SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

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)

Show Answers Only
  1. `35\ 500\ text{m}^2`
  2. `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}` `=67\ 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)

Show Answers Only

`27\ text(cm²)`

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{cm²  (nearest whole)}`

Measurement, STD2 M1 SM-Bank 2 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 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)`

`= pi(25 − 9)`

`= 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 – 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`

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 – 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 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(cm²)`
  2. `33\ text(cm²)`
  3. `132\ text(cm²)`
  4. `143\ text(cm²)`
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(cm²)\ \ \ text{(nearest cm²)}` 

 
`=>\ 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 – Area of rectangle)`
  `=1/4 pi r^2-(6xx2)`
  `=1/4 pi xx9^2-12`
  `=51.617…\ text(cm²)`

`=>\ B`