Practice Set 2 – Solutions ചുറ്റളവ് കണക്കുകൾ
Five questions on circumference. In each, the real work is finding the radius or diameter – usually with Pythagoras – and then c = πd = 2πr. Answers are left in terms of π, as the textbook suggests.
- c = πd = 2πr (page 7).
- A regular hexagon in a circle has side = radius; a square or rectangle in a circle has its diagonal as a diameter.
- The perpendicular from the centre to a chord (ഞാൺ) cuts it in half; then r² = (distance)² + (half chord)².
Question 1 · Circles around polygons
ചുവടെ വരച്ചിരിക്കുന്ന വൃത്തങ്ങളുടെയെല്ലാം ചുറ്റളവ് കണക്കാക്കുക.
Q1. Find the circumference of each circle: (a) a regular hexagon of side 2 cm drawn in it (b) a square of side 2 cm drawn in it (c) a rectangle 2 cm × 1.5 cm drawn in it.
(a) The hexagon is six equilateral triangles. (b), (c) The diagonal of the square or rectangle passes through the centre.
(a) radius 2 cm ⇒ c = 2π × 2 = 4π cm (≈ 12.57 cm). (b) diameter = diagonal = 2√2 cm ⇒ c = 2√2π cm (≈ 8.89 cm). (c) diameter = √(2² + 1.5²) = √6.25 = 2.5 cm ⇒ c = 2.5π cm (≈ 7.85 cm).
Question 2 · A chord and its distance
Q2. A chord 6 cm long is 4 cm away from the centre of a circle. Find the circumference.
Draw the perpendicular from the centre: it meets the chord at its middle, making a right triangle with sides 3 and 4.
Half chord 3, distance 4 ⇒ r = √(3² + 4²) = 5 cm ⇒ c = 2π × 5 = 10π cm (≈ 31.42 cm).
Question 3 · An isosceles triangle in a circle
Q3. An isosceles triangle with base 4 cm and height 4 cm is drawn with its three vertices on a circle. Find the circumference.
By symmetry the centre lies on the height. Call its distance from the base k; it is r from the base ends and r from the top vertex.
Distance from centre to top = 4 − k; to a base end = √(2² + k²). Both are r: (4 − k)² = 4 + k² ⇒ 16 − 8k = 4 ⇒ k = 1.5, r = 2.5 cm ⇒ c = 5π cm (≈ 15.71 cm).
Question 4 · Circles along a diameter
Q4. In each figure the centres of the circles are on one line, and the small circles fill the diameter of the big circle (two equal, three equal, and three unequal). Prove that in every figure the sum of the circumferences of the small circles equals the circumference of the big circle.
Write each circumference as π × its diameter, then add.
If the small diameters are d₁, d₂, d₃, … then d₁ + d₂ + d₃ + … = D, the big diameter. So πd₁ + πd₂ + πd₃ + … = π(d₁ + d₂ + d₃ + …) = πD, the big circumference. ∎
Question 5 · One centimetre more
Q5. Two circles have the same centre; the radius of the bigger one is 1 cm more than that of the smaller. How much more is the bigger circumference?
Radii r and r + 1. Subtract 2πr from 2π(r + 1).
The difference is always 2π ≈ 6.28 cm. A rope round the Earth's equator, lifted 1 metre all round, needs only 2π ≈ 6.28 m more rope!
2π(r + 1) − 2πr = 2π: the bigger circumference is 2π cm ≈ 6.28 cm more, whatever the radii.
Check yourself
- Hexagon side = radius; square/rectangle diagonal = diameter. (a) 4π (b) 2√2π (c) 2.5π cm.
- Chord 6 at distance 4 ⇒ r = 5 ⇒ 10π cm. Isosceles triangle base 4, height 4 ⇒ r = 2.5 ⇒ 5π cm.
- Small circles filling a diameter: circumferences add up to the big one. Radii 1 cm apart ⇒ circumferences 2π cm apart.
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