SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

Area, SM-Bank 136

The diagram shows an annulus.
 

 
Calculate the area of the shaded region (annulus), correct to 2 decimal places.  (2 marks)

Show Answers Only

\(\approx 50.27\ \text{cm}^2\ (2\text{ d.p.})\)

Show Worked Solution

\(\text{Method 1:}\)

\(\text{Radius small circle}\ (r)=3\)

\(\text{Radius large circle}\ (R)=\dfrac{10}{2}=5\)

\(\text{Shaded region}\) \(=\text{Area large circle}-\text{Area small circle}\)
  \(=\pi\times 5^2-\pi \times 3^2\)
  \(=25\pi-9\pi\)
  \(=16\pi\)
  \(=50.27\ \text{cm}^2\ (2\text{ d.p.})\)

 
\(\text{Method 2:}\)

\(\text{Area of annulus}\)

\(=\pi(R^2 − r^2)\)

\(=\pi(5^2 − 3^2)\)

\(=\pi(25 − 9)\)

\(=16\pi\)

\(=50.27\ \text{cm}^2\ (2\text{ d.p.})\)

Area, SM-Bank 135

A shape consisting of a quadrant and a right-angled triangle is shown.
 

  1. Use Pythagoras' Theorem to calculate the radius of the quadrant.  (2 marks)

  2. What is the area of this shape, correct to one decimal place?  (2 marks)
Show Answers Only

a.    \(8\ \text{cm}\)

b.    \(74.3\ \text{cm}^2\ (1\text{d.p.})\)

Show Worked Solution

a.    \(\text{Using Pythagoras to find radius}\ (r):\)

\(a^2+b^2\) \(=c^2\)
\(r^2+6^2\) \(=10^2\)
\(r^2\) \(=10^2-6^2\)
\(r\) \(=\sqrt{64}\)
  \(=8\ \text{cm}\)

 

b.    \(\text{Total area}\) \(=\text{Area of triangle}+\text{Area of quadrant}\)
    \(=\dfrac{1}{2}\times 8\times 6+\dfrac{1}{4}\times \pi\times 8^2\)
    \(=24+50.265\dots\)
    \(=74.265\dots\)
    \(\approx 74.3\ \text{cm}^2\ (1\text{d.p.})\)

Area, SM-Bank 126

Calculate the area of the shaded region in the following composite shape, giving your answer correct to one decimal place.   (2 marks)
 

Show Answers Only

\(2789.0\ \text{m}^2\ (1 \text{ d.p.})\)

Show Worked Solution

\(\text{Diameter semi-cirle}=114\ \text{cm}\)

\(\text{Radius semi-cirle}(r)=57\ \text{cm}\)

\(\text{Total area}=\text{Area square}-\text{Area semi-cirle}\)

\(A\) \(=s^2-\pi r^2\)
  \(=114^2-\pi\times 57^2\)
  \(=12\ 996-10\ 207.034\dots\)
  \(=2788.965\dots\approx 2789.0\ \text{m}^2\ (1 \text{ d.p.})\)

Area, SM-Bank 125

Calculate the area of the shaded region in the following composite shape, giving your answer correct to one decimal place.   (2 marks)
 

Show Answers Only

\(22.0\ \text{m}^2\ (1 \text{ d.p.})\)

Show Worked Solution

\(\text{Radius small cirle}(r)=3\ \text{m}\)

\(\text{Radius large cirle}(R)=4\ \text{m}\)

\(\text{Total area}=\text{Area large cirle}-\text{Area small cirle}\)

\(A\) \(=\pi R^2-\pi r^2\)
  \(=\pi\times 4^2-\pi\times 3^2\)
  \(=50.265\dots-28.274\dots\)
  \(=21.991\dots\approx 22.0\ \text{m}^2\ (1 \text{ d.p.})\)

Area, SM-Bank 124

Calculate the area of the following composite shape, giving your answer correct to one decimal place.   (2 marks)
 

Show Answers Only

\(321.0\ \text{cm}^2\ (1 \text{ d.p.})\)

Show Worked Solution

\(\text{Radius small semi-cirle}(r)=4.5\ \text{mm}\)

\(\text{Radius large semi-cirle}(R)=9\ \text{mm}\)

\(\text{Total area}=\text{Area small semi-cirle}+\text{Area large semi-cirle}+\text{Area rectangle}\)

\(A\) \(=\dfrac{1}{2}\times \pi r^2+\dfrac{1}{2}\times \pi R^2+lb\)
  \(=\dfrac{1}{2}\times \pi\times 4.5^2+\dfrac{1}{2}\times \pi\times 9^2+18\times 9\)
  \(=31.808\dots+127.234\dots+162\)
  \(=321.043\dots\approx 321.0\ \text{cm}^2\ (1 \text{ d.p.})\)

Area, SM-Bank 123

Calculate the area of the following composite shape, giving your answer correct to one decimal place.   (2 marks)
 

Show Answers Only

\(178.3\ \text{cm}^2\ (1 \text{ d.p.})\)

Show Worked Solution

\(\text{Radius}=8\ \text{cm}\)

\(\text{Rectangle length}=24-8=16\ \text{cm}\)

\(\text{Total area}=\text{Area Quadrant}+\text{Area rectangle}\)

\(A\) \(=\dfrac{1}{4}\times \pi r^2+lb\)
  \(=\dfrac{1}{4}\times \pi\times 8^2+16\times 8\)
  \(=50.265\dots+128\)
  \(=178.265\dots\approx 178.3\ \text{cm}^2\ (1 \text{ d.p.})\)

Area, SM-Bank 120

A one-on-one basketball court is a composite shape made up of a rectangle and a semicircle, as shown below.
 

Calculate the area of the court, giving your answer correct to one decimal place.   (2 marks)

Show Answers Only

\(104.5\ \text{m}^2\ (1 \text{ d.p.})\)

Show Worked Solution

\(\text{Diameter}=12\ \text{m}\)

\(\therefore\ \text{Radius}=6\ \text{m}\)

\(\text{Total area}=\text{Area semi-circle}+\text{Area rectangle}\)

\(A\) \(=\dfrac{1}{2}\times \pi r^2+lb\)
  \(=\dfrac{1}{2}\times \pi\times 6^2+12\times 4\)
  \(=104.548\dots\)
  \(=104.5\ \text{m}^2\ (1 \text{ d.p.})\)

Area, SM-Bank 064 MC

A circular pool is located in a square lawn, as shown below.
 

The sides of the square lawn are 10 m in length.

The pool has a radius of 3 m.

The area of the lawn surrounding the pool, in square metres, is closest to

  1. \(59\)
  2. \(72\)
  3. \(81\)
  4. \(128\)
Show Answers Only

\(B\)

Show Worked Solution
\(\text{Area of square}\) \(=\text{side}^2\)
  \(=10^2\)
  \(=100\ \text{m}^2\)

   

\(\text{Area of pool}\) \(=\pi r^2\)
  \(=\pi \times 3^2\)
  \(= 28.27\dots\ \text{m}^2\)

 

\(\therefore\ \text{Area of lawn}\) \(=100-28.27\dots\)
  \(=71.72\dots\ \text{m}^2\)

 
\(\Rightarrow B\)

Measurement, STD1 M1 2019 HSC 15

The diagram shows a shape made up of a square of side length 8 cm and a semicircle.
  


 

Find the area of the shape to the nearest square centimetre.  (3 marks)

Show Answers Only

`89\ text(cm²  (nearest cm²))`

Show Worked Solution

♦♦ Mean mark 26%.

`text(Area)` `=\ text(Area of square + Area of semicircle)`
  `= 8 xx 8 + 1/2 xx pi xx 4^2`
  `= 89.13…`
  `= 89\ text(cm²  (nearest cm²))`

Measurement, STD2 M1 2008 HSC 11 MC

The diagram shows the floor of a shower. The drain in the floor is a circle with a diameter of 10 cm.

What is the area of the shower floor, excluding the drain?
 

 
 

  1. 9686 cm²
  2. 9921 cm²
  3. 9969 cm²
  4. 10 000 cm²
Show Answers Only

`B`

Show Worked Solution
COMMENT: Students should see that answers are all in cm², and therefore use cm as the base unit for their calculations. 
`text(Area)` `=\ text(Square – Circle)`
  `= (100 xx 100)-(pi xx 5^2)`
  `= 10\ 000-78.5398…`
  `= 9921.46…\ text(cm²)`

 
`=>  B`

Measurement, STD2 M1 2014 HSC 12 MC

A path 1.5  metres wide surrounds a circular lawn of radius 3 metres. 

What is the approximate area of the path?

  1. 7.1 m²
  2. 21.2 m²
  3. 35.3 m²
  4. 56.5 m²
Show Answers Only

`C`

Show Worked Solution

`text(Area of annulus)`

`= pi (R^2-r^2)`

`= pi (4.5^2-3^2)`

`= pi (11.25)`

`=35.3\ text{m²  (1 d.p.)}`
 

`=>  C`