Measuring Space: Perimeter and Area | EOT

Question 18

Use the above to make a conjecture about the area occupied by circles fitted into a rectangle in the manner shown. Test your conjecture for particular cases: 10 circles; 20 circles; 50 circles. Then prove your conjecture!

Solution

Understand the Question

When identical circles of radius $r$ are fitted in a regular grid inside a rectangle, each circle is enclosed in a square bounding box of side length equal to its diameter $d = 2r$ .
The area of each circle is $\pi r^2$ , while the area of each enclosing square unit is $(2r)^2 = 4r^2$ .
Since the number of circles matches the number of square units, the ratio of the total circle area to the rectangle area is expected to remain constant at $\dfrac{\pi}{4} \approx 78.54\%$ , regardless of the number of circles $n$ .

Step 1 · State the Conjecture

Conjecture: The total area occupied by $n$ identical circles fitted in a grid inside a rectangle is always equal to $\dfrac{\pi}{4}$ of the area of the rectangle: $\text{Area of circles} = \dfrac{\pi}{4} \times \text{Area of rectangle}$ That is, the circles always occupy $\dfrac{\pi}{4} \approx 78.54\%$ of the total area of the rectangle, regardless of the number of circles.

Step 2 · Test the Conjecture for Particular Cases

Let the radius of each circle be $r$ . Area of one circle $= \pi r^2$ Area of the enclosing square cell for one circle $= (2r)^2 = 4r^2$

Case 1: $10$ circles ( $n = 10$ ) $\text{Area of circles} = 10 \times \pi r^2 = 10\pi r^2$ $\text{Area of rectangle} = 10 \times 4r^2 = 40r^2$ $\text{Ratio} = \dfrac{10\pi r^2}{40r^2} = \dfrac{\pi}{4}$

Case 2: $20$ circles ( $n = 20$ ) $\text{Area of circles} = 20 \times \pi r^2 = 20\pi r^2$ $\text{Area of rectangle} = 20 \times 4r^2 = 80r^2$ $\text{Ratio} = \dfrac{20\pi r^2}{80r^2} = \dfrac{\pi}{4}$

Case 3: $50$ circles ( $n = 50$ ) $\text{Area of circles} = 50 \times \pi r^2 = 50\pi r^2$ $\text{Area of rectangle} = 50 \times 4r^2 = 200r^2$ $\text{Ratio} = \dfrac{50\pi r^2}{200r^2} = \dfrac{\pi}{4}$

In all cases, the area occupied by the circles is $\dfrac{\pi}{4}$ of the area of the rectangle.

Step 3 · Prove the Conjecture Generally

Let $n$ identical circles of radius $r$ be arranged in a rectangular grid of $p$ rows and $q$ columns, where $n = p \times q$ .

The dimensions of the rectangle containing the circles are: $\text{Length} = q \times 2r = 2qr$ $\text{Breadth} = p \times 2r = 2pr$

Calculate the area of the rectangle:

\begin{aligned} \text{Area of rectangle} &= \text{Length} \times \text{Breadth} \\ &= (2qr) \times (2pr) \\ &= 4(pq)r^2 \\ &= 4nr^2 \end{aligned}

Calculate the total area occupied by the $n$ circles: $\text{Area of } n \text{ circles} = n \times \pi r^2 = n\pi r^2$

Divide the area of the circles by the area of the rectangle:

\begin{aligned} \dfrac{\text{Area of circles}}{\text{Area of rectangle}} &= \dfrac{n\pi r^2}{4nr^2} \\[0.6em] &= \dfrac{\pi}{4} \end{aligned}

$\therefore \text{Area of circles} = \dfrac{\pi}{4} \times \text{Area of rectangle}$

Hence, the conjecture is proved for any number of circles $n$ .

Answer

$\text{Area of circles} = \dfrac{\pi}{4} \times \text{Area of rectangle} \approx 78.54\% \text{ of the rectangle's area}$

Common Mistakes

Area of Bounding Cell: Taking the area of each enclosing cell as $r^2$ instead of $(2r)^2 = 4r^2$ .
Assuming Dependence on $n$ : Expecting the packing efficiency ratio to change as the number of circles increases; because both total circle area and rectangle area scale linearly with $n$ , the variable $n$ cancels out completely.
Radius vs Diameter Confusion: Using the diameter $d$ in the formula $\pi r^2$ without dividing by $2$ .

More questions in EOT

Identities in algebra can sometimes be shown as area relationships. For example:

The figure shown corresponds to the identity $(a + b)^2 = a^2 + 2ab + b^2$

Do you see how?

Draw figures corresponding to the identities $(a + b)(a - b) = a^2 - b^2$ and $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$ .

An isosceles triangle has perimeter $40\text{ cm}$ ; the equal sides are $15\text{ cm}$ each. Find the area of the triangle.

An isosceles triangle has base $10\text{ cm}$ , and its area is $60\text{ cm}^2$ . What are the lengths of the equal sides?

The area of a right-angled triangle is $54\text{ sq. cm}$ . One of its legs has length $12\text{ cm}$ . Find its perimeter.

The sides of a triangle are in the ratio $2 : 3 : 4$ , and its perimeter is $45\text{ cm}$ . Find its area.

The sides of a triangle have lengths $7\text{ cm}$ , $24\text{ cm}$ , $25\text{ cm}$ . Find the area of the triangle in two different ways.

If the wheel of a bicycle has a diameter of $60\text{ cm}$ , find how far a cyclist will have travelled after the wheel has rotated $100$ times.

Find the area of a quadrant of a circle whose circumference is $66 \text{ cm}$ .

The wheel of a car has an outer radius of $28\text{ cm}$ . Calculate how far the car travels after one complete turn of the wheel, and how many times the wheel turns during a journey of $1\text{ km}$ .

Q10

Two rectangles have the same area and the same perimeter. Does this mean that they are congruent to each other?

Q11

You know that the area of a parallelogram is base $\times$ height. Using this and the figure, show that the area of a trapezium is half the sum of the parallel sides $\times$ height, i.e., $\dfrac{1}{2}(a + b)h$ .

Q12

By dividing a trapezium into two triangles show that its area is, half the sum of the parallel sides multiplied by the height (the same formula as the one given above).

Q13

Show how we can use two identical copies of a trapezium to make a parallelogram. How will this give us the formula for the area of a trapezium?

Q14

Show that the area of a kite is half the product of its diagonals. Show this: (i) using algebra, and (ii) using geometry.

Q15

Three problems about fitting congruent shapes together:

(i) Rectangle ABCD has sides $a$ , $b$ , and rectangle PQRS has sides $2a$ , $2b$ . Show that PQRS has 4 times the area of ABCD. Does this mean that 4 copies of rectangle ABCD will fit into rectangle PQRS? Check and see!

(ii) $\Delta ABC$ has sides $a$ , $b$ , $c$ , and $\Delta PQR$ has sides $2a$ , $2b$ , $2c$ . Show that $\Delta PQR$ has 4 times the area of $\Delta ABC$ . Does this mean that 4 copies of $\Delta ABC$ will fit into $\Delta PQR$ ? Check and see!

(iii) $\Delta ABC$ has sides $a$ , $b$ , $c$ , and $\Delta PQR$ has sides $3a$ , $3b$ , $3c$ . Show that $\Delta PQR$ has 9 times the area of $\Delta ABC$ . Does this mean that 9 copies of $\Delta ABC$ will fit into $\Delta PQR$ ? Check and see!

Q16

Find the fraction of the shaded region in each of the following figures:

(i) Fig. 6.43: What fraction of the triangle is shaded? (ii) Fig. 6.44: What fraction of the square is shaded?

Q17

Find the fraction of the rectangle covered by the circles in each of the following figures:

(i) Fig. 6.45: What fraction of the rectangle is covered by the circles?

(ii) Fig. 6.46: What fraction of the rectangle is covered by the circles?

Q18

Q19

*19. The figure shows nine identical rectangles fitted together to make a large rectangle whose area is $72\text{ cm}^2$ . Find the perimeter of each small rectangle.

Q20

Show that the areas of the shaded blue triangle and the shaded red triangle are equal.

Find a way of cutting up the blue triangle into some number of pieces and rearranging the pieces to cover the red triangle.

Q21

The figure shows a quarter circle in a square. Its centre is at one vertex, and it passes through two adjacent vertices. There are two semicircles on two adjacent sides as diameters. They create the shaded regions $A$ and $B$ .

Show that $A$ and $B$ have equal area.

Q22

In Fig. 6.50, four semicircles have been drawn within the given square whose side is $2 \text{ units}$ . The centres of these semicircles are the midpoints of the sides. They create a 4-petalled flower (shown in blue). Find the perimeter and the area of this flower.

Q23

In Fig. 6.51 we see two concentric circles with a common centre $O$ . A chord $BC$ of the larger circle is drawn, touching the smaller circle at $A$ . The length of $BC$ is $l$ . Show that the area of the green region enclosed between the two circles is $\dfrac{1}{4} \pi l^2$ .

Q24

In Fig. 6.52, semicircles have been drawn on all the sides of a right-angled triangle as shown. Show that $\text{Area (A)} + \text{Area (B)} = \text{Area (C)}$ .

Q25

Fig. 6.53 shows two circles passing through each other's centres. Find the area of the region enclosed by the two circles in terms of the common radius $r$ .

Q26

In Fig. 6.54, we see three triangles within a rectangle. The areas of the triangles are $A, B, C$ , as marked. Show that the area of the rectangle is

$\dfrac{2(A + C)(B + C)}{C}$

Q27

In the figure we see two shaded regions formed by a quarter circle, a semicircle, and a triangle.

Show that the areas of the two shaded regions are equal.