SmarterEd

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Area, SMB-036

Find the surface area of the solid pictured below which is composed of a right cone with a hemisphere attached to the base. Give your answer to the nearest square centimetre.  (3 marks)
  

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\(\ 616\ \text{cm}^2\)

Show Worked Solution
\(\text{SA (hemisphere)}\) \(= \dfrac{1}{2} \times 4 \pi r^2\)  
  \(= \dfrac{1}{2} \times 4 \pi \times 7^2\)  
  \(=307.87…\ \text{cm}^2\)  

 

\(\text{SA (cone)}\) \(= \pi rl\)  
  \(= \pi \times 7 \times 14\)  
  \(=307.87…\ \text{cm}^2\)  

 
\(\text{Total SA}\ = 2 \times 307.87… = 616\ \text{cm}^2\ \ (\text{nearest cm}^2) \)

Area, SMB-018

A right cone with perpendicular height of 8 cm, slant height of 10 cm and a base diameter of 12 cm is pictured below.
 

Find the total surface area of the cone, including its base, correct to two decimal places.   (3 marks)

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\(301.59\ \text{cm}^2 \)

Show Worked Solution

\(r = 6\ \text{cm}, \ l =  10\ \text{cm} \)

\(\text{SA}\) \(= \pi r^2 + \pi r l\)  
  \(=  \pi \times 6^2 + \pi \times 6 \times 10 \)  
  \(=301.592… \)  
  \(=301.59\ \text{cm}^2\ \text{(2 d.p.)}\)  

Area, SMB-017

A square pyramid with a slant height of 15 centimetres is pictured below.
 

Find the total surface area of the pyramid, including its base.   (2 marks)

Show Answers Only

\(400\ \text{cm}^2 \)

Show Worked Solution
\(\text{SA}\) \(= (10 \times 10) + 4 \times (\dfrac{1}{2} \times 10 \times 15)\)  
  \(=100 + 4(75) \)  
  \(=400\ \text{cm}^2 \)  

Area, SMB-016

A square pyramid with a perpendicular height of 8 metres is pictured below.
 

  1. Using Pythagoras, calculate the slant height of the pyramid.   (2 marks)
  1. Find the total surface area of the pyramid, including its base, to the nearest square metre.   (2 marks)
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i.    \(10\ \text{m} \)

ii.    \(384\ \text{m}^2 \)

Show Worked Solution

i.    \(\text{Let}\ \ s =\ \text{slant height}\)

\(s^2\) \(=6^2 + 8^2\)  
  \(=100\)  
\(s\) \(=10\ \text{m}\)  

 

ii.    \(\text{SA}\) \(= (12 \times 12) + 4 \times (\dfrac{1}{2} \times 12 \times 10)\)  
  \(=144 + 4(60) \)  
  \(=384\ \text{m}^2 \)  

Area, SMB-015

A letterbox, which has been constructed by joining a half cylinder and rectangular prism, needs to be painted on each external surface.
 

Find the surface area of the letterbox, giving your answer correct to the nearest square centimetre.   (3 marks)

Show Answers Only

`9061\ text(cm)^2`

Show Worked Solution
`text{Surface Area (half cylinder)}` `= \frac{1}{2} (2pi r^2 + 2pi rh)`
  `= \frac{1}{2} (2pi xx 17.5^2 + 2pi xx 17.5 xx 40)`
  `= 3161.227…\ text{cm}^2`

 

`text{Area (rest)}` `= (35 xx 40)+2(35 xx 30) + 2(40 xx 30)`  
  `=1400+2(1050)+2(1200)`  
  `=5900\ text{cm}^2`  

 
`text{SA (total)}\ = 3161 + 5900 = 9061\ text{cm}^2`

Area, SMB-014

Calculate the surface area of the half-cylinder below (including the base), giving your answer correct to two decimal places.   (3 marks)
 

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`848.29\ text(m)^2`

Show Worked Solution
`text(Surface Area)` `= \frac{1}{2} (2pi r^2 + 2pi rh) + (7 xx 45)`
  `= \frac{1}{2} (2pi xx 3.5^2 + 2pi xx 3.5 xx 45) + 315`
  `= 848.285…`
  `= 848.29\ \text{m}^2\ \ text{(2 d.p.)}`

Area, SMB-013

Calculate the surface area of the cylinder below, giving your answer correct to two decimal places.   (2 marks)
 

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`58.31\ text(m)^2`

Show Worked Solution
`text(Surface Area)` `= 2pi r^2 + 2pi rh`
  `= 2pi (1.6)^2 + 2 pi xx 1.6 xx 4.2`
  `= 58.307…`
  `= 58.31\ \text{m}^2\ \ text{(2 d.p.)}`

Area, SMB-035

The surface area of a brick is given by the rule:

total surface area = 2 × [(width × height) + (width × length) + (height × length)]

The brick shown has a total surface area of 982 square centimetres.
 

What is the width of the brick?   (2 marks)

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`10.5\ text(cm)`

Show Worked Solution

`text(Let)\ \ w = text(width of the brick)`

`text(Surface Area)` `= 2 xx [8w + 22w + (8 xx 22)]`
`982` `= 2(30w + 176)`
`982` `= 60w + 352`
`60w` `= 630`
`:. w` `= 10.5\ text(cm)`

Area, SMB-034

The total surface area of a cube is 150 cm².
 

 How long is an edge of the cube?   (2 marks)

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`text(5 cm)`

Show Worked Solution

`text(Surface area of 1 face of the cube)`

`=150/6`

`=25\ text(cm)^2`
 

`text(Let)\ \ x=\ text(length of 1 side)`

`x^2` `=25`
`:.\ x` `=5\ text(cm)`

Area, SMB-033

The rectangular prism, shown below, is cut in half to form 2 equal cubes.
 

What is the ratio of the surface area of the rectangular prism to one of the cubes?   (3 marks)

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`5 : 3`

Show Worked Solution

`text(S.A. of rectangular prism)`

`= 2 xx (h xx h) + 4 xx (2h xx h)`

`= 2h^2 + 8h^2`

`= 10 h^2`
 

`text(S.A. of cube)\ = 6 xx (h xx h) = 6h^2`
 

`:.\ text(Ratio of rectangular prism to cube)`

`= 10h^2 : 6h^2`

`= 5 : 3`

Area, SMB-032

A net of a cube is pictured below.
  

The cube has a volume of 512 cubic centimetres.

What is the height of the net in centimetres?   (3 marks)

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`24\ text(cm)`

Show Worked Solution
`text(Let)\ \ s` `=\ text(length of 1 side of the cube)`
`s^3` `= 512`
`s` `= 8\ text(cm)`

 

`:.\ text(Height of net)` `= 3 xx 8`
  `= 24\ text(cm)`

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`

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

Area, SMB-030

A glass aviary is made up of four triangles and a square, as shown in the diagram below.

Harry is hired to clean the interior sides of the aviary, not including the floor.

What is the area that Harry will need to clean?   (2 marks)

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`402\ text(m)^2`

Show Worked Solution

`text(Area to clean)`

`= 4 xx 1/2 bh`

`= 4 xx 1/2 xx 15 xx 13.4`

`= 402\ text(m)^2`

Area, SMB-029 MC

The plan of a square area of land is drawn below.

A mining licence is granted over part of the land.

All measurements are given in kilometres.
 

The area of land the mining licence covers is

  1. `x^2-20`
  2. `x^2 + 20`
  3. `2x^2-9`
  4. `2x^2-20`
Show Answers Only

`A`

Show Worked Solution

`text{Area of original land}\ = x xx x = x^2\ text{km}^2`

`text{Size of non-mining area}\ = 4 xx 5 = 20\ text{km}^2`

`:.\ text(Area of mining)\ = (x^2-20)\ text{km}^2`

`=>A`

Area, SMB-027

Kate owns a farm with a square paddock. She increases the area of the paddock by changing it into the shape of trapezium.
 

What is the area of the new paddock, in square metres?   (2 marks)

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`1505\ text(m²)`

Show Worked Solution
`A` `= 1/2h(a + b)`
  `= 1/2 xx 35 xx (35 + 51)`
  `= 1505\ text(m²)`

Area, SMB-026

Clancy needs to paint all sides of a triangular prism.

The area of each triangular face is 12 cm².

What is the total area Clancy needs to paint?   (2 marks)

Show Answers Only

`78\ text(cm²)`

Show Worked Solution
`text(Total Area)` `= (2 xx 12) + 2 xx (3 xx 5) + (8 xx 3)`
  `= 78\ text(cm²)`

Area, SMB-025

An irregular shaped lawn is pictured below, with all measurements in metres.
  

 

What is the area of the lawn in square metres?   (3 marks)

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`65\ \text{m}^2`

Show Worked Solution

`text(Split the lawn into 3 areas)`

`text(Area 1) = 9 xx 3 = 27\ text(m²)`

`text(Area 2) = 5 xx 2 = 10\ text(m²)`

`text(Area 3:)`

`text(Height) = 5+2 = 7\ text(m)`

`text(Width)= 12-(5 + 3)= 4\ text(m)`

`=> text(Area 3) = 7 xx 4 = 28\ text(m²)`

 

`:.\ text(Area of lawn)` `= 27 + 10 + 28`
  `= 65\ text(m²)`

Area, SMB-024

A plan of Zev's backyard is shown.
  

 

Zev wants to fertilise the lawn in his backyard and needs to know the total area.

What is the total area of Zev's lawn, in square metres?   (2 marks)

Show Answers Only

`text(26 m²)`

Show Worked Solution

`text(Area)` `=\ text(square + triangle)`
  `= (4 xx 4) + (1/2 xx 4 xx 5)`
  `= 16 + 10`
  `= 26\ text(m²)`

Area, SMB-023 MC

Harry has an irregularly shaped back yard.
  

What is the area of Harry's back yard?

  1. `text(93 m)²`
  2. `text(384 m)²`
  3. `text(544 m)²`
  4. `text(768 m)²`
Show Answers Only

`B`

Show Worked Solution
`text(Area)` `= (1/2 xx 14 xx 32) + (1/2 xx 10 xx 32)`
  `= 224 + 160`
  `= 384\ text(m)²`

 
`=>B`

Area, SMB-022

A cube has a side length of 6 cm.

Two smaller cubes of side length 3 cm are attached to the larger cube as shown in the diagram below.
  

What is the surface area of the new object?   (3 marks)

Show Answers Only

`270\ text(cm²)`

Show Worked Solution

`text(One strategy:)`

`text{Calculate the surface area (S.A.) of the 2 objects joined}`

`text(separately and deduct the area where they join.)`
 

`text{S.A. (large cube)} = 6 xx 6xx 6 = 216\ text(cm²)`

`text{S.A. (2 smaller cubes joined together)}`

`=2 xx (3 xx 3) + 4 xx (6 xx 3)`

`=90\ text(cm²)`

`text{S.A. to be deducted (where cubes are attached)}`

`= 2 xx (6 xx 3)`

`= 36\ text(cm²)`

 

`:.\ text{S.A. (new object)}` `= 216 + 90-36`
  `=270\ text(cm²)`

Area, SMB-021 MC

Pablo cuts a square out of a rectangular piece of paper, as shown below.

The square Pablo cut out has a side length of 6 cm.

Which of these expressions gives the area of Pablo's piece of paper after cutting out the square?

  1. `(28 xx 20)-(6 xx 6)`
  2. `(22 xx 14) + (6 xx 6)`
  3. `(28 + 20)-(6 + 6)`
  4. `(28 + 20) + (6 xx 6)`
Show Answers Only

`A`

Show Worked Solution
`text(Area)` `=\ text(Area of larger rectangle − Area of square)`
  `= (28 xx 20)-(6 xx 6)`

 
`=>A`

Area, SMB-020

A trapezium is constructed on a grid of 10 rectangles.

Each rectangle measures  3 cm × 7 cm.

 

Find the area of the trapezium?   (2 marks)

Show Answers Only

`168\ text(cm²)`

Show Worked Solution
`text(Area 1 rectangle)` `= 3 xx 7`
  `= 21\ text(cm²)`

 

`:.\ text(Total Area)` `= (6 xx 21) + 2\ text(triangles)`
  `= 6 xx 21 + 2 xx (1/2 xx 3 xx 14)`
  `= 126 + 42`
  `= 168\ text(cm²)`

Area, SMB-019 MC

Olive drew this plan of her lawn.

Which expression gives the area of Olive's lawn?

  1. `(a xx b) + (c xx d)`
  2. `(a xx b) xx (c xx d)`
  3. `(a + b) + (c + d)`
  4. `(a + b) xx (c + d)`
Show Answers Only

`A`

Show Worked Solution

`text(Total Area)` `= text(Area 1) + text(Area 2)`
  `= (a xx b) + (c xx d)`

`=> A`

Area, SMB-012

Find the surface area of the solid below where all measurements are in metres.  (2 marks)
 

Show Answers Only

\(104\ \text{m}^2\)

Show Worked Solution

\(\text{Area (front side)} = (4 \times 2) + (2 \times 2) = 12\ \text{m}^2 \)

\(\text{S.A.}\) \(=2 \times 12 + 2 \times (5 \times 4) + 4 \times (5 \times 2) \)  
  \(= 24 + 2(20) + 4(10) \)  
  \(=104\ \text{m}^2\)  

Area, SMB-011

A triangular prism has the following dimensions (all measurements are in metres).
 

  1. Draw a net of the solid   (1 mark)
  1. Find the surface area of the prism.  (2 marks)
Show Answers Only

i. 

ii.    \(348\ \text{m}^2\)

Show Worked Solution

i.    

ii.    \(\text{S.A.}\) \(=2 \times (5 \times 18) + (8 \times 18) + 2 \times (\dfrac{1}{2} \times 8 \times 3) \)  
  \(= 2(90) + 144 + 2(12) \)  
  \(=348\ \text{m}^2\)  

Area, SMB-010

A triangular prism has the following dimensions (all measurements are in centimetres).
 

  1. Draw a net for this solid.   (1 mark)
  1. Find the surface area of the prism.  (2 marks)
Show Answers Only

i.  

 

ii.    \(408\ \text{cm}^2\)

Show Worked Solution

i.   

ii.    \(\text{S.A.}\) \(=(15 \times 6) + (15 \times 10) + (15 \times 8) + 2 \times (\dfrac{1}{2} \times 8 \times 6) \)  
  \(=90+150+120+2(24) \)  
  \(=408\ \text{cm}^2\)  

Area, SMB-009

Find the surface area of the rectangular prism pictured below, correct to 1 decimal place. All measurements are in metres.  (2 marks)
 

Show Answers Only

\(68.0\ \text{m}^2\)

Show Worked Solution
\(\text{S.A.}\) \(=2 \times (2.8 \times 1.7) + 2 \times (2.8 \times 6.5) + 2 \times (1.7 \times 6.5) \)  
  \(= 2(4.76 + 18.2 + 11.05) \)  
  \(=68.02\)  
  \(=68.0\ \text{m}^2\ \ \text{(1 d.p.)}\)  

Area, SMB-008

Find the surface area of the rectangular prism pictured below. All measurements are in metres.  (2 marks)
 

Show Answers Only

\(368\ \text{m}^2\)

Show Worked Solution
\(\text{S.A.}\) \(=2 \times (6 \times 14) + 2 \times (6 \times 5) + 2 \times (5 \times 14) \)  
  \(=2(84+30+70)\)  
  \(=368\ \text{m}^2\)  

Area, SMB-007

The diagram shows the shape of Carmel’s garden bed. All measurements are in metres.
  

Show that the area of the garden bed is 57 square metres.  (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution
`text(Area of)\ Delta ABC` `= 1/2 xx b xx h`
  `= 1/2 xx 10 xx 5.1`
  `= 25.5\ text(m²)`
`text(Area of)\ Delta ACD` `= 1/2 xx 10 xx 6.3`
  `= 31.5\ text(m²)`

 

`:.\ text(Total Area)` `= 25.5 + 31.5`
  `= 57\ text(m² … as required)`

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)
 

Show Answers Only

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

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

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`
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`

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)

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 …`
  `=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)

Show Answers Only

`35\ text{m²}`

Show Worked Solution

`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)
 

Show Answers Only

`9921\ text(cm²)`

Show Worked Solution
`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)

Show Answers Only

`52\ text(cm²)`

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…`
  `=52\ text(cm²)`

Area, SMB-002

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

Show Answers Only

`42\ text{cm²}`

Show Worked Solution
`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 2022 HSC 34

A composite solid is shown. The top section is a cylinder with a height of 3 cm and a diameter of 4 cm. The bottom section is a hemisphere with a diameter of 6 cm. The cylinder is centred on the flat surface of the hemisphere.
 


 

Find the total surface area of the composite solid in cm², correct to 1 decimal place.  (4 marks)

Show Answers Only

`122.5\ text{cm}^2`

Show Worked Solution
`text{S.A. of Cylinder}` `=pir^2+2pirh`  
  `=pi(2^2)+2pi(2)(3)`  
  `=16pi\ text{cm}^2`  

 

`text{S.A. of Hemisphere}` `=1/2 xx 4pir^2`  
  `=2pi(3^2)`  
  `=18pi\ text{cm}^2`  

 

`text{Area of Annulus}` `=piR^2-pir^2`  
  `=pi(3^2)-pi(2^2)`  
  `=5pi\ text{cm}^2`  

 

`text{Total S.A.}` `=16pi+18pi+5pi`  
  `=39pi`  
  `=122.522…`  
  `=122.5\ text{cm}^2\ \ text{(to 1 d.p.)}`  

♦ Mean mark 50%.

Measurement, STD2 M1 2020 HSC 25

A composite solid consists of a triangular prism which fits exactly on top of a cube, as shown.
 

Find the surface area of the composite solid.   (3 marks)

Show Answers Only

`424 \ text{cm}^2`

Show Worked Solution

`text{S.A. of 1 face of cube} = 8 xx 8 = 64 \ text{cm}^2`

`text{Height of triangle} = 11 – 8 = 3 \ text{cm}`

`therefore \ text{S.A. (triangular prism)}` `= 2 xx ( frac{1}{2} xx 8 xx 3 ) + 2 xx (5 xx 8)`
  `= 24 + 80`
  `= 104 \ text{cm}^2`

 

`therefore \ text{Total S.A.}` `= 5 xx 64 + 104`
  `= 424 \ text{cm}^2`

Measurement, STD2 M1 2018 HSC 27c

A shade shelter is to be constructed in the shape of half a cylinder with open ends. The diameter is 3.8 m and the length is 10 m.
 

 
The curved roof is to be made of plastic sheeting.

What area of plastic sheeting is required, to the nearest m²?  (2 marks)

Show Answers Only

`60\ text(m²  (nearest m²))`

Show Worked Solution

`text(Flatten out the half cylinder,)`

`text(Width)` `= 1/2 xx text(circumference)`
  `= 1/2 xx pi xx 3.8`
  `= 5.969…`

 

`:.\ text(Sheeting required)` `= 10 xx 5.969…`
  `= 59.69…`
  `= 60\ text(m²  (nearest m²))`

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`

Area, SMB-028

 

What is the area of the shaded part of the figure?   (2 marks)

Show Answers Only

`8.5\ text(cm)^2`

Show Worked Solution

`text(Area of large triangle)`

`= 1/2 xx 5 xx 5`

`= 12.5\ text(cm)^2`

`text(Area of smaller triangle)`

`= 1/2 xx 2 xx 2`

`= 2\ text(cm)^2`

 

`:.\ text(Shaded area)`

`= 12.5-(2 xx 2)`

`= 8.5\ text(cm)^2`

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 2016 HSC 26a

Calculate the surface area of a sphere with a radius of 5 cm, correct to the nearest whole number.  (1 mark)
 

Show Answers Only

`314\ text{cm²}`

Show Worked Solution
`SA` `= 4pir^2`
  `= 4 xx pi xx 5^2`
  `= 314.15…`
  `= 314\ text{cm²  (nearest cm²)}`

Measurement, STD2 M1 2014 HSC 25 MC

A grain silo is made up of a cylinder with a hemisphere (half a sphere) on top. The outside of the silo is to be painted.
  

 What is the area to be painted?

  1. `8143\ text(m²)`
  2. `11\ 762\ text(m²)`
  3. `12\ 667\ text(m²)`
  4. `23\ 524\ text(m²)`
Show Answers Only

`A`

Show Worked Solution

`text(Total Area) = text(Area of cylinder) + text(½ sphere)`

♦ Mean mark 40%
`text(Area of cylinder)` `= 2 pi rh`
  `= 2pi xx 24 xx 30`
  `= 4523.9`
`text(Area of ½ sphere)` `= 1/2 xx 4 pi r^2`
  `= 1/2 xx 4 pi xx 24^2`
  `= 3619.1`
`:.\ text(Total area)` `= 4523.9 + 3619.1`
  `= 8143\ text(m²)`

`=>  A`

Measurement, STD2 M1 2009 HSC 23c

The diagram shows the shape and dimensions of a terrace which is to be tiled.
 

  1. Find the area of the terrace.   (2 marks)

  2. Tiles are sold in boxes. Each box holds one square metre of tiles and costs $55. When buying the tiles, 10% more tiles are needed, due to cutting and wastage.

     

    Find the total cost of the boxes of tiles required for the terrace.   (2 marks)

Show Answers Only

i.    `13.77\ text(m²)`

ii.   `$880`

Show Worked Solution
i.
`text(Area)` `=\ text(Area of big square – Area of 2 cut-out squares`
  `= (2.7 + 1.8) xx (2.7 + 1.8)\-2 xx (1.8 xx 1.8)`
  `= 20.25\-6.48`
  `= 13.77\ text(m²)`

 

ii. `text(Tiles required)` `= (13.77 +10 text{%}) xx 13.77`
    `= 15.147\ text(m²)`

 

 `=>\ text(16 boxes are needed)`

`:.\ text(Total cost of boxes)` `=16 xx $55`
  `= $880`

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`