$ are drawn perpendicular to a diameter $AB$ . Prove that the segment $MM'$ joining the midpoints of the chords $CD$ and $C'D'$ is perpendicular to $AB$ .

Q26

How would you use the following figure to justify the statement that the sum of the opposite angles of a cyclic quadrilateral is $180^\circ$ ?