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!
Step-by-step solution
Idea: The answers for 3 and 4 circles were the same, so guess the fraction never changes. A proof must work for every number n.
- Conjecture: for any number of equal circles fitted in a row inside a rectangle, the fraction covered is π4 (about 0.785; 1114 with π = 227).1 mark
- Tests (radius r): 10 circles: rectangle 20r × 2r = 40r2, circles 10πr2, fraction π4. 20 circles: 80r2 and 20πr2, fraction π4. 50 circles: 200r2 and 50πr2, fraction π4.1 mark
- Proof: with n circles of radius r in a row, the rectangle has length n × 2r = 2nr and width 2r, so area 4nr2. The circles have area nπr2.1 mark
- Fraction = nπr24nr2 = π4. Both n and r cancel, so the fraction is π4 for every number of circles of any size. Proved.1 mark
Answer to write in the exam
Conjecture: fraction covered = π4 for any number of circles
n = 10: 10πr240r2 = π4; n = 20: 20πr280r2 = π4; n = 50: 50πr2200r2 = π4
General: rectangle = 2nr × 2r = 4nr2; circles = nπr2
∴ Fraction = nπr24nr2 = π4 for every n
Common mistakes that cost marks
- Checking only a few cases and calling it a proof. The proof must use a general n.
- Expecting the fraction to approach 1 as the number of circles grows.
How this can come in the exam
Assertion (A): 25 equal circles fitted in a row in a rectangle cover π4 of it.
Reason (R): Each circle covers π4 of the 2r × 2r square it sits in.
- Both A and R are true, and R is the correct explanation of A.
- Both A and R are true, but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.
Show answer
(A) Both A and R are true, and R is the correct explanation of A.
The rectangle is made of 25 such squares, each π4 covered, so A follows from R.
Try one yourself
6 circles of radius 3.5 cm are fitted in a row in a rectangle. Find the area of the rectangle not covered. (Use π = 227.)
Show answer
Rectangle 42 × 7 = 294; circles 6 × 38.5 = 231; uncovered 63 cm2 (= 314 of 294).
More questions like this
- The figure shows nine identical rectangles fitted together to make a large rectangle whose area is 72 cm2. Find the perimeter of each small rectangle.
- 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.
- 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.
- In the figure, four semicircles have been drawn within the given square whose side is 2 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.
- In the figure 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 14πl2.