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v1 Measurement, STD2 M1 2009 HSC 11 MC

 What is the area of the shaded part of this quadrant, to the nearest square centimetre?  

  1. 68 m²
  2. 73 m²
  3. 95 m²
  4. 193 m²
Show Answers Only

`C`

Show Worked Solution
`text(Area)` `=\ text(Area of Sector – Area of triangle)`
  `= (theta/360 xx pi r^2)-(1/2 xx bh)`
  `= (90/360 xx pi xx 12^2)-(1/2 xx 6 xx 6)`
  `= 113.097…-18`
  `= 95.097…\ text(m²)`

`=> C`

v1 Measurement, STD1 M1 2019 HSC 25

The diagram illustrates a sector formed by a central angle of 105°, taken from a circle with a radius of 15 metres.

What is the perimeter of the sector? Write your answer correct to 1 decimal place.  (3 marks)

Show Answers Only

`57.5\ \ (text(1 d. p.))`

Show Worked Solution
`text(Arc length)` `= 105/360 xx 2 xx pi xx 15`
  `= 27.49`

 

`:.\ text(Perimeter)` `= 27.49 + 2 xx 15`
  `= 57.49`
  `= 57.5\ \ (text(1 d. p.))`

v1 Measurement, STD2 M1 2016 HSC 30c

A landscape artist was commissioned to design a garden consisting of part of a circle, with centre `O`, and a rectangle, as shown in the diagram. The radius `OC` of the circle is 20 m, the width `BC` of the rectangle is 10 m, and `DOC` is 100°.
 

What is the area of the whole garden, correct to the nearest square metre?  (5 marks)

Show Answers Only

`6281\ text{m²  (nearest m²)}`

Show Worked Solution

`text(In)\ \triangle ODC,`

`sin50^@` `= (ED)/20`
`ED` `= 20 \times \sin50^@`
  `= 34.472`
`:. DC` `= 2 \times 34.472 = 68.944\ \text{m}`

 

`cos50^@` `= (OE)/20`
`:. OE` `= 20 \times \cos50^@ = 28.925`

 

`text(Area of)\ \triangle ODC`

`= \frac{1}{2} \times 68.944 \times 28.925 = 997.12\ \text{m}^2`

 

`text(Area of rectangle ABCD)` `= 10 \times 68.944 = 689.44\ \text{m}^2`

 

`text(Area of major sector DOAC)`

`= \pi \times 20^2 \times \frac{260}{360} = 4594.58\ \text{m}^2`

 

`:.\ \text{Area of garden}`

`= 997.12 + 689.44 + 4594.58 = 6281.14`

`= 6281\ \text{m² (nearest m²)}`

v1 Measurement, STD2 M1 2015 HSC 28a

The diagram shows a circular garden bed with a circular path around it.
 

The radius of the entire structure (garden + path) is 6 m, and the radius of the inner garden is 4 m.

Calculate the area of the path. (1 mark)

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`≈ 62.83…\ text(m²)`

Show Worked Solution

`text(Area of path)`

`= pi(R^2 − r^2)`

`= pi(6^2 − 4^2)`

`= pi(36 − 16)`

`= 20pi\ \text(m²)`

`≈ 62.83…\ \text(m²)`

Measurement, STD2 M1 2024 HSC 34

A container for soccer balls is made using two half spheres joined to each end of a cylindrical body.
 

Three soccer balls fit exactly inside the container. Each ball has a diameter of 23 cm.

The hemispherical ends of the container just touch the surface of the soccer balls.

What is the total surface area of the container? Give your answer in square metres correct to one decimal place.   (4 marks)

Show Answers Only

\(\text{0.5 m}^2\)

Show Worked Solution

\(\text{Ball diameter = 23 cm}\ \Rightarrow \ r=11.5\ \text{cm}\)

\(\text{S.A. (2 half spheres)}\) \(=2 \times \dfrac{1}{2} \times 4 \pi \times 11.5^2\)
  \(=1662\ \text{cm}^2\)
♦ Mean mark 41%.

\(\text{Cylinder width}\ =\ \text{ball circumference}= 2 \times \pi \times 11.5\)

\(\text{Cylinder height}\ = 2 \times 23 = 46\ \text{cm}\)

\(\text{S.A. (cylinder)}\ = 2\pi rh = 2 \times \pi \times 11.5 \times 46 = 3324\ \text{cm}^2\)
 

\(\text{Note:}\ 1\ \text{m}^2= 100\ \text{cm} \times 100\ \text{cm}\ = 10\,000\ \text{cm}^2\)

\(\text{S.A. (container)}\) \(=1662 + 3324=4986\ \text{cm}^2\)
  \(=\dfrac{4986}{10\,000}\ \text{m}^2=0.5\ \text{m}^2\ \text{(1 d.p.)}\)

Measurement, STD1 M1 2019 HSC 25

The diagram shows a sector with an angle of 120° cut from a circle with radius 10 m.

What is the perimeter of the sector? Write your answer correct to 1 decimal place.   (3 marks)

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`40.9\ text{m (1 d. p.)}`

Show Worked Solution

`text(Arc length)= 120/360 xx 2 xx pi xx 10= 20.94\ text(m)`

`:.\ text(Perimeter)= 20.94 + 2 xx 10= 40.94= 40.9\ text{m (1 d. p.)}`

Measurement, STD2 M1 2022 HSC 32

The diagram shows a park consisting of a rectangle and a semicircle. The semicircle has a radius of 100 m. The dimensions of the rectangle are 200 m and 250 m.

A lake occupies a section of the park as shown. The rest of the park is a grassed section. Some measurements from the end of the grassed section to the edge of the lake are also shown.
 

  1. Using two applications of the trapezoidal rule, calculate the approximate area of the grassed section.   (2 marks)

  2. Hence calculate the approximate area of the lake, to the nearest square metre.   (2 marks)

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a.   `35\ 500\ text{m}^2`

b.   `30\ 208\ text{m}^2`

Show Worked Solution

a.   `h=100\ text{m}`

`A` `~~h/2(x_1+x_2)+h/2(x_2+x_3)`  
  `~~100/2(160+150)+100/2(150+250)`  
  `~~50xx310+50xx400~~35\ 500\ text{m}^2`  

  

b.    `text{Total Area}` `=\ text{Area of Rectangle + Area of Semi-Circle}`
    `= (250 xx 200) + 1/2 xx pi xx 100^2`
    `=65\ 707.96\ text{m}^2`

 
`text{Area of Lake} =65\ 708-35\ 500=30\ 208\ text{m}^2`


♦ Mean mark (b) 44%.

Measurement, STD2 M1 2021 FUR1 6

A child's toy has the following design.
 

Find the area of the shaded region to the nearest square centimetre.   (2 marks)

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`27\ text(cm)^2`

Show Worked Solution

`text{Circle radius = 3 cm}`

`text(Consider the rectangle starting from the middle of the left circle.)`

`text{Shaded Area}` `= text{Area rectangle}-3.5 xx text{Area circle}`
  `= (21 xx 6)-3.5 xx pi xx 3^2`
  `= 27.03 …\ text{cm}^2`
  `=27\ text{cm}^2\ text{(nearest whole)}`

Measurement, STD2 M1 EQ-Bank 14 MC

Four identical circles of radius `r` are drawn inside a square, as shown in the diagram below

The region enclosed by the circles has been shaded in the diagram.
 

The shaded area can be found using

  1. `4 r^2-2 pi r`
  2. `4 r^2-pi r^2`
  3. `4 r-pi r^2`
  4. `2 r^2-pi r^2`
Show Answers Only

`B`

Show Worked Solution

`text(Area of square) = 4r xx 4r = 16r^2`

`text(Area of circles) = 4 pi r^2`

`text(Shaded Area) = 1/4 xx (16r^2-4 pi r^2)= 4 r^2-pi r^2`

`=>  B`

Measurement, STD2 M1 2018 HSC 24 MC

The coordinates of city A are (39°N, 75°W). City B lies on the same longitude and is 5700 km south of city A.

What is the latitude of city B, given the earth's radius is 6400 km?

  1. 51°N
  2. 51°S
  3. 12°N
  4. 12°S
Show Answers Only

`D`

Show Worked Solution

`text(Circumference) = 2 xx pi xx 6400`

♦ Mean mark 45%.
COMMENT: If the earth’s radius is stated, this past question is arguably within the scope of the new syllabus.

`theta/360` `= 5700/(2pi xx 6400)`
`theta` `= (5700 xx 360)/(2pi xx 6400)~~ 51^(circ)`

    
`:. text(City)\ B\ text(is 12° South.)`

`=>D`

Measurement, STD2 M1 2018 HSC 22 MC

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

 
What is the perimeter of this shape, correct to one decimal place?

  1. 28.6 cm
  2. 36.6 cm
  3. 66.3 cm
  4. 74.3 cm
Show Answers Only

`B`

Show Worked Solution

`text(Using Pythagoras to find radius)\ (r):`

`r= sqrt(10^2-6^2)= sqrt64= 8\ text(cm)`

`text(Arc length)= 1/4 xx 2 pi r= 1/4 xx 2 xx pi xx 8= 12.56…\ text(cm)`

`:.\ text(Perimeter)= 8 + 6 + 10 + 12.56…= 36.57…`

  
`=> B`

Measurement, STD2 M1 2017 HSC 25 MC

In the circle, centre `O`, the area of the quadrant is 100 cm².
 


 

What is the arc length `l`, correct to one decimal place?

  1. 8.9 cm
  2. 11.3 cm
  3. 17.7 cm
  4. 25.1 cm
Show Answers Only

`C`

Show Worked Solution

`text(Find)\ r:`

♦ Mean mark 44%.
`text(Area)` `= 1/4 pir^2`
`100` `= 1/4 pir^2`
`r^2` `= 400/pi`
`:. r` `= 11.283…\ text(cm)`

 

`text(Arc length)` `= theta/360 xx 2pir= 90/360 xx 2pi xx 11.283…`
  `= 17.724…= 17.7\ text(cm)`

 
`=> C`

Measurement, STD2 M1 2016 HSC 30c

A school playground consists of part of a circle, with centre `O`, and a rectangle as shown in the diagram. The radius `OB` of the circle is 45 m, the width `BC` of the rectangle is 20 m and `AOB` is 100°.
 

What is the area of the whole playground, correct to the nearest square metre?   (5 marks)

Show Answers Only

`6971\ text{m²  (nearest m²)}`

Show Worked Solution

`text(In)\ DeltaOEB,`

`sin50^@` `= (EB)/45`
`EB` `= 45 xx sin50^@= 34.47…`

  
`:. AB= 2 xx 34.47…= 68.944\ \ (text(3 d.p.))`

`cos50^@` `= (OE)/45`
`:. OE` `= 45 xx cos50^@= 28.925\ \ (text(3 d.p.))`

  
`text(Area of)\ DeltaOAB= 1/2 xx AB xx OE= 1/2 xx 68.944 xx 28.925`

`= 997.12\ text(m²)`

  
`text(Area)\ ABCD= 20 xx 68.944= 1378.88\ text(m²)`

  
`text(Area of major sector)\ OAB= pi xx 45^2 xx 260/360`

`= 4594.58\ text(m²)`

  
`:.\ text(Area of playground)= 997.12 + 1378.88 + 4594.58`

`= 6970.58= 6971\ text{m²  (nearest m²)}`

Measurement, STD2 M1 2015 HSC 28a

The diagram shows an annulus.
 

2015 28a

 
Calculate the area of the annulus.   (1 mark)

Show Answers Only

`≈ 50.26…\ text(cm²)`

Show Worked Solution

`text(Area of annulus)`

`= pi(R^2-r^2)= pi(5^2-3^2)`

`= 16pi\ text(cm²)≈ 50.26…\ text(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)-text(Circle)`
  `= (100 xx 100)-(pi xx 5^2)= 10\ 000-78.5398…`
  `= 9921.46…\ text(cm)^2`

 
`=>  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`

Measurement, STD2 M1 2009 HSC 11 MC

 What is the area of the shaded part of this quadrant, to the nearest square centimetre?  

  1. 34 cm²
  2. 42 cm²
  3. 50 cm²
  4. 193 cm²
Show Answers Only

`B`

Show Worked Solution
`text(Area)` `=text(Area of Sector)-text(Area of triangle)`
  `= (theta/360 xx pi r^2)-(1/2 xx bh)`
  `= (90/360 xx pi xx 8^2)-(1/2 xx 4 xx 4)`
  `= 50.2654…-8= 42.265…\ text(cm²)`

  
`=>  B`

Measurement, STD2 M1 2012 HSC 27b

The sector shown has a radius of 13 cm and an angle of 230°. 
 

 2012 27b
 

 What is the perimeter of the sector to the nearest centimetre?   (2 marks) 

Show Answers Only

`text{78 cm (nearest cm)}`

Show Worked Solution
♦ Mean mark 39%
MARKER’S COMMENT: The formula of the length of an arc is given in the formula sheet.

`text(Perimeter)= 2 xx text(radius) + text(arc length)`

`text(Arc length)` `= theta/360 xx 2 xx pi xx r`
  `= 230/360 xx 2 xx pi xx 13`
  `= 52.1853…`

  
`:.\ text(Perimeter)= 2 xx 13 + 52.1853…= 78.1853…=text{78 cm (nearest cm)}`

Measurement, STD2 M1 2013 HSC 19 MC

A logo is designed using half of an annulus.
  

What is the area of the logo, to the nearest cm²?

  1. `25\ text(cm)^2`
  2. `33\ text(cm)^2`
  3. `132\ text(cm)^2`
  4. `143\ text(cm)^2`
Show Answers Only

`B`

Show Worked Solution
`text(Area)` `=1/2xxpi(R^2-r^2)=1/2xxpi(5^2-2^2)`
  `=21/2pi=33\ text(cm)^2\ text{(nearest cm}^2 text{)}`

 
`=> B`

Measurement, STD2 M1 2013 HSC 16 MC

The shaded region shows a quadrant with a rectangle removed.
  

What is the area of the shaded region, to the nearest cm2?

  1. 38 cm²
  2. 52 cm²
  3. 61 cm²
  4. 70 cm²
Show Answers Only

`B`

Show Worked Solution
`text(Shaded area)` `=text(Area of segment)-text(Area of rectangle)`
  `=1/4 pi r^2-(6xx2)=1/4 pi xx9^2-12`
  `=51.617…\ text(cm²)`

`=> B`