Circles · word problems

Circle Heuristics

10 question types · PSLE Paper 1 and 2 · one figure that follows the deduction

01

Perimeter and Area of a Quadrant

propertyA quadrant is a quarter of the circle: a quarter of the circumference, plus two radii

OAB is a quadrant of a circle with centre O and radius 14 cm. Take π = 227. Find the perimeter and the area of the quadrant.

OAB14 cm
A quarter of a circle of radius 14 cm.
Circumference of the full circle = 2 × 227 × 14 = 88 cm.
step 1 of 5

The curved edge is a quarter of the full circumference. The perimeter is that arc and the two straight radii; the area is a quarter of the circle's.

  1. Circumference of the full circle = 2 × 227 × 14 = 88 cm.
  2. Arc AB = 14 × 88 = 22 cm.
  3. Perimeter = 22 + 14 + 14 = 50 cm.
  4. Area of the full circle = 227 × 14 × 14 = 616 cm2.
  5. Area of the quadrant = 14 × 616 = 154 cm2.

answer50 cm; 154 cm2

Common pitfalls

  • Giving the arc alone, 22 cm, as the perimeter: the two radii are edges of the shape too.
  • Quartering the radius instead of the area: a quadrant of radius 14 is not a circle of radius 3.5.
02

Perimeter of a Rectangle with a Semicircle

propertyOnly the outside edges count: the side the semicircle sits on is inside the shape

A shape is made of a rectangle 20 cm by 14 cm with a semicircle attached to one of its 14 cm sides. Take π = 227. Find the perimeter and the area of the shape.

20 cm14 cm7
The semicircle sits on a 14 cm side, so its radius is 7 cm.
Radius of the semicircle = 14 ÷ 2 = 7 cm.
step 1 of 5

The semicircle's diameter is 14 cm, so its radius is 7 cm. Its arc replaces one side of the rectangle in the outline; its area adds to the rectangle's.

  1. Radius of the semicircle = 14 ÷ 2 = 7 cm.
  2. Arc length = 12 × 2 × 227 × 7 = 22 cm.
  3. The outline: two 20 cm sides, one 14 cm side, and the arc. The other 14 cm side is inside the shape.
  4. Perimeter = 20 + 20 + 14 + 22 = 76 cm.
  5. Area = 20 × 14 + 12 × 227 × 7 × 7 = 280 + 77 = 357 cm2.

answer76 cm; 357 cm2

Common pitfalls

  • Adding all four sides of the rectangle and the arc: the side under the semicircle is not on the outline.
  • Using 14 as the radius of the semicircle; it is the diameter.
03

A Ring Between Two Circles

propertyBig circle minus small circle; the ring's width is the difference of the radii

Two circles have the same centre. The inner circle has radius 7 cm and the ring between the circles is 7 cm wide. Take π = 227. Find the area of the ring and the circumference of the outer circle.

7714
Inner radius 7 cm plus a 7 cm ring: outer radius 14 cm.
Outer radius = 7 + 7 = 14 cm.
step 1 of 4

The outer radius is the inner radius plus the ring's width. The ring is what is left when the inner circle is taken out of the outer one.

  1. Outer radius = 7 + 7 = 14 cm.
  2. Outer area = 227 × 14 × 14 = 616 cm2; inner area = 227 × 7 × 7 = 154 cm2.
  3. Ring = 616 − 154 = 462 cm2.
  4. Outer circumference = 2 × 227 × 14 = 88 cm.

answer462 cm2; 88 cm

Common pitfalls

  • Squaring the difference of the radii: π × 72 is the inner circle, not the ring.
  • Taking the ring's width as the outer radius.
04

Semicircle on the Hypotenuse

propertyA right-angled corner on the arc: the hypotenuse is the diameter, and the region outside the triangle is semicircle minus triangle

Triangle ABC has a right angle at C, with AC = 6 cm, BC = 8 cm and AB = 10 cm. A semicircle is drawn with AB as diameter, passing through C. Take π = 3.14. Find the area of the semicircle and the area inside the semicircle but outside the triangle.

ABC6810 cm5
AB is the diameter, so the radius is 5 cm and C sits on the arc.
Radius = 10 ÷ 2 = 5 cm.
step 1 of 4

The diameter is the hypotenuse, so the radius is half of it. The triangle sits inside the semicircle; subtract its area from the semicircle's.

  1. Radius = 10 ÷ 2 = 5 cm.
  2. Area of the semicircle = 12 × 3.14 × 5 × 5 = 39.25 cm2.
  3. Area of the triangle = 12 × 6 × 8 = 24 cm2, using the two legs as base and height.
  4. Inside the semicircle, outside the triangle: 39.25 − 24 = 15.25 cm2.

answer39.25 cm2; 15.25 cm2

Common pitfalls

  • Using 10 cm as the radius.
  • Finding the triangle's area with the hypotenuse as base and a leg as height; the height to AB is neither leg.
05

Semicircles on a Shared Diameter

propertySubtract the two small semicircles from the large one; every boundary here is an arc

A semicircle has diameter AB = 28 cm. Two smaller semicircles, each of diameter 14 cm, are drawn inside it on AB, touching at the midpoint. Take π = 227. Find the area of the region inside the large semicircle but outside the two small ones, and the total length of its curved boundary.

AB28 cm14
The large semicircle has radius 14 cm.
Large semicircle: radius 14 cm, area = 12 × 227 × 14 × 14 = 308 cm2.
step 1 of 5

The region is the large semicircle with two small semicircles removed. Its boundary is three arcs and no straight line at all, because the two small semicircles cover the whole diameter.

  1. Large semicircle: radius 14 cm, area = 12 × 227 × 14 × 14 = 308 cm2.
  2. Each small semicircle: radius 7 cm, area = 12 × 227 × 7 × 7 = 77 cm2.
  3. Region = 308 − 77 − 77 = 154 cm2.
  4. Curved boundary: large arc = 12 × 2 × 227 × 14 = 44 cm; each small arc = 12 × 2 × 227 × 7 = 22 cm.
  5. Total boundary = 44 + 22 + 22 = 88 cm.

answer154 cm2; 88 cm

Common pitfalls

  • Adding the 28 cm diameter to the boundary: the two small arcs run along it, so no part of AB is an edge of the region.
  • Using one small semicircle where there are two.
06

Semicircle Inside a Quadrant

propertyQuadrant minus the semicircle on its radius; the boundary is two arcs and one radius

OAB is a quadrant of radius 14 cm. A semicircle with OA as diameter is drawn inside it. Take π = 227. Find the area of the region inside the quadrant but outside the semicircle, and the perimeter of that region.

OAB14 cm
Quadrant of radius 14: 154 cm².
Quadrant area = 14 × 227 × 14 × 14 = 154 cm2.
step 1 of 5

The semicircle's diameter is the radius of the quadrant, so its radius is 7 cm. The region's outline is the quadrant's arc, the semicircle's arc, and the radius OB.

  1. Quadrant area = 14 × 227 × 14 × 14 = 154 cm2.
  2. Semicircle: radius 7 cm, area = 12 × 227 × 7 × 7 = 77 cm2.
  3. Region = 154 − 77 = 77 cm2.
  4. Quadrant arc = 14 × 2 × 227 × 14 = 22 cm; semicircle arc = 12 × 2 × 227 × 7 = 22 cm.
  5. Perimeter = 22 + 22 + 14 = 58 cm, the 14 being the radius OB.

answer77 cm2; 58 cm

Common pitfalls

  • Including OA in the perimeter: the semicircle's arc runs from O to A, so OA is inside the region.
  • Taking the semicircle's radius as 14 cm.
07

Two Circles Through Each Other's Centres

propertyCentre, centre and a crossing point are all one radius apart: an equilateral triangle

Two circles of equal radius, with centres O and P, each pass through the other's centre. They cross at A and B. Find ∠ AOP and ∠ AOB.

OPAB
OP is a radius of both circles; so are OA and PA of one each.
OA = OP (radii of the circle with centre O) and PA = PO (radii of the circle with centre P).
step 1 of 4

OA and OP are radii of the first circle, PA and PO radii of the second, so all three sides of triangle OAP are equal. The same holds below the line of centres.

  1. OA = OP (radii of the circle with centre O) and PA = PO (radii of the circle with centre P).
  2. So OA = OP = PA: triangle OAP is equilateral and ∠ AOP = 60°.
  3. By the same argument triangle OBP is equilateral: ∠ BOP = 60°.
  4. ∠ AOB = 60° + 60° = 120°.

answer∠ AOP = 60°; ∠ AOB = 120°

Common pitfalls

  • Assuming ∠ AOB = 90° from the symmetry of the picture: the symmetry gives equal halves, not their size.
  • Missing that OP is a radius of both circles, which is the fact that makes the triangle equilateral.
08

Three Touching Circles and the Gap Between

propertyJoining the centres gives an equilateral triangle; the gap's edges are three 60° arcs

Three identical circles of radius 7 cm touch each other. Their centres are joined to make a triangle. Take π = 227. Find the angle of that triangle at each centre, and the length of the boundary of the curved gap enclosed between the three circles.

714
Touching circles: centres 7 + 7 = 14 cm apart, all three ways.
Touching circles: the distance between two centres is 7 + 7 = 14 cm, so the centre triangle is equilateral.
step 1 of 4

Two touching circles have their centres one diameter apart, so all three sides of the centre triangle are 14 cm and every angle is 60°. Each side of the gap is the arc of one circle cut off by that 60°.

  1. Touching circles: the distance between two centres is 7 + 7 = 14 cm, so the centre triangle is equilateral.
  2. Each angle at a centre = 60°.
  3. Each arc of the gap is a 60° piece of a circle: 60360 = 16 of the circumference 2 × 227 × 7 = 44 cm, so 446 = 713 cm.
  4. Three arcs: 3 × 713 = 22 cm.

answer60°; 22 cm

Common pitfalls

  • Measuring the gap's boundary with straight lines between the touching points.
  • Using one quarter of the circumference for each arc: the angle at the centre is 60°, not 90°.
09

A Rolling Wheel

propertyOne turn moves the wheel one circumference along the ground

A wheel has radius 35 cm. Take π = 227. How far does it travel in 100 turns? How many turns does it make over 44 m?

35
The rim is 2 × 227 × 35 = 220 cm round.
Circumference = 2 × 227 × 35 = 220 cm.
step 1 of 4

Each full turn lays the rim's whole length on the ground once. Distance is turns times circumference; turns is distance divided by circumference, in the same units.

  1. Circumference = 2 × 227 × 35 = 220 cm.
  2. 100 turns: 100 × 220 = 22 000 cm = 220 m.
  3. 44 m = 4400 cm.
  4. Turns = 4400 ÷ 220 = 20.

answer220 m; 20 turns

Common pitfalls

  • Dividing 44 by 220 without changing metres to centimetres first.
  • Using the radius or the diameter as the distance per turn.
10

A Rope Tied at a Corner

propertyThe rope sweeps a circle, and the wall blocks a quarter of it

A goat is tied by a 14 m rope to the outside corner of a large shed with straight walls, longer than 14 m in both directions. Take π = 227. Find the area the goat can graze, and the length of the curved edge of that area.

shed14 m
Without the shed the rope would sweep a full circle: 616 m².
Full circle area = 227 × 14 × 14 = 616 m2.
step 1 of 4

With no wall, the rope would sweep a full circle of radius 14 m. The two walls meet at a right angle at the corner and block exactly one quarter of it.

  1. Full circle area = 227 × 14 × 14 = 616 m2.
  2. The walls block a quarter: grazing area = 34 × 616 = 462 m2.
  3. Full circumference = 2 × 227 × 14 = 88 m.
  4. Curved edge = 34 × 88 = 66 m.

answer462 m2; 66 m

Common pitfalls

  • Giving a semicircle, as if the goat were tied to a flat wall rather than a corner.
  • Adding the two 14 m rope lines along the walls to a perimeter that asked only for the curved edge.
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