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Area, SM-Bank 052

The following shape has a perimeter of 36 metres. Calculate its' area.  (2 marks)

 

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\(81\ \text{m}^2\)

Show Worked Solution
\(\text{Perimeter}\) \(=36\ \text{m}\)
\(\therefore\ \text{Side}\) \(=\dfrac{36}{4}\)
  \(=9\ \text{m}\)

 

\(\text{Area}\) \(=s^2\)
  \(=9^2\)
  \(=81\ \text{m}^2\)

Area, SM-Bank 051

Calculate the area of the following squares.

  1.  
      (2 marks)

  2.  
      (2 marks)

Show Answers Only

a.    \(151.29\ \text{cm}^2\)

b.    \(3.24\ \text{m}^2\)

Show Worked Solution
a.    \(\text{Area}\) \(=s^2\)
    \(=12.3^2\)
    \(=151.29\ \text{cm}^2\)

 

b.    \(\text{Area}\) \(=s^2\)
    \(=1.8^2\)
    \(=3.24\ \text{m}^2\)

Area, SM-Bank 050

Calculate the area of the following squares.

  1.  
        (2 marks)

     
  2.  
       (2 marks)

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a.    \(64\ \text{cm}^2\)

b.    \(20\ 164\ \text{mm}^2\)

Show Worked Solution
a.    \(\text{Area}\) \(=s^2\)
    \(=8^2\)
    \(=64\ \text{cm}^2\)

 

b.    \(\text{Area}\) \(=s^2\)
    \(=142^2\)
    \(=20\ 164\ \text{mm}^2\)

Area SM-Bank 049

A rectangle has an area of 24 square centimetres.

  1. One possible pair of integer dimensions for this rectangle is \(2\ \text{cm}\times 12\ \text{cm}\).
    Write down all possible pairs of integer dimensions for a rectangle with an area of 24 square centimetres.  (2 marks)

  2. Using the given dimensions and your answers from (a), calculate the largest possible perimeter for a rectangle with an area of 24 square centimetres.  (2 marks)


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a.    \(1\ \text{cm}\times 24\ \text{cm}, 2\ \text{cm}\times 12\ \text{cm}, 3\ \text{cm}\times 8\ \text{cm}, 4\ \text{cm}\times 6\ \text{cm}\)

b.    \(50\ \text{cm}\)

Show Worked Solution

a.    \(\text{All possible integer dimensions:}\)

\(1\ \text{cm}\times 24\ \text{cm},\ \ 2\ \text{cm}\times 12\ \text{cm},\ \ 3\ \text{cm}\times 8\ \text{cm},\ \ 4\ \text{cm}\times 6\ \text{cm}\)

b.    \(\text{Perimeters}\)

\(\text{P}_{1}=2\times 1+2\times 24=50\ \text{cm}\)

\(\text{P}_{2}=2\times 2+2\times 12=28\ \text{cm}\)

\(\text{P}_{3}=2\times 3+2\times 8=22\ \text{cm}\)

\(\text{P}_{4}=2\times 4+2\times 6=20\ \text{cm}\)

\(\therefore\ \text{Largest possible perimeter}= 50\ \text{cm}\)

Area, SM-Bank 048

Jocasta is sewing a quilt in the shape of a rectangle, as shown below. She knows the length of one side, and the length of diagonal of the quilt.

  1. Calculate the length of the other side of the quilt, giving your answer in exact surd form.  (2 marks)

  2. Using your answer from (a) calculate the area of the quilt in square metres, correct to 1 decimal place?  (2 marks)


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a.    \(\sqrt{2.05}\ \text{m}\)

b.    \(2.6\ \text{m}^2\ (1\ \text{d.p.})\)

Show Worked Solution

a.    \(\text{Using Pythagoras to find the shorter side:}\)

\(a^2+b^2\) \(=c^2\)
\(a^2+1.8^2\) \(=2.3^2\)
 \(a^2\) \(=2.3^2-1.8^2\)
\(a^2\) \(=2.05\)
 \(a\) \(=\sqrt{2.05}\ \text{m}\)

 

b.    \(\text{Area}\) \(=l\times b\)
    \(=1.8\times \sqrt{2.05}\)
    \(=2.577\dots\)
    \(=2.6\ \text{m}^2\ (1\ \text{d.p.})\)

Area, SM-Bank 047

Jordy is tiling the rectangular living area pictured below.

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

  2. What is the cost of tiling the living area if tiles cost $45 per square metre?  (2 marks)

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a.    \(25.84\ \text{m}^2\)

b.    \($1162.80\)

Show Worked Solution
a.    \(\text{Area}\) \(=l\times b\)
    \(=7.6\times 3.2\)
    \(=25.84\ \text{m}^2\)

 

b.    \(\text{Cost}\) \(=\text{price per square metre}\times \text{number of square metres}\)
    \(=$45\times 25.84\)
    \(=$1162.80\)

Area, SM-Bank 046

A rectangular paddock has dimensions 1.2 kilometres by 1.4 kilometres. Calculate the area of the paddock in square kilometres.  (2 marks)

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\(1.68\ \text{km}^2\)

Show Worked Solution
\(\text{Area}\) \(=l\times b\)
  \(=1.2\times 1.4\)
  \(=1.68\ \text{km}^2\)

Area, SM-Bank 045

Calculate the area of the following rectangles, correct to 1 decimal place.

  1.  
        (2 marks)

  2.  
       (2 marks)

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a.    \(82.1\ \text{m}^2\ (1\ \text{d.p.})\)

b.    \(1023.8\ \text{mm}^2\ (1\ \text{d.p.})\)

Show Worked Solution
a.    \(\text{Area}\) \(=l\times b\)
    \(=7.2\times 11.4\)
    \(=82.08\)
    \(=82.1\ \text{m}^2\ (1\ \text{d.p.})\)

 

b.    \(\text{Area}\) \(=l\times b\)
    \(=67.8\times 15.1\)
    \(=1023.78\)
    \(=1023.8\ \text{mm}^2\ (1\ \text{d.p.})\)

Area, SM-Bank 044

Calculate the area of the following shapes in square centimetres.

  1.   
       (2 marks)

  2.   
       (2 marks)

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a.    \(78\ \text{cm}^2\)

b.    \(274\ \text{cm}^2\)

Show Worked Solution
a.    \(\text{Area}\) \(=l\times b\)
    \(=13\times 6\)
    \(=78\ \text{cm}^2\)

 

b.    \(\text{Area}\) \(=l\times b\)
    \(=10\times 27.4\)
    \(=274\ \text{cm}^2\)

Area, SM-Bank 043

Calculate the area of the following shapes in square units.

  1.  
              (1 mark)

     

  2.       
            (1 mark)

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a.    \(12\ \text{square units}\)

b.    \(10\ \text{square units}\)

Show Worked Solution

a.    \(\text{Area}=4\times 3=12\ \text{square units}\)

b.    \(\text{Area}\) \(=\text{Triangle }1+\text{Triangle }2\)
    \(=\dfrac{1}{2}\times 15+\dfrac{1}{2}\times 5\)
    \(=10\ \text{square units}\)

 

Area, SM-Bank 041

A golf course has a sprinkler system that waters the grass in the shape of a sector, as shown in the diagram below. 
 

A sprinkler is positioned at point \(L\) and can turn through an angle of 100°.

The section of grass that is watered is 4.5 m wide at all points.

Water can reach a maximum of 12 m from the sprinkler at \(L\).

What is the area of grass that this sprinkler will water?

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

NOTE:  \(\text{Sector Area}=\dfrac{\theta}{360}\times \pi r^2\)

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`199\ text{m²  (nearest m²)}`

Show Worked Solution
\(\text{Area}\) \(=\dfrac{360-\theta}{360}\times \pi\times R^2-\dfrac{360-\theta}{360}\times \pi\times r^2\)
  \(=\dfrac{260}{360}\times \pi\times 12^2-\dfrac{260}{360}\times \pi\times 7.5^2\)
  \(= 199.09\dots\)
  \(= 199\ \text{m²  (nearest m}^2)\)

 

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 039

The game of squash is played indoors on a court with a front wall, a back wall and two side walls, as shown in the image below.
 

 
Each side wall has the following dimensions.
 

The shaded region in the diagram above is considered part of the playing area.

Calculate the area, in square metres, of the shaded region in the diagram above. Round your answer to two decimal places.  (2 marks)

Show Answers Only

\(32.66 \ \text{m}^2\)

Show Worked Solution
\(\text{Shaded Area (trapezium)}\) \(=\dfrac{h}{2}(a+b)\)
  \(=\dfrac{9.75}{2}\times (4.57 + 2.13)\)
  \(=32.6625\)
  \(= 32.66\ \text{m}^2 \ (2\ \text{d.p.)}\)

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 037

The pie chart below displays the results of a survey.
 


 

Eighty per cent of the people surveyed selected ‘agree’.

Twenty per cent of the people surveyed selected ‘disagree’.

The radius of the pie chart is 16 mm.

Calculate the area of the "agree" sector, correct to the nearest square millimetre.  (2 marks)

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\(643\ \text{mm}^2\)

Show Worked Solution

\(\text{Agree represents }80\text{% of the circle}\)

\(\text{Area}\) \(=80\text{%}\times \pi r^2\)
  \(=0.8\times \pi\times 16^2\)
  \(=643.39\dots\)
  \(\approx 643\ \text{mm}^2\)

 

Area, SM-Bank 036

A windscreen wiper blade can clean a large area of windscreen glass, as shown by the shaded area in the diagram below.
 

 

The windscreen wiper blade is \(30\) cm long and it is attached to a \(9\) cm long arm.

The arm and blade move back and forth in a circular arc with an angle of \(110^\circ\) at the centre.

Calculate the area cleaned by this blade, in square centimetres, correct to one decimal place.  (2 marks)

NOTE:  \(\text{Sector area}=\dfrac{\theta}{360}\times \pi r^2\)

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\(1382.3\ \text{cm}^2\)

Show Worked Solution

\(\theta=110^\circ,\ \text{Radius large sector}=39,\ \text{Radius small sector}=9\)

\(\text{Area cleaned}\) \(\ =\ \text{large sector − small sector}\)
  \(=\dfrac{110}{360}\times \pi\times 39^2-\dfrac{110}{360}\times \pi\times 9^2\)
  \(=1382.300\dots\)
  \(\approx 1382.3\ \text{cm}^2\)

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 034

The radius of a circle is 6.5 centimetres.

A square has the same area as this circle.

Calculate the side length of the square, in centimetres correct to one decimal place.   (3 marks)

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\(11.5\ \text{cm}\)

Show Worked Solution
\(\text{Area of circle}\) \(=\pi r^2\)
  \(=\pi\times 6.5^2\)
  \(= 132.73\dots\ \text{cm}^2\)

 
\(\text{Area of square}\ =s^2=\ \text{Area of circle}\)

\(s^2\) \(= 132.73\dots\)
\(s\) \(=\sqrt{132.73\dots}\)
  \(= 11.5\ \text{cm (1 d.p.)}\)

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 032 MC

A flag consists of three different coloured sections: red, white and blue.

The flag is 3 m long and 2 m wide, as shown in the diagram below.
 

The blue section is an isosceles triangle that extends to half the length of the flag.

The area of the blue section, in square metres, is

  1.  1.5
  2.  2
  3.  3
  4.  6
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\(A\)

Show Worked Solution

\(\text{Perpendicular height}\ \Delta=\dfrac{1}{2}\times 3=1.5\ \text{m}\)

\(\text{Area }\Delta\) \(=\dfrac{1}{2}\times bh\)
  \(=\dfrac{1}{2}\times 2\times 1.5\)
  \(= 1.5\ \text{m}^2\)

 
\(\Rightarrow A\)

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)

Show Answers Only

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

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)

Show Answers Only

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

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)

Show Answers Only

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

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 026 MC

David's backyard is in shape of a rectangle and has an area of \(50\ \text{m}^2\).

Tim's backyard is also rectangular but with side lengths that are double those of David's backyard.

What is the area of Tim's backyard.

  1. \(25\ \text{m}^2\)
  2. \(100\ \text{m}^2\)
  3. \(200\ \text{m}^2\)
  4. \(300\ \text{m}^2\)
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Let the dimensions of David’s yard be:}\ \ x, y\)

\(\text{Area} =x\times y\)

  
\(\Longrightarrow\ \text{Tim’s yard’s dimensions are:}\ \ 2x, 2y\)

\(\text{Area}\) \(=2x\times 2y\)
  \(=4\times xy\)
  \(= 4\times 50\)
  \(= 200\ \text{m}^2\)

 
\(\Rightarrow C\)

Area, SM-Bank 025 MC

Mark decides to make the paddock on his farm bigger.

The paddock is in the shape of a rectangle.

Mark makes both the length and the width of the paddock 3 times longer.

How many times bigger is the area of the new paddock than the original paddock?

  1. \(3\)
  2. \(4\)
  3. \(6\)
  4. \(9\)
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Area of original paddock}=l\times b\)

\(\text{Area of new paddock}\) \(=3l\times 3b\)
  \(=9\times (l\times b)\)
  \(=9\times \text{Area of original paddock}\)

\(\Rightarrow D\)

Area, SM-Bank 024

A rectangle has a length of 25 cm and a width of 20 cm.

A square has the same perimeter as this rectangle.

What is the area of this square in square centimetres?  (2 marks)

Show Answers Only

\(506.25\ \text{cm}^2\)

Show Worked Solution
\(\text{Perimeter of rectangle}\) \(=(2\times 25)+(2\times 20)\)
  \(=90\ \text{cm}\)

 

\(\therefore\ \text{Side length of square}\) \(=\dfrac{90}{4}\)
  \(=22.5\ \text{cm}\)

 

\(\therefore\ \text{Area of square}\) \(=\text{(side)}^2\)
  \(=22.5^2\)
  \(=506.25\ \text{cm}^2\)

Area, SM-Bank 023

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

 

Show Answers Only

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

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 022 MC

Bruce has a square and a rectangle.

He cuts the rectangle along the dotted line as shown below.
 


 

Bryce then joins these 3 pieces to make a parallelogram.
 


 

Which of these calculations could Bryce use to find the area of the parallelogram?

  1. \(3\ \text{cm}\times 6\ \text{cm}\)
  2. \(6\ \text{cm}\times 9\ \text{cm}\)
  3. \(12\ \text{cm}\times 6\ \text{cm}\)
  4. \(21\ \text{cm}\times 3\ \text{cm}\)
Show Answers Only

\(B\)

Show Worked Solution
\(\text{Area}\) \(=\text{area of square}+\text{area of rectangle}\)
  \(=6\times 6+3\times 6\)
  \(=6\times (6+3)\)
  \(=6\times 9\)

 
\(\Rightarrow B\)

Area, SM-Bank 021

A square has an area of \(81\ \text{cm}^2\).

A rectangle has the same perimeter as the square and has a width of \(15\ \text{cm}\).

What is the length of the rectangle in centimetres?  (2 marks)

Show Answers Only

\(3\ \text{cm}\)

Show Worked Solution

\(\text{Area of square} = 81\ \text{cm}^2\)

\(\therefore\ \text{Side length of square}=\sqrt{81}=9\ \text{cm}\)

\(\therefore\ \text{Perimeter of square}=4\times 9=36\ \text{cm}\)

\(\therefore\ \text{Length of rectangle}\) \(=\dfrac{36-(2\times 15)}{2}\)
  \(=\dfrac{6}{2}\)
  \(=3\ \text{cm}\)

Area, SM-Bank 020

Dave has a backyard in the shape of a rectangle.
 

   
 

The longer side is  \(1\dfrac{1}{3}\)  times longer than the shorter side.

What is the area of Dave’s backyard?  (2 marks)

Show Answers Only

\(588\ \text{m}^2\)

Show Worked Solution
\(\text{Longer side}\) \(=21\times \dfrac{4}{3}\)
  \(=28\ \text{m}\)

 

\(\therefore\ \text{Area}\) \(=21\times 28\)
  \(=588\ \text{m}^2\)

Area, SM-Bank 019 MC

Brock decided to cut out a small rectangle from a piece of patterned rectangular paper.

The rectangle cut out has a length of 60 mm and a height of 40 mm.

Which of the following expressions gives the area of patterned paper that was left after cutting out the smaller rectangle?
 

  1. \((120\times 200)-(60\times 160)\)
  2. \((60\times 160)\)
  3. \((120\times 200)-(40\times 160)\)
  4. \((120\times 200)-(60\times 40)\)

Show Answers Only

\(D\)

Show Worked Solution
\(\text{Area}\) \(=\text{Area of large rectangle}-\text{Area of small rectangle}\)
  \(=(120\times 200)-(60\times 40)\)

 
\(\Rightarrow D\)

Area, SM-Bank 018

Airships are a form of aircraft.

An airship has a cabin in which the pilots and passengers travel, and cargo is carried. This is shown in the simplified diagram below.
 

The floor of the cabin is a rectangle, with a length of 9 m and a width of 2.5 m.

The cockpit occupies an area 1.5 m by 2.5 m at the front of the cabin. This is shown shaded in the diagram below.

The remainder of the floor space is available for passengers and cargo.
 

Calculate the area available for passengers and cargo, in square metres.  (2 marks)

Show Answers Only

\(18.75\ \text{m}^2\)

Show Worked Solution
\(\text{Area}\) \(=9\times 2.5-1.5\times 2.5\)
  \(=22.5-3.75\)
  \(=18.75\ \text{m}^2\)

Area, SM-Bank 017

Sequoia owns a farm with a rectangular paddock.

She increases the area of the paddock by adding land that changes it into the shape of a trapezium.
 

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

Show Answers Only

\(2599\ \text{m}^2\)

Show Worked Solution
\(\text{Area trapezium}\) \(=\dfrac{h}{2}(a+b)\)
  \(=\dfrac{46}{2}(71+42)\)
  \(=2599\ \text{m}^2\)

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 015

The floor of a chicken coop is in the shape of a trapezium.

The floor, \(ABCD\), and the chicken coop are shown below.

 

\(AB = 3\ \text{m}, BC = 2\ \text{m and}\ \ CD = 5\ \text{m.}\)
 

  1. What is the area of the floor of the chicken coop?

     

    Write your answer in square metres.  (2 marks)

  2. What is the perimeter of the floor of the chicken coop?

     

    Write your answer in metres, correct to one decimal place.  (2 marks)

Show Answers Only

a.    \(8\ \text{m}^2\)

b.    \(12.8\ \text{m}\)

Show Worked Solution
a.    \(A\) \(=\dfrac{h}{2}(a+b)\)
    \(=\dfrac{2}{2}\times (3 + 5)\)
    \(=8\ \text{m}^2\)

 

b.  

\(\text{Using Pythagoras,}\)

\(AD^2\) \(=2^2+2^2\)
\(AD^2\) \(=8\)
\(AD\) \(=\sqrt{8}\)
  \(=2.82\dots\ \text{m}\)

 

\(\therefore\ \text{Perimeter}\) \(=3+2+5+2.82\dots\)
  \(=12.8\ \text{m  (1 d.p.)}\)

Area, SM-Bank 014 MC

The top of a table is in the shape of a trapezium, as shown below.

The area of the tabletop, in square centimetres, is

  1. \(200\)
  2. \(4200\)
  3. \(4800\)
  4. \(288\ 000\)
Show Answers Only

\(B\)

Show Worked Solution
\(\text{Area}\) \(=\dfrac{h}{2}(a+b)\)
  \(=\dfrac{60}{2}(80+60)\)
  \(=4200\)

 
\(\Rightarrow B\)

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)

Show Answers Only

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

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)

Show Answers Only

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

Show Worked Solution

\(\text{Method 1:  Composite}\)

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

 

\(\text{Method 2:  Trapezium}\)

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

Area, SM-Bank 011

This triangle was made by cutting a square in half.
 

 
The perimeter of the triangle is 51.21 cm.

What is the area of the triangle?  (2 marks)

Show Answers Only

\(112.5\ \text{cm}^2\)

Show Worked Solution

\(\text{Triangle is isosceles with perimeter}=51.21\ \text{cm.}\)

\(\Rightarrow\ \text{Length of triangle side}\)

\(=\dfrac{1}{2}\times (51.21 – 21.21)\)

\(=\dfrac{1}{2}\times 30\)

\(=15\ \text{cm}\)

\(\therefore\ \text{Area}\) \(=\dfrac{1}{2}\times bh\)
  \(=\dfrac{1}{2}\times 15\times 15\)
  \(=112.5\ \text{cm}^2\)

Area, SM-Bank 010

A square has an area of 169 square centimetres.

What is the perimeter?  (2 marks)

Show Answers Only

\(52\ \text{cm}\)

Show Worked Solution

\(\text{Let}\ \ s=\text{side length of the square}\)

\(\text{Area:}\rightarrow\ \ s^2\) \(=169\)
\(s\) \(=\sqrt{169}\)
  \(=13\ \text{cm}\)

 

\(\therefore\ \text{Perimeter}\) \(=4\times 13\)
  \(=52\ \text{cm}\)

Area, SM-Bank 009

Jill is playing in her parents' rectangular courtyard.

The courtyard is measured at 6 metres by 10 metres.

Jill draws a triangle on the courtyard with chalk, pictured below.

What is the area of this triangle?  (2 marks)
 

Show Answers Only

\(14\ \text{m}^2\)

Show Worked Solution

\(\text{Base}=7\ \text{m}, \text{ Height}=4\ \text{m}\)

\(\text{Area of}\ \Delta\) \(=\dfrac{1}{2}\times bh\)
  \(=\dfrac{1}{2}\times 7\times 4\)
  \(=14\ \text{m}^2\)

Area, SM-Bank 008

The length of this rectangle is one and a half times its width.
 

 
The perimeter of the rectangle is 50 centimetres.

What is the area of the rectangle?  (2 marks)

Show Answers Only

\(150\ \text{cm}^2\)

Show Worked Solution

\(\text{Let}\ x=\text{width, then}\ \ 1.5x=\text{length}\)

\(\text{Perimeter:}\ \rightarrow\ \) \(\ \ 2\times x+2\times 1.5 x\) \(=50\)
  \(5x\) \(=50\)
  \(x\) \(=10\)

\(\therefore\ \text{width}=10\ \text{cm,  length}=15\ \text{cm}\)

\(\therefore\ \text{Area}\) \(=10\times 15\)
  \(=150\ \text{cm}^2\)

Area, SM-Bank 007 MC

A triangle is drawn on grid paper.
 

 What is the area of the triangle?

  1. \(6\ \text{square units}\)
  2. \(8\ \text{square units}\)
  3. \(12\ \text{square units}\)
  4. \(16\ \text{square units}\)
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Area}\) \(=\dfrac{1}{2}\times bh\)
  \(=\dfrac{1}{2}\times 4\times 4\)
  \(= 8\ \text{square units}\)

 
\(\Rightarrow B\)

Area, SM-Bank 006 MC

A triangle is drawn on grid paper.
  

 What is the area of the triangle?

  1. \(15\ \text{square units}\)
  2. \(16\ \text{square units}\)
  3. \(18\ \text{square units}\)
  4. \(20\ \text{square units}\)
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Area}\) \(=\dfrac{1}{2}\times bh\)
  \(=\dfrac{1}{2}\times 6\times 5\)
  \(= 15\ \text{square units}\)

 
\(\Rightarrow A\)

Area, SM-Bank 005

The area of the shaded rectangle below is \(84\ \text{cm}^2\).
 

 
What is the length of the shaded rectangle?  (2 marks)

Show Answers Only

\(14\ \text{cm}\)

Show Worked Solution

\(\text{Let}\ \ l =\text{length of rectangle}\)

\(l\times 6\) \(=84\)
\(\therefore\ l\) \(=\dfrac{84}{6}\)
  \(=14\ \text{cm}\)

Area, SM-Bank 004 MC

A neighbourhood soccer oval is marked out with the dimensions shown below.
 

 
What is the area of the field?

  1. \(298\ \text{m}^2\)
  2. \(2842.75\ \text{m}^2\)
  3. \(5261.25\ \text{m}^2\)
  4. \(8123.25\ \text{m}^2\)
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Area of the soccer field}\)

\(=57.5\times 91.5\)

\(= 5261.25\ \text{m}^2\)

\(\Rightarrow C\)

Area, SM-Bank 003 MC

A resort has 4 pools.

Which pool has the largest surface area?
 

Show Answers Only

\(D\)

Show Worked Solution

\(\text{Consider the surface area of each pool:}\)

\(\text{Pool A}=6\times 19=114\ \text{m}^2\)

\(\text{Pool B}=7\times 18=126\ \text{m}^2\)

\(\text{Pool C}=10\times 15=150\ \text{m}^2\)

\(\text{Pool D}=12.5\times 12.5=156.25\ \text{m}^2\)

 
\(\therefore\ \text{Pool D,}\ 12.5\times 12.5,\text{ has the largest}\)

\(\text{surface area.}\)

\(\Rightarrow D\)

Area, SM-Bank 002 MC

Bob had a mini tennis court in his backyard, as shown in the diagram below.
 

What is the area of the mini tennis court?

  1. \(6\ \text{m}^2\)
  2. \(12\ \text{m}^2\)
  3. \(24\ \text{m}^2\)
  4. \(48\ \text{m}^2\)
Show Answers Only

\(C\)

Show Worked Solution

\(\text{The mini tennis court has dimensions}\ \ 6\times 4\)

\(\therefore\ \text{Area}\) \(=6\times 4\)
  \(=24\ \text{m}^2\)

 
\(\Rightarrow C\)

Area, SM-Bank 001 MC

Which triangle has an area greater than 4 square units?
 

  1. \(\text{Triangle A}\)
  2. \(\text{Triangle B}\)
  3. \(\text{Triangle C}\)
  4. \(\text{Triangle D}\)
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Consider All options:}\)

\(\text{Triangle A}\) \(=\dfrac{1}{2}\times 3\times 2=3\ \text{square units}\)
\(\text{Triangle B}\) \(=\dfrac{1}{2}\times 3\times 3=4.5\ \text{square units}\ \checkmark\)
\(\text{Triangle C}\) \(=\dfrac{1}{2}\times 2\times 4=4\ \text{square units}\)
\(\text{Triangle D}\) \(=\dfrac{1}{2}\times 1\times 6=3\ \text{square units}\)

 
\(\Rightarrow B\)

Right-angled Triangles, SM-Bank 053

  1. Use Pythagoras' Theorem to calculate the length of the hypotenuse in the isosceles triangle below. Give your answer correct in exact surd form.  (2 marks)
     
           

  2. Using your result from part (a) above, calculate the perimeter of the shape below, correct to 1 decimal place.  (2 marks)
     
         

Show Answers Only

a.    \(\sqrt{50}\ \text{cm (exact surd form)}\)

b.    \(30.6\ \text{cm (1 d.p.)}\)

Show Worked Solution

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

\(\text{Let }a=5\ \text{and }b=5\)

\(\text{Then}\ \ c^2\) \(=5^2+5^2\)
\(c^2\) \(=50\)
\(c\) \(=\sqrt{50}\ \text{cm (exact surd form)}\)

 

b.    \(\text{Perimeter}\) \(=\text{chord (a)}\ +\dfrac{3}{4}\times\text{circumference}\)
    \(=\sqrt{50}+\dfrac{3}{4}\times 2\pi r\)
    \(=\sqrt{50}+\dfrac{3}{4}\times 2\pi\times 5\)
    \(=30.633\dots\)
    \(\approx 30.6\ \text{cm (1 d.p.)}\)

Right-angled Triangles, SM-Bank 052

Use Pythagoras' Theorem to calculate the perimeter of the isosceles triangle below, correct to the nearest centimetre.  (3 marks)

Show Answers Only

\(29\ \text{cm}\ (\text{nearest cm})\)

Show Worked Solution

\(\text{Find length of the equal sides of the triangle.}\)

\(\text{Pythagoras’ Theorem states:  }c^2=a^2+b^2\)

\(\text{Let }a=5\ \text{and }b=8\)

\(\text{Then}\ \ c^2\) \(=5^2+8^2\)
\(c^2\) \(=89\)
\(c\) \(=\sqrt{89}\)
\(c\) \(=9.433\dots\)

 
\(\text{Perimeter}\)

\(=2\times 9.433\dots+10\)

\(=28.867\dots\approx 29\ \text{cm}\ (\text{nearest cm})\)

Right-angled Triangles, SM-Bank 051

Use Pythagoras' Theorem to calculate the perimeter of the trapezium below, correct to 1 decimal place.  (3 marks)

Show Answers Only

\(31.1\ \text{mm}\ (1\ \text{d.p.})\)

Show Worked Solution

\(\text{Find length of sloped side of trapezium.}\)

\(\text{Pythagoras’ Theorem states:  }c^2=a^2+b^2\)

\(\text{Let }a=4\ \text{and }b=7)

\(\text{Then}\ \ c^2\) \(=4^2+7^2\)
\(c^2\) \(=65\)
\(c\) \(=\sqrt{65}\)
\(c\) \(=8.062\dots\)

 
\(\text{Perimeter}\)

\(=6+7+10+8.062\dots\)

\(=31.062\dots\approx 31.1\ \text{mm}\ (1\ \text{d.p.})\)

Right-angled Triangles, SM-Bank 050

Use Pythagoras' Theorem to calculate the perimeter of the rectangle below, correct to 1 decimal place.  (3 marks)

Show Answers Only

\(13.0\ \text{m}\ (1\ \text{d.p.})\)

Show Worked Solution

\(\text{Find side length of rectangle.}\)

\(\text{Pythagoras’ Theorem states:  }a^2+b^2=c^2\)

\(\text{Let }b=3\ \text{and }c=4.6\)

\(\text{Then}\ \ a^2+3^2\) \(=4.6^2\)
\(a^2\) \(=4.6^2-3^2\)
\(a\) \(=\sqrt{12.16}\)
\(a\) \(=3.487\dots\)

 
\(\text{Perimeter}\)

\(=2\times 3.487\dots+ 2\times 3\)

\(=12.974\dots\approx 13.0\ \text{m}\ (1\ \text{d.p.})\)

Right-angled Triangles, SM-Bank 049

  1. Calculate the length of the hypotenuse in the following right-angled triangle. Give your answer correct to the nearest metre.   (2 marks)
     

  2. Find the perimeter of the triangle.  (1 mark)
Show Answers Only

a.    \(7\ \text{m}\)

b.    \(18\ \text{m}\)

Show Worked Solution

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

\(\text{Let }a=5\ \text{and }b=6\)

\(\text{Then}\ \ c^2\) \(=5^2+6^2\)
\(c^2\) \(=61\)
\(c\) \(=\sqrt{61}\)
\(c\) \(=7.141\dots\approx 7\)

 
\(\text{The hypotenuse is }7\ \text{metres (nearest metre})\)

b.    \(\text{Perimeter}\)

\(=7+5+6\)

\(=18\ \text{m}\)

Right-angled Triangles, SM-Bank 048

The size of a computer monitor is determined by the length of the diagonal of the screen. Use Pythagoras' Theorem to calculate the size of the monitor in the diagram below.  (2 marks)

Show Answers Only

\(51\ \text{centimetres}\)

Show Worked Solution

\(\text{Pythagoras’ Theorem states:  }c^2=a^2+b^2\)

\(\text{Let }a=24\ \text{and }b=45\)

\(\text{Then}\ \ c^2\) \(=24^2+45^2\)
\(c^2\) \(=2601\)
\(c\) \(=\sqrt{2601}\)
\(c\) \(=51\)

 
\(\text{The size of the monitor is }51\ \text{centimetres}\)

Right-angled Triangles, SM-Bank 047

Use Pythagoras' Theorem to calculate the length of the diagonal in the rectangle below, correct to 1 decimal place.  (2 marks)

Show Answers Only

\(8.4\ \text{centimetres}\ (1\ \text{d.p.})\)

Show Worked Solution

\(\text{Pythagoras’ Theorem states:  }c^2=a^2+b^2\)

\(\text{Let }a=4.2\ \text{and }b=7.3\)

\(\text{Then}\ \ c^2\) \(=4.2^2+7.3^2\)
\(c^2\) \(=70.98\)
\(c\) \(=\sqrt{70.98}\)
\(c\) \(=8.4219\dots\approx 8.4\ (1\ \text{d.p.})\)

 
\(\text{The length of the string is }8.4\ \text{centimetres}\ (1\ \text{d.p.})\)

Right-angled Triangles, SM-Bank 046

Lewis is holding a string attached to a kite. Lewis is 8 metres from the kite and the kite is flying 4 metres above his hand. Use Pythagoras' Theorem to calculate the length of the string attached to the kite, correct to 2 decimal places.  (2 marks)

Show Answers Only

\(8.94\ \text{metres}\ (2\ \text{d.p.})\)

Show Worked Solution

\(\text{Pythagoras’ Theorem states:  }c^2=a^2+b^2\)

\(\text{Let }a=4\ \text{and }b=8\)

\(\text{Then}\ \ c^2\) \(=4^2+8^2\)
\(c^2\) \(=80\)
\(c\) \(=\sqrt{80}\)
\(c\) \(=8.9442\dots\approx 8.94\ (2\ \text{d.p.})\)

 
\(\text{The length of the string is }8.94\ \text{metres}\ (2\ \text{d.p.})\)