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Area, SMB-031

A target is drawn as follows

What is the area of the entire target?

Round your answer to the nearest square centimetre.   (2 marks)

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`284\ text(cm)^2`

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`text(Area)` `= pi r^2`
  `= pi (2 + 1.5 xx 5)^2`
  `= pi (9.5)^2`
  `= 283.52…`
  `= 284\ text(cm)^2`

Area, SMB-001

Find the area of the shaded part in the diagram below, giving your answer in square metres to 1 decimal place.  (3 marks)
 

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\(62.8\ \text{m}^2 \)

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\(\text{Shaded area}\) \(=\ \text{Area of large sector – Area of small sector}\)  
  \(= \dfrac{30}{360} \times \pi R^2-\dfrac{30}{360} \times \pi r^2 \)  
  \(= \dfrac{1}{12} \pi (16^2-4^2) \)  
  \(= 62.83… \)  
  \(=62.8\ \text{m}^2\ \ \text{(1 d.p.)} \)  

Area, SMB-008 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`
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`B`

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`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`

Area, SMB-006

A child's toy has the following design.
 

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

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`27\ text(cm²)`

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`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 …`
  `=27\ text{cm²  (nearest whole)}`

Area, SMB-005

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

Find the area of the path, correct to the nearest square metre.   (2 marks)

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`35\ text{m²}`

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`text(Area of annulus)`

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

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

`= pi (11.25)`

`=35.3…`

`=35\ text{m²  (nearest m²)}`

Area, SMB-004

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

Find the area of the shower floor, excluding the drain, to the nearest square centimetre.   (3 marks)
 

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`9921\ text(cm²)`

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`text(Area)` `=\ text(Square – Circle)`
  `= (100 xx 100)-(pi xx 5^2)`
  `= 10\ 000-78.5398…`
  `= 9921.46…\ text(cm²)`
  `= 9921\ text{cm² (nearest cm²)}`

Area, SMB-003

The shaded region shows a quadrant with a rectangle removed.
  

Find the area of the shaded region, to the nearest cm2.   (2 marks)

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`52\ text(cm²)`

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`text(Shaded area)` `=\ text(Area of segment – Area of rectangle)`
  `=1/4 pi r^2-(6xx2)`
  `=1/4 pi xx9^2-12`
  `=51.617…`
  `=52\ text(cm²)`

Area, SMB-002

 What is the area of the shaded part of this quadrant, to the nearest square centimetre?  (3 marks)

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`42\ text{cm²}`

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`text(Area)` `=\ text(Area of Sector – Area of triangle)`
  `= (theta/360 xx pi r^2)-(1/2 xx l xx h)`
  `= (90/360 xx pi xx 8^2)-(1/2 xx 4 xx 4)`
  `= 50.2654…-8`
  `= 42.265…\ text(cm²)`
  `= 42\ text{cm²  (nearest cm²)}`

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 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)`
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`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`