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Measurement, STD1 M1 2024 HSC 18

A garden is made up of a right-angled triangle and a semicircle as shown.
 

What is the area of the garden, correct to the nearest square metre?   (3 marks)

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\(35\ \text{m}^2\)

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\(\text{Base of triangle }=2\times\ \text{radius} = 6\ \text{metres}\)

\(\text{Total Area }\) \(=\text{ Area of triangle + area of semicircle}\)
  \(=\dfrac{1}{2}\times 6\times 7 + \dfrac{1}{2}\times \pi\times 3^2\)
  \(=21+14.137\dots\)
  \(=35.137\dots\)
  \(=35\ \text{m}^2\ (\text{nearest m}^{2})\)

Measurement, STD1 M1 2022 HSC 4 MC

The area `(A)` of a circle is given by the formula `A=\pi r^2`, where `r` is the radius.

What is the value of `A`, correct to three significant figures, if `r=3.55` ?

  1. 39.5
  2. 39.6
  3. 39.591
  4. 39.592
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`B`

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`A` `=pi\r^2`  
  `=pi xx3.55^2`  
  `=39.591…`  
  `=39.6\ \ text{(3 sig fig)}`  

  
`=>B`


♦♦ Mean mark 30%.
COMMENT: Significant figures warrant attention.

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(1 d. p.))`

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

 

♦♦♦ Mean mark 16%.

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

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)

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`89\ text(cm²  (nearest cm²))`

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♦♦ 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 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
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`text(B)`

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`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…`

`=>\ text(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
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`C`

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`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 2015 HSC 28a

The diagram shows an annulus.
 

2015 28a

 
Calculate the area of the annulus.  (1 mark)

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`≈ 50.26…\ text(cm²)`

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`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
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`B`

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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
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`C`

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`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
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`B`

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`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 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) 

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`text(78 cm)\ \ \ \  text((nearest cm))`

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♦ 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)\ \ \ \ text((nearest cm))`