Find the perimeters of the following shapes (taking the arcs to be quarter or half or three-quarters of a circle, as appropriate) (the figure (i) to (ix)):
- (i) rectangle 80 m by 60 m with a semicircle on each 60 m end
- (ii) half ring: outer diameter 12 cm, inner diameter 8 cm
- (iii) square of side 10 cm with a semicircle on each side
- (iv) equilateral triangle of side 12 cm with a semicircle on each side
- (v) cross of five squares of side 14 cm with semicircles and quarter circles
- (vi) semicircle on 28 cm with four small semicircles along the diameter
- (vii) right triangle with legs 8 cm and 6 cm and a semicircle on each side
- (viii) semicircle on 12 cm with three semicircles on 4 cm parts inside
- (ix) semicircle on 20 cm with semicircles on the two 10 cm halves
Step-by-step solution
Idea: Walk round the outside edge only. Dashed lines are inside the shape and do not count. A semicircle on a diameter d has length πd2; a quarter circle of radius r has length πr2.
(i) rectangle 80 m by 60 m with a semicircle on each 60 m end
- Boundary: the two 80 m sides, plus a semicircle of diameter 60 m at each end (the dashed 60 m lines are inside, so they are not counted).
- Two semicircles of diameter 60 m = one circle: π × 60 = 227 × 60 = 13207 ≈ 188.57 m. Perimeter = 2 × 80 + 188.57 = 348.57 m.½ mark
(ii) half ring: outer diameter 12 cm, inner diameter 8 cm
- Boundary: outer semicircle (radius 6 cm), inner semicircle (radius 4 cm) and two flat ends, each (12 − 8) ÷ 2 = 2 cm.
- Perimeter = π × 6 + π × 4 + 2 + 2 = 10π + 4 = 2207 + 4 = 2487 ≈ 35.43 cm.½ mark
(iii) square of side 10 cm with a semicircle on each side
- Boundary: four semicircles, each on a 10 cm side (radius 5 cm). The square’s sides are inside.
- Perimeter = 4 × π × 5 = 20π = 4407 ≈ 62.86 cm.½ mark
(iv) equilateral triangle of side 12 cm with a semicircle on each side
- Boundary: three semicircles, each on a 12 cm side (radius 6 cm).
- Perimeter = 3 × π × 6 = 18π = 3967 ≈ 56.57 cm.½ mark
(v) cross of five squares of side 14 cm with semicircles and quarter circles
- All the small segments are equal (14 cm). The boundary has 4 small semicircles (diameter 14 cm, on the four end squares) and 4 big quarter circles (radius 14 cm, one at each corner between the arms).
- 4 semicircles: 4 × 227 × 7 = 4 × 22 = 88 cm. 4 quarter circles = one full circle of radius 14: 2 × 227 × 14 = 88 cm.½ mark
- Perimeter = 88 + 88 = 176 cm.½ mark
(vi) semicircle on 28 cm with four small semicircles along the diameter
- Boundary: the big semicircle (radius 14 cm) and four small semicircles on equal parts of 28 ÷ 4 = 7 cm (radius 3.5 cm), two below and two above the line.
- Perimeter = π × 14 + 4 × π × 3.5 = 44 + 44 = 88 cm.½ mark
(vii) right triangle with legs 8 cm and 6 cm and a semicircle on each side
- Hypotenuse = √(82 + 62) = √100 = 10 cm. The boundary is three semicircles, on diameters 8, 6 and 10 cm.
- Perimeter = π2(8 + 6 + 10) = 12π = 2647 ≈ 37.71 cm.½ mark
(viii) semicircle on 12 cm with three semicircles on 4 cm parts inside
- Boundary: the big semicircle (diameter 12 cm) and three small semicircles (diameter 4 cm each).
- Perimeter = π2 × 12 + 3 × π2 × 4 = 6π + 6π = 12π = 2647 ≈ 37.71 cm.½ mark
(ix) semicircle on 20 cm with semicircles on the two 10 cm halves
- Boundary: the big semicircle (diameter 20 cm), a small semicircle above the left half and a small semicircle below the right half (diameter 10 cm each).
- Perimeter = π2 × 20 + 2 × π2 × 10 = 10π + 10π = 20π = 4407 ≈ 62.86 cm.½ mark
Check: In (vi), (viii) and (ix) the small semicircles always add up to the same as one semicircle on the whole line they sit on, because a semicircle’s length is proportional to its diameter.
Answer to write in the exam
(i)
Perimeter = 2 × 80 + 2 × 12π × 60
= 160 + 227 × 60 = 160 + 188.57
∴ Perimeter ≈ 348.57 m
(ii)
Outer arc = π × 6, inner arc = π × 4
Flat ends = 2 × 12 − 82 = 4 cm
Perimeter = 10π + 4 = 2207 + 4
∴ Perimeter ≈ 35.43 cm
(iii)
Perimeter = 4 × (π × 5) = 20π
= 20 × 227
∴ Perimeter ≈ 62.86 cm
(iv)
Perimeter = 3 × (π × 6) = 18π
= 18 × 227
∴ Perimeter ≈ 56.57 cm
(v)
4 semicircles (r = 7) = 4 × π × 7 = 88 cm
4 quarter circles (r = 14) = 2π × 14 = 88 cm
∴ Perimeter = 88 + 88 = 176 cm
(vi)
Big semicircle = π × 14 = 44 cm
4 small semicircles = 4 × π × 3.5 = 44 cm
∴ Perimeter = 88 cm
(vii)
Hypotenuse = √(82 + 62) = 10 cm
Perimeter = π2 × 8 + π2 × 6 + π2 × 10 = 12π
∴ Perimeter = 2647 ≈ 37.71 cm
(viii)
Big semicircle = π2 × 12 = 6π
3 small semicircles = 3 × π2 × 4 = 6π
∴ Perimeter = 12π ≈ 37.71 cm
(ix)
Big semicircle = π2 × 20 = 10π
2 small semicircles = 2 × π2 × 10 = 10π
∴ Perimeter = 20π ≈ 62.86 cm
Common mistakes that cost marks
- Adding the dashed lines (the 60 m ends in (i), the square’s sides in (iii)) to the perimeter. They are inside the shape.
- In (ii), forgetting the two flat 2 cm pieces at the ends of the half ring.
- Using the radius where the diameter is needed: a semicircle on a diameter of 12 cm has length 6π, not 12π.
How this can come in the exam
A square of side 7 cm has a semicircle drawn outward on each side. The perimeter of the shape is (use π = 227)
- 28 cm
- 44 cm
- 66 cm
- 88 cm
Show answer
(B) 44 cm
4 × π × 3.5 = 14π = 44 cm.
A running field is a rectangle 120 m long with a semicircle on each 77 m end. Find the length of its boundary. (Use π = 227.)
Show answer
Two semicircles = one circle of diameter 77 m: 227 × 77 = 242 m (1 mark). Boundary = 2 × 120 + 242 = 482 m (1 mark).Try one yourself
A shape is made of a semicircle on a 14 cm line, with two semicircles on the 7 cm halves, one above and one below. Find its perimeter. (Use π = 227.)
Show answer
π2 × 14 + 2 × π2 × 7 = 14π = 44 cm.
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