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Area, SM-Bank 138

A path 1.8 m wide is being built around a rectangular garden. The garden is 8.4 m long and 5.4 m wide. The path is shaded in the diagram.
 

 
 

Calculate the area of the path in square metres.  (2 marks)

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\(62.64\ \text{m}^2\)

Show Worked Solution

\(\text{Length of large rectangle}=1.8+8.4+1.8=12\ \text{m}\)

\(\text{Width of large rectangle}=1.8+5.4+1.8=9\ \text{m}\)

\(\text{Shaded Area}\) \(=\text{Large rectangle}-\text{garden area}\)
  \(=12\times 9-8.4\times5.4\)
  \(=108-45.36\)
  \(=62.64\ \text{m}^2\)

Area, SM-Bank 122

Tim sketched a plot of land with the following measurements in metres.

What is the area of the land in square metres?  (2 marks)

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\(487\ \text{m}^2\)

Show Worked Solution

\(\text{Total Area}=\text{Area Rectangle}+\text{Area trapezium}\)

\(\text{Total Area}\) \(=lb+\dfrac{h}{2}(a+b)\)
  \(=(12\times 25)+\dfrac{11}{2}(24+10)\)
  \(=300+187\)
  \(=487\ \text{m}^2\)

Area, SM-Bank 104

Calculate the area of the following composite figure in square centimetres   (2 marks)
 

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\(15.5\ \text{cm}^2\)

Show Worked Solution
\(\text{Area}\) \(=1\times \text{triangles}+1\times\text{trapezium}\)
  \(=\dfrac{1}{2}\times bh +\dfrac{h}{2}(a+b)\)
  \(=\dfrac{1}{2}\times 5\times 3+\dfrac{2}{2}\times (3+5)\)
  \(=7.5+8\)
  \(=15.5\ \text{cm}^2\)

Area, SM-Bank 103

Calculate the area of the following composite figure in metres squared.   (2 marks)
 

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\(1116\ \text{m}^2\)

Show Worked Solution
\(\text{Area}\) \(=3\times \text{triangles}+1\times\text{square}\)
  \(=\dfrac{1}{2}\times 24\times 12+\dfrac{1}{2}\times 24\times 9+\dfrac{1}{2}\times 24\times 24+24^2\)
  \(=144+108+288+576\)
  \(=1116\ \text{m}^2\)

Area, SM-Bank 102

Calculate the area of the following composite figure in square centimetres.  (2 marks)
 

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\(182\ \text{cm}^2\)

Show Worked Solution
\(\text{Area}\) \(=\text{Area triangle 1}+\text{Area triangle 2}\)
  \(=\dfrac{1}{2}\times 14\times 12+\dfrac{1}{2}\times 14\times 14\)
  \(=84+98\)
  \(=182\ \text{cm}^2\)

Area, SM-Bank 096

Luke builds a rectangular wooden deck in his backyard, with dimension 12 metres by 5 metres.
 

Luke is going to create a 0.5 metre wide path around the full perimeter of his deck.

  1. What is the total area of the path in square metres?  (2 marks)

  2. He is creating the path using pavers at a cost of $92 per square metre. Calculate the cost of the pavers.  (1 mark)

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a.    \(18\ \text{m}^2\)

b.    \($1656\)

Show Worked Solution
a.    \(\text{Area of path}\) \(=2\times (12\times 0.5)+2\times (5\times 0.5)+4\times (0.5^2)\)
    \(=12+5+1\)
    \(=18\ \text{m}^2\)

 

b.    \(\text{Cost of pavers}\) \(=18\times $92\)
    \(=$1656\)

Area, SM-Bank 095

A cement slab is laid in Yvette's backyard that forms an 8 metre by 4 metre rectangle.
 

Yvette is going to lay a 0.25 metre wide path around the full perimeter of her slab.

  1. What is the total area of the perimeter path in square metres?  (2 marks)

  2. She is covering the path with artificial grass at a cost of $45 per square metre. Calculate the cost of laying turf on the path.  (1 mark)

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a.    \(6.25\ \text{m}^2\)

b.    \($281.25\)

Show Worked Solution
a.    \(\text{Area of path}\) \(=2\times (8\times 0.25)+2\times (4\times 0.25)+4\times (0.25^2)\)
    \(=4+2+0.25\)
    \(=6.25\ \text{m}^2\)

 

b.    \(\text{Cost of artificial turf}\) \(=6.25\times $45\)
    \(=$281.25\)

Area, SM-Bank 093 MC

Ken puts two cardboard squares together, as shown in the diagram below.

The squares have areas of 4 cm² and 25 cm².

Ken draws a line from the bottom left to top right, and shades the region above the line.
 

What is the area of the shaded region?

  1. \(13.5\ \text{cm}^2\)
  2. \(14.5\ \text{cm}^2\)
  3. \(17.5\ \text{cm}^2\)
  4. \(19\ \text{cm}^2\)
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\(C\)

Show Worked Solution

\(\text{Small square }\rightarrow 2\ \text{cm sides}\)

\(\text{Large square }\rightarrow 5\ \text{cm sides}\)
 

 
 

\(\text{Shaded Area}\) \(=\dfrac{1}{2}\times bh\)
  \(=\dfrac{1}{2}\times 5\times 7\)
  \(=17.5\ \text{cm}^2\)

 
\(\Rightarrow C\)

Area, SM-Bank 092

Anthony is tiling one wall of a bathroom.

The wall has 2 identical windows as shown in the diagram below.
 

What is the total area Anthony has to tile?  (2 marks)

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\(12.9\ \text{m}^2\)

Show Worked Solution
\(\text{Area}\) \(=(5.3\times 3)-2\times (1\times 1.5)\)
  \(=15.9-3\)
  \(=12.9\ \text{m}^2\)

Area, SM-Bank 071

A holiday unit is shaped like a hexagon.

The dimensions of its floor plan are shown below.

What is the total area of the holiday unit in square metres? (2 marks)

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\(153\ \text{m}^2\)

Show Worked Solution
\(\text{Holiday unit area}\) \(=\text{Area of rectangle}+2\times \text{Area of triangle}\)
  \(=(9\times 14)+2\times\bigg(\dfrac{1}{2}\times 9\times 3\bigg)\)
  \(=126+27\)
  \(=153\ \text{m}^2\)

Area, SM-Bank 070

A swimming pool is shaped like a hexagon.

The dimensions are given from the top view of the swimming pool.
 

What is the total area of the swimming pool in square metres?   (2 marks)

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\(27\ \text{m}^2\)

Show Worked Solution
\(\text{Pool area}\) \(=\text{Area of rectangle}+2\times \text{Area of triangle}\)
  \(=(3\times 4)+2\times\bigg(\dfrac{1}{2}\times 3\times 5\bigg)\)
  \(=12+15\)
  \(=27\ \text{m}^2\)

Area, SM-Bank 069

Binky used the paver pictured below to pave her pool area.

Altogether, she used 50 tiles.

What is the total area of Binky's pool area in square metres? (2 marks)

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\(13.5\ \text{m}^2\)

Show Worked Solution

\(\text{Convert cm to metres:}\)

\(\rightarrow\ \ 60\ \text{cm}=0.6\ \text{m}\)

\(\rightarrow\ \ 30\ \text{cm}=0.3\ \text{m}\)

\(\text{Area of 1 paver}\) \(=0.6^2-0.3^2\)
  \(=0.36-0.09\)
  \(=0.27\ \text{m}^2\)

 

\(\text{Total pool area paved}\) \(=0.27\times 50\)
  \(=13.5\ \text{m}^2\)

Area, SM-Bank 068

A plan of Bob's outdoor area is shown below.

  1. Calculate the area of Bob's outdoor area in square metres.   (2 marks)

  2. What is the cost of tiling Bob's outdoor area, if tiles cost $42.50 per square metre?   (2 marks)

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a.    \(180\ \text{m}^2\)

b.    \($7650\)

Show Worked Solution

a.   \(\text{Outdoor area}\)

\(\text{Total area}\) \(=5\times 8+7\times 20\)
  \(=40+140\)
  \(=180\ \text{m}^2\)

 

b.    \(\text{Cost of tiling}\) \(=180\times $42.50\)
    \(=$7650\)

Area, SM-Bank 067 MC

Bernie drew this plan of his timber deck.
 

Which expression gives the area of Bernie's timber deck?

  1. \((c+d)-(a+b)\)
  2. \((c\times d)-(a\times b)\)
  3. \((c\times d)\times (a\times b)\)
  4. \((c\times d)+(a\times b)\)
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\(B\)

Show Worked Solution

\(\text{Total area}\) \(=\text{Area}\ 1-\text{Area}\ 2\)
  \(=(c\times d)-(a\times b)\)

\(\Rightarrow B\)

Area, SM-Bank 066 MC

Vera drew this plan of her entertaining area.

Which expression gives the area of Vera's entertaining area?

  1. \((e\times f)\times (a\times b)\times (c\times d)\)
  2. \((e\times f)+(a\times b)+(c\times d)\)
  3. \((e+f)+(a+b)+(c+d)\)
  4. \((e\times f)+(a\times (b+d))+(c\times d)\)
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\(D\)

Show Worked Solution

\(\text{Total area}\) \(=\text{Area}\ 1+\text{Area}\ 2+\text{Area}\ 3\)
  \(=(e\times f)+(a\times (d+b))+(c\times d)\)

\(\Rightarrow D\)

Area, SM-Bank 065 MC

Olive drew this plan of her lawn.

Which expression gives the area of Olive's lawn?

  1. \((a\times b)+(c\times d)\)
  2. \((a\times b)\times (c\times d)\)
  3. \((a+b)+(c+d)\)
  4. \((a+b)\times (c+d)\)
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\(A\)

Show Worked Solution

\(\text{Total area}\) \(=\text{Area}\ 1+\text{Area}\ 2\)
  \(=(a\times b)+(c\times d)\)

\(\Rightarrow A\)

Area, SM-Bank 040

Cabins are being built at a camp site.

The dimensions of the front of each cabin are shown in the diagram below.
 

The walls of each cabin are 2.4 m high.

The sloping edges of the roof of each cabin are 2.4 m long.

The front of each cabin is 4 m wide.

The pependicular height the triangular shaped roof is `h` metres.

  1. Use Pythagoras to show that the value of \(h\) is 1.33 m, correct to two decimal places.  (2 marks)

  2. Calculate the total area of the front of the cabin.  (2 marks)

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a.    \(1.33\ \text{m}\)

b.    \(12.26\ \text{m}^2\)

Show Worked Solution

a.    \(\text{Using Pythagoras:  }a^2+b^2=c^2\)

\(h^2+2^2\) \(=2.4^2\)
\(h^2\) \(=2.4^2-2^2\)
\(h^2\) \(=1.76\)
\(h\) \(=\sqrt{1.76}\)
  \(=1.326\dots\)
  \(\approx 1.33\ \text{m}\ (2\ \text{d.p.}\)

 

b.   \(\text{Area of walls and roof}\)

\(=\text{Area of Rectangle}+\text{Area of Triangle}\)

\(=4\times 2.4+\dfrac{1}{2}\times 4\times 1.33\)

\(=12.26\ \text{m}^2\)

Area, SM-Bank 038

The following diagram shows a cargo ship viewed from above.
 

The shaded region illustrates the part of the deck on which shipping containers are stored.

What is the area, in square metres, of the shaded region?  (2 marks)

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\(6700\ \text{m}^2\)

Show Worked Solution
\(\text{Area}\) \(= 160\times 40+12\times 25\)
  \(=6700\ \text{m}^2\)

Area, SM-Bank 035

\(PQRS\) is a square of side length 4 m as shown in the diagram below.

The distance \(ST\) is 1 m.

Calculate the shaded area \(PQTS\) in square metres.  (2 marks)

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\(10\ \text{m}^2\)

Show Worked Solution

\(\text{Method 1:}\)

\(\text{Area of}\ \Delta QRT\) \(=\dfrac{1}{2}\times RT\times QR\)
  \(=\dfrac{1}{2}\times 3\times 4\)
  \(=6\ \text{m}^2\)

 
\(\therefore\ \text{Shaded Area}\ =4\times 4-6 =10\ \text{m}^2\)
 

\(\text{Method 2:}\)

\(\text{Area of Trapezium }PSQT\) \(=\dfrac{PS}{2}(ST+PQ)\)
  \(=\dfrac{4}{2}(1+4)\)
  \(=10\ \text{m}^2\)

Area, SM-Bank 033 MC

A piece of cardboard is shown in the diagram below.
 

     

The area of the cardboard, in square centimetres, is

  1.   4
  2. 21
  3. 25
  4. 29
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\(B\)

Show Worked Solution
\(\text{Area}\) \(=\text{Large square}-4\times\text{Corner squares}\)
  \(=(5\times 5)-4\times (1\times 1)\)
  \(=21\ \text{cm}^2\)

 
\(\Rightarrow B\)

Area, SM-Bank 031 MC

Consider the diagram below.
 

The shaded area, in square centimetres, is

  1. \(35\)
  2. \(45\)
  3. \(60\)
  4. \(95\)
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\(A\)

Show Worked Solution
\(\text{Shaded Area}\) \(=\text{Area of Triangle}-\text{Area of Square}\)
  \(=\Bigg(\dfrac{1}{2}\times 12\times 10\Bigg)-(5\times5)\)
  \(=60-25\)
  \(=35\ \text{cm}^2\)

\(\Rightarrow A\)

Area, SM-Bank 042

Calculate the area of the composite figure below.  (2 marks)
 

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\(37\ \text{cm}^2\)

Show Worked Solution
\(\text{Area}\) \(=\text{Area of upper rectangle}+\text{Area of lower rectangle}\)
  \(=9\times 3+5\times 2\)
  \(=37\ \text{cm}^2\)

Area, SM-Bank 030

Calculate the area of the composite figure below.  (2 marks)
 

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\(95\ \text{cm}^2\)

Show Worked Solution
\(\text{Area}\) \(=\text{Area of large rectangle}-\text{Area of small rectangle}\)
  \(=16\times 11-9\times 9\)
  \(=95\ \text{cm}^2\)

Area, SM-Bank 029

Mitchell lays rubber matting in his gym, as shown below.
 

What is the area of his gym in square metres?   (2 marks)

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\(41.9\ \text{m}^2\)

Show Worked Solution

\(\text{Area of large rectangle}\) \(=3.7\times 11\)
  \(=40.7\ \text{m}^2\)

 

\(\text{Area of small rectangle}\) \(=0.8\times 1.5\)
  \(= 1.2\ \text{m}^2\)

\(\therefore\ \text{Total area of gym}\)

\(=40.7+1.2\)

\(=41.9\ \text{m}^2\)

Area, SM-Bank 028

Lilo lays turf on his terrace, as shown below.

What is the area of his terrace in square metres?  (2 marks)

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\(368.56\ \text{m}^2\)

Show Worked Solution

\(\text{Area of rectangle 1}+\text{Area of rectangle 2}\) \(=15\times 5+35.8\times 8.2\)
  \(= 75+293.56\)
  \(=368.56\ \text{m}^2\)

Area, SM-Bank 027

 

A regular hexagon has side length 3.0 cm and height 5.2 cm as shown in the diagram above.

Calculate the area of the hexagon, giving your answer correct to one decimal place.   (3 marks)

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\(23.4\ \text{cm}^2\)

Show Worked Solution

 GEOMETRY, FUR1 2008 VCAA 8 MC Answer

\(\text{Area of rectangle}\) \(=3.0\times 5.2\)
  \(= 15.6\ \text{cm}^2\)

 
\(\text{Using Pythagoras to find}\ h:\)

\(3.0^2\) \(=2.6^2+h^2\)
 \(h^2\) \(=9-6.76\)
\(h^2\) \(=2.24\)
 \(h\) \(=1.496\dots\)

 

\(\text{Area of}\ \Delta ABC\)

\(=\dfrac{1}{2}\times bh\)

\(=\dfrac{1}{2}\times 5.2\times 1.496\dots\)

\(= 3.891\dots\ \text{cm}^2\)

 

\(\therefore\ \text{Area of hexagon}\)

\(=15.6+(2\times 3.891\dots)\)

\(=23.382\dots\approx 23.4\ \text{cm}^2\)

Area, SM-Bank 023

Calculate the area of this composite shape.   (2 marks)
 

 

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\(44\ \text{cm}^2\)

Show Worked Solution

\(\text{Method 1:  Subtraction}\)

\(\text{Area}\) \(=\text{Area of large rectangle}-\text{Area of cut-out rectangle}\)
  \(=(10\times 6)-(8\times 2)\)
  \(=44\ \text{cm}^2\)

 
\(\text{Method 2:  Addition}\)

\(\text{Area}\) \(=\text{Area of large rectangle}+\text{Area of small rectangle}\)
  \(=(10\times 4)+(2\times 2)\)
  \(=44\ \text{cm}^2\)

Area, SM-Bank 016 MC

A trapezium is constructed on a grid of 10 rectangles.

Each rectangle measures  3 cm × 7 cm.
 


   

What is the area of the trapezium?

  1. \(150\ \text{cm}^2\)
  2. \(168\ \text{cm}^2\)
  3. \(189\ \text{cm}^2\)
  4. \(210\ \text{cm}^2\)
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Method 1: Composite}\)

\(\therefore\ \text{Total Area}\) \(=\text{Area 1 rectangle}+2\times\ \text{Area of triangle}\)
  \(=6\times 21+2\times\Bigg(\dfrac{1}{2}\times 3\times 14\Bigg)\)
  \(=126+42\)
  \(=168\ \text{cm}^2\)

  
\(\text{Method 2: Trapezium}\)

\(\text{Area}\) \(=\dfrac{h}{2}(a+b)\)
  \(=\dfrac{14}{2}(15+9)\)
  \(=168\ \text{cm}^2\)

 
\(\Rightarrow B\)

Area, SM-Bank 013

Lucy designs an outdoor table that is in the shape of a trapezium.

The dimensions of the table top are shown in the picture below.

What is the area of Lucy's table top?   (2 marks)

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\(2600\ \text{cm}^2\)

Show Worked Solution

\(\text{Method 1:  Composite}\)

\(\text{Area}\) \(=\text{Area of rectangle}+2\times \text{Area of triangle}\)
  \(=(50\times 40) + 2\times\Bigg(\dfrac{1}{2}\times 15\times 40\Bigg)\)
  \(=2000 + 600\)
  \(=2600\ \text{cm}^2\)

 

\(\text{Method 2:  Trapezium}\)

\(\text{Area}\) \(=\dfrac{h}{2}(a+b)\)
  \(=\dfrac{40}{2}(80+50)\)
  \(=2600\ \text{cm}^2\)

Area, SM-Bank 012

Luke designs a table that is in the shape of a trapezium.

The dimensions of the table top are shown in the picture below.
 

What is the area of Luke's table top?   (2 marks)

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\(880\ \text{cm}^2\)

Show Worked Solution

\(\text{Method 1:  Composite}\)

\(\text{Area}\) \(=\text{Area of rectangle}+2\times \text{Area of triangle}\)
  \(=(38\times 20) + 2\times\Bigg(\dfrac{1}{2}\times 6\times 20\Bigg)\)
  \(=760 + 120\)
  \(=880\ \text{cm}^2\)

 

\(\text{Method 2:  Trapezium}\)

\(\text{Area}\) \(=\dfrac{h}{2}(a+b)\)
  \(=\dfrac{20}{2}(38+50)\)
  \(=880\ \text{cm}^2\)