Question 42 of 42advanced✨ CreateLong Answer5 marks

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!

Correct Answer

Conjecture: When identical circles are fitted into a rectangle in a grid pattern (touching each other and the sides of the rectangle), the fraction of the area of the rectangle occupied by the circles is always π/4\pi/4, regardless of the number of circles.

Testing for 10 circles: Let the circles be arranged in 2 rows and 5 columns with radius rr. The area of 10 circles is 10πr210\pi r^2. The rectangle's length is 5×2r=10r5 \times 2r = 10r and breadth is 2×2r=4r2 \times 2r = 4r. The area of the rectangle is 10r×4r=40r210r \times 4r = 40r^2. The fraction occupied is:

10πr240r2=π4\frac{10\pi r^2}{40r^2} = \frac{\pi}{4}

Testing for 20 circles: Let the circles be arranged in 4 rows and 5 columns. The area of 20 circles is 20πr220\pi r^2. The rectangle's length is 10r10r and breadth is 8r8r. The area of the rectangle is 80r280r^2. The fraction occupied is:

20πr280r2=π4\frac{20\pi r^2}{80r^2} = \frac{\pi}{4}

Testing for 50 circles: Let the circles be arranged in 5 rows and 10 columns. The area of 50 circles is 50πr250\pi r^2. The rectangle's length is 20r20r and breadth is 10r10r. The area of the rectangle is 200r2200r^2. The fraction occupied is:

50πr2200r2=π4\frac{50\pi r^2}{200r^2} = \frac{\pi}{4}

Proof of the conjecture: Let nn circles be arranged in mm rows and kk columns, so n=m×kn = m \times k. The radius of each circle is rr. The total area of the circles is nπr2n\pi r^2. The length of the rectangle is 2kr2kr and the breadth is 2mr2mr. The area of the rectangle is:

2kr×2mr=4mkr2=4nr22kr \times 2mr = 4mkr^2 = 4nr^2

The fraction of area occupied is:

nπr24nr2=π4\frac{n\pi r^2}{4nr^2} = \frac{\pi}{4}

This proves that the fraction is always π/4\pi/4.

Exercise: END-OF-CHAPTER EXERCISES | Q: 18 | (Page 34)
For More Understanding

Explanation

The question requires formulating and proving a conjecture about the area occupied by circles packed in a rectangular grid. By testing specific cases (10, 20, and 50 circles), a pattern emerges where the fraction of the rectangle's area covered by the circles consistently equals π/4\pi/4.

The general proof confirms this by showing that the total area of the circles (nπr2n\pi r^2) divided by the area of the bounding rectangle (4nr24nr^2) simplifies to π/4\pi/4, as the number of rows and columns cancel out.

Solution Steps

  1. Step 1: State the conjecture that the fraction of the rectangle's area covered by the circles is π/4\pi/4.

  2. Step 2: Test for 10 circles (e.g., 2 rows, 5 cols): Area of circles = 10πr210\pi r^2, Area of rectangle = 40r240r^2, Fraction = π/4\pi/4.

  3. Step 3: Test for 20 circles (e.g., 4 rows, 5 cols): Area of circles = 20πr220\pi r^2, Area of rectangle = 80r280r^2, Fraction = π/4\pi/4.

  4. Step 4: Test for 50 circles (e.g., 5 rows, 10 cols): Area of circles = 50πr250\pi r^2, Area of rectangle = 200r2200r^2, Fraction = π/4\pi/4.

  5. Step 5: Prove generally for nn circles (mm rows, kk cols): Area of circles = nπr2n\pi r^2, Area of rectangle = 4mkr2=4nr24mkr^2 = 4nr^2, Fraction = nπr24nr2=π/4\dfrac{n\pi r^2}{4nr^2} = \pi/4.

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