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Area, Surface Area and Volume: Question 1

Syllabus C5.2, E5.2, C5.4, E5.4

Structured Core 6 marks

A landscape designer is planning a flower bed for a town square. The flower bed is made by joining a semicircle to one of the shorter sides of a rectangle, so the two shapes form a single flat bed.

The rectangle measures 10 m10\text{ m} by 6 m6\text{ m}. The straight side that the rectangle and the semicircle share is 6 m6\text{ m} long, so the semicircle has a diameter of 6 m6\text{ m} and bulges outwards from that end of the rectangle.

Take π=3.142\pi = 3.142.

(a) Work out the total area of the flower bed. Give your answer in m2\text{m}^2. [3]

(b) The designer will fit a low metal edging around the complete outside of the flower bed. Work out the total length of edging needed. [3]

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Worked solution

Part (a): Total area

Split the flower bed into a rectangle and a semicircle, then add the two areas.

Rectangle Arect=length×width=10×6=60 m2A_{\text{rect}} = \text{length} \times \text{width} = 10 \times 6 = 60\ \text{m}^2

Semicircle

The diameter is 6 m6\text{ m}, so the radius is r=62=3 m.r = \frac{6}{2} = 3\ \text{m}.

A semicircle is half of a full circle, so Asemi=12πr2=12×3.142×32=12×3.142×9=14.139 m2.A_{\text{semi}} = \frac{1}{2}\pi r^{2} = \frac{1}{2}\times 3.142 \times 3^{2} = \frac{1}{2}\times 3.142 \times 9 = 14.139\ \text{m}^2.

Total area A=60+14.139=74.139 m2.A = 60 + 14.139 = 74.139\ \text{m}^2.

A≈74.1 m2 (3 s.f.)\boxed{A \approx 74.1\ \text{m}^2 \ (3\text{ s.f.})}

Part (b): Length of edging (perimeter)

The edging follows the outside boundary only. Going around the shape, this is made up of:

  • the two long sides of the rectangle,
  • the one short side of the rectangle (the end without the semicircle),
  • the curved arc of the semicircle.

Important: the 6 m6\text{ m} side where the semicircle is attached lies inside the flower bed, so it is not part of the outside edging.

Straight parts 10+10⏟two long sides+6⏟one short side=26 m\underbrace{10 + 10}_{\text{two long sides}} + \underbrace{6}_{\text{one short side}} = 26\ \text{m}

Curved part (half the circumference)

The full circumference is πd\pi d, so the arc of the semicircle is arc=12πd=12×3.142×6=9.426 m.\text{arc} = \frac{1}{2}\pi d = \frac{1}{2}\times 3.142 \times 6 = 9.426\ \text{m}.

(Equivalently πr=3.142×3=9.426 m\pi r = 3.142 \times 3 = 9.426\ \text{m}.)

Total perimeter P=26+9.426=35.426 m.P = 26 + 9.426 = 35.426\ \text{m}.

P≈35.4 m (3 s.f.)\boxed{P \approx 35.4\ \text{m} \ (3\text{ s.f.})}

Quick check

A useful sanity check on part (b): the curved edge (≈9.4 m\approx 9.4\text{ m}) is a bit longer than the straight 6 m6\text{ m} end it replaces, which makes sense, because a curved path bulging outwards is always longer than the straight line across it.