SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

Area, SM-Bank 112

Calculate the area of the following quadrant, giving your answer as an exact value in terms of \(\pi\).  (2 marks)

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\(\dfrac{225\pi}{4}\ \text{cm}^2\)

Show Worked Solution
\(\text{Area quadrant}\) \(=\dfrac{1}{4}\times\pi r^2\)
  \(=\dfrac{1}{4}\times\pi\times 15^2\)
  \(=\dfrac{225}{4}\pi\)
  \(=\dfrac{225\pi}{4}\ \text{cm}^2\)

Area, SM-Bank 111

Calculate the area of the following semi-circle, giving your answer as an exact value in terms of \(\pi\).  (2 marks)
 

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\(1250\pi\ \text{m}^2\)

Show Worked Solution

\(\text{Diameter}=100\ \text{m}\)

\(\therefore\ \text{Radius}=50\ \text{m}\)

\(\text{Area semi-circle}\) \(=\dfrac{1}{2}\times\pi r^2\)
  \(=\dfrac{1}{2}\times\pi\times 50^2\)
  \(=1250\pi\ \text{m}^2\)

Area, SM-Bank 110

Calculate the area of the following semi-circle, giving your answer to 2 decimal places.   (2 marks)
 

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\(0.57\ \text{mm}^2\ (2\ \text{d.p.})\)

Show Worked Solution

\(\text{Diameter}=1.2\ \text{mm}\)

\(\therefore\ \text{Radius}=0.6\ \text{mm}\)

\(\text{Area semi-circle}\) \(=\dfrac{1}{2}\times\pi r^2\)
  \(=\dfrac{1}{2}\times\pi\times 0.6^2\)
  \(=0.5654\dots\)
  \(\approx 0.57\ \text{mm}^2\ (2\ \text{d.p.})\)

Area, SM-Bank 109

Calculate the area of the following semi-circle, giving your answer to 2 decimal places.   (2 marks)
 

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\(173.18\ \text{m}^2\ (\text{2 d.p.})\)

Show Worked Solution

\(\text{Diameter}=21\ \text{m}\)

\(\therefore\ \text{Radius}=10.5\ \text{m}\)

\(\text{Area semi-circle}\) \(=\dfrac{1}{2}\times\pi r^2\)
  \(=\dfrac{1}{2}\times\pi\times 10.5^2\)
  \(=173.1802\dots\)
  \(\approx 173.18\ \text{m}^2\ (2\ \text{d.p.})\)

Area, SM-Bank 108

Calculate the area of the following circle, giving your answer as an exact value in terms of \(\pi\).   (2 marks)
 

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\(25\pi\ \text{m}^2\)

Show Worked Solution

\(\text{Diameter}=10\ \text{m}\)

\(\therefore\ \text{Radius}=5\ \text{m}\)

\(\text{Area}\) \(=\pi r^2\)
  \(=\pi\times 5^2\)
  \(=25\pi\ \text{m}^2\)

Area, SM-Bank 107

Calculate the area of the following circle, correct to one decimal place.  (2 marks)
 

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\(422.7\ \text{cm}^2\ (1\ \text{d.p.})\)

Show Worked Solution

\(\text{Diameter}=23.2\ \text{cm}\)

\(\therefore\ \text{Radius}=11.6\ \text{cm}\)

\(\text{Area}\) \(=\pi r^2\)
  \(=\pi\times 11.6^2\)
  \(=422.7327\dots\)
  \(\approx 422.7\ \text{cm}^2\ (1\ \text{d.p.})\)

Area, SM-Bank 106

Calculate the area of the following circle, correct to one decimal place.   (2 marks)
 

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\(50.3\ \text{m}^2\ (1\ \text{d.p.})\)

Show Worked Solution
\(\text{Area}\) \(=\pi r^2\)
  \(=\pi\times 4^2\)
  \(=50.2654\dots\)
  \(\approx 50.3\ \text{m}^2\ (1\ \text{d.p.})\)

Area, SM-Bank 105

Calculate the area of the following circle, correct to one decimal place.  (2 marks)
 

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\(514.7\ \text{cm}^2\ (1\ \text{d.p.})\)

Show Worked Solution
\(\text{Area}\) \(=\pi r^2\)
  \(=\pi\times 12.8^2\)
  \(=514.7185\dots\)
  \(\approx 514.7\ \text{cm}^2\ (1\ \text{d.p.})\)

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 101

The triangle below is isosceles.

  1. Use Pythagoras' Theorem to calculate the perpendicular height (\(h\)) of the triangle.  (2 marks)

  2. Using your answer from (a), find the area of the triangle.  (2 marks)

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a.    \(12\ \text{m}\)

b.    \(60\ \text{m}^2\)

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a.   

\(a^2+b^2\) \(=c^2\)
\(h^2+5^2\) \(=13^2\)
\(h^2\) \(=13^2-5^2\)
\(h^2\) \(=144\)
\(h\) \(=12\)

 
\(\therefore\ \text{The perpendicular height of the triangle is }12\ \text{m}\)
 

b.    \(\text{Area}\) \(=\dfrac{1}{2}\times bh\)
    \(=\dfrac{1}{2}\times 10\times 12\)
    \(=60\ \text{m}^2\)

Area, SM-Bank 100

Calculate the area of the following triangles.

  1.  
    (2 marks)

  2.   
    (2 marks)

  3.   
        (2 marks)

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a.    \(87.42\ \text{m}^2\)

b.    \(1995\ \text{mm}^2\)

c.    \(6650\ \text{m}^2\ \text{or}\ 0.665\ \text{m}^2\)

Show Worked Solution
a.    \(\text{Area}\) \(=\dfrac{1}{2}\times bh\)
    \(=\dfrac{1}{2}\times 12.4\times 14.1\)
    \(=87.42\ \text{m}^2\)

 

b.    \(\text{Area}\) \(=\dfrac{1}{2}\times bh\)
    \(=\dfrac{1}{2}\times 42\times 95\)
    \(=1995\ \text{mm}^2\)

 

c.    \(\text{Area in (cm)}^2\) \(=\dfrac{1}{2}\times bh\)
    \(=\dfrac{1}{2}\times 95\times 140\)
    \(=6650\ \text{cm}^2\)
 

 
\(\text{Area in (m)}^2\)		  
\(=\dfrac{1}{2}\times 0.95\times 1.40\)
    \(=0.665\ \text{m}^2\)

 

Area, SM-Bank 099

Calculate the area of the following triangles.

  1.  
    (2 marks)

  2.  
    (2 marks)

  3.  
      (2 marks)

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a.    \(42\ \text{cm}^2\)

b.    \(17.5\ \text{m}^2\)

c.    \(24\ \text{mm}^2\)

Show Worked Solution
a.    \(\text{Area}\) \(=\dfrac{1}{2}\times bh\)
    \(=\dfrac{1}{2}\times 12\times 7\)
    \(=42\ \text{cm}^2\)

 

b.    \(\text{Area}\) \(=\dfrac{1}{2}\times bh\)
    \(=\dfrac{1}{2}\times 5\times 7\)
    \(=17.5\ \text{m}^2\)

 

c.    \(\text{Area}\) \(=\dfrac{1}{2}\times bh\)
    \(=\dfrac{1}{2}\times 6\times 8\)
    \(=24\ \text{mm}^2\)

Area, SM-Bank 098

Label the base (\(b\)) and draw and label a line indicating the perpendicular height (\(h\)), on the following triangles.

a.   (1 mark)

                      b.    (1 mark)

                   

c.   (1 mark)
   
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a. 

                     b. 

                   

c. 

   
Show Worked Solution

a. 

                     b. 

                   

c. 

   
         
         

Area, SM-Bank 097

Identify the base (\(b\)) and the perpendicular height (\(h\)), by labelling the following triangles.

a.   (1 mark)

                      b.    (1 mark)

                   

c.   (1 mark)
   
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a. 

                     b. 

                   

c. 

   
Show Worked Solution

a. 

                     b. 

                   

c. 

   

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 094

A poster has an area of 5250 square centimetres.

Find the area of the poster in square millimetres?   (2 marks)

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\(525\ 000\ \text{mm}^{2}\)

Show Worked Solution

\(\text{1 cm}^{2}\ = 10\ \text{mm}\ \times 10\ \text{mm} = 100\ \text{mm}^{2} \)

\(\text{5250 cm}^{2} = 5250 \times 100 = 525\ 000\ \text{mm}^{2}\)

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 091

A walled city has a land area of 950 hectares.

Express the area of the city in square kilometres.  (2 marks)

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\(9.5\ \text{km}^2\)

Show Worked Solution

\(\text{1 hectare}\ = 10\ 000\ \text{m}^{2}\)

\(\text{950 hectares}\ = 950 \times 10\ 000 = 9\ 500\ 000\ \text{m}^{2} \)

\(\text{1 km}^{2}\ = 1000\ \text{m} \times 1000\ \text{m}\ = 1\ 000\ 000\ \text{m}^{2}\)

\(\text{950 hectares}\ = \dfrac{9\ 500\ 000}{1\ 000\ 000} = 9.5\ \text{km}^{2} \)

Area, SM-Bank 090

Jim has a hobby farm with an area of 7 hectares.

What is the size of Jim's farm in square metres.   (1 mark)

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\(70\ 000\ \text{m}^2\)

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\(\text{1 hectare}\ = 10\ 000\ \text{m}^{2} \)

\(\text{7 hectares}\ = 7 \times 10\ 000 = 70\ 000\ \text{m}^{2} \)

Area, SM-Bank 089

Express an area of 0.003 square metres in square millimetres.  (2 marks)

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\(3000\ \text{mm}^2\)

Show Worked Solution

\(\text{1 m}^{2}\ =1000\ \text{mm}\times 1000\ \text{mm} =1\ 000\ 000\ \text{mm}^2\)

\(\text{0.003 m}^{2}\ = 0.003 \times 1\ 000\ 000 = 3000\ \text{mm}^{2} \)

Area, SM-Bank 088

Convert an area of 9 300 000 square metres into square kilometres.  (1 mark)

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\(9.3\ \text{km}^2\)

Show Worked Solution

\(\text{1 km}^{2} =1000\ \text{m}\times 1000\ \text{m}\ =1\ 000\ 000\ \text{m}^2\)

\(\text{9 300 000 m}^{2} =\dfrac{9\ 300\ 000}{1\ 000\ 000}=9.3\ \text{km}^2\)

Area, SM-Bank 087

A dining table has an area of 35 700 square centimetres.

Express this area in square metres.  (1 mark)

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\(3.57\ \text{m}^2\)

Show Worked Solution

\(\text{1 m}^{2}\ = 100\ \text{cm}\times 100\ \text{cm}\ = 10\ 000\ \text{cm}^{2} \)

\(\text{35 700 cm}^{2} = \dfrac{35\ 700}{10\ 000} = 3.57\ \text{m}^2\)

Area, SM-Bank 086

Convert an area of 500 square millimetres to square centimetres.  (1 mark)

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\(5\ \text{cm}^2\)

Show Worked Solution

\(\text{1 cm}^{2}\ =10\ \text{mm}\times 10\ \text{mm}\ =100\ \text{mm}^2\)

\(500\ \text{mm}^2 =\dfrac{500}{100} =5\ \text{cm}^2\)

Area, SM-Bank 085

A remote island has a land area of 4.6 square kilometres.

Convert this area into square metres.  (1 mark)

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\(4\ 600\ 000\ \text{m}^2\)

Show Worked Solution

\( \text{1 km}^{2}\ =1000\ \text{m} \times 1000\ \text{m} = 1\ 000\ 000\ \text{m}^2 \)

\(\text{4.6 km}^{2}\ = 4.6\times 1\ 000\ 000=4\ 600\ 000\ \text{m}^2 \)

Area, SM-Bank 084

Convert an area of 15 square metres to square centimetres.  (1 mark)

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\(150\ 000\ \text{cm}^2\)

Show Worked Solution

\(\text{1 m}^{2} = 100\ \text{cm}\times 100\ \text{cm}=10\ 000\ \text{cm}^2\)

\(\text{15 m}^{2}\ = 15 \times 10\ 000 = 150\ 000\ \text{cm}^2\)

Area, SM-Bank 083

Convert an area of 3 square centimetres to square millimetres.   (1 mark)

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\(300\ \text{mm}^2\)

Show Worked Solution

\(\text{1 cm}^{2} = 10\ \text{mm}\times 10\ \text{mm}\ =100\ \text{mm}^2\)

\(\text{3 cm}^{2} = 3 \times 100 = 300\ \text{mm}^2\)

 

Area, SM-Bank 082

Julia bought a kitchen rug with an area of 0.85 square metres.

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

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\(8500\ \text{cm}^{2} \) 

Show Worked Solution

\(\text{1 m}^{2} = 100\ \text{cm} \times 100\ \text{cm}\ =10\ 000\ \text{cm}^{2} \)

\(\text{0.85 m}^{2} = 0.85 \times 10\ 000 = 8500\ \text{cm}^{2} \) 

Area, SM-Bank 081 MC

Ben bought a dog mat with an area of 0.5 square metres.

What is the area of the dog mat in square centimetres?

  1. \(2.5\ \text{cm}^2\)
  2. \(5\ \text{cm}^2\)
  3. \(5000\ \text{cm}^2\)
  4. \(25\ 000\ \text{cm}^2\)
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\(C\)

Show Worked Solution

\(\text{1 m}^{2} =100\ \text{cm}\times 100\ \text{cm} = 10\ 000\ \text{cm}^{2} \)

\(\text{0.5 m}^{2} = 0.5\times 10\ 000 =5000\ \text{cm}^2\)  

\(\Rightarrow C\)

Area, SM-Bank 080

Bobby used 3 litres of varnish to paint the loungeroom floor.

The floor was a square with sides 6 metres long.

How many litres of varnish would he need to paint a rectangular floor which is 6 metres long and 10 metres wide?  (2 marks)

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\(5\ \text{litres}\)

Show Worked Solution

\(\text{Area of square floor}\)

\(=6^2\)

\(=36\ \text{m}^2\)

\(\text{Area of rectangular floor}\)

\(=6\times 10\)

\(=60\ \text{m}^2\)

\(\text{Paint needed for rectangular wall}\)

\(=\dfrac{60}{36}\times 3\)

\(=5\ \text{litres}\)

Area, SM-Bank 079

Shinji used 8 litres of paint to paint a wall.

The wall was a square with sides 4 metres long.

How many litres of paint would he need to paint a rectangular wall which is 3 metres high and 10 metres wide?  (2 marks)

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\(15\ \text{litres}\)

Show Worked Solution

\(\text{Area of square wall}\)

\(=4^2\)

\(=16\ \text{m}^2\)

\(\text{Area of rectangular wall}\)

\(=3\times 10\)

\(=30\ \text{m}^2\)

\(\text{Paint needed for rectangular wall}\)

\(=\dfrac{30}{16}\times 8\)

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

Area, SM-Bank 078

Shinji used 8 litres of paint to paint a wall.

The wall was a square with sides 4 metres long.

How many litres of paint would he need to paint a rectangular wall which is 3 metres high and 10 metres wide?  (2 marks)

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\(15\ \text{litres}\)

Show Worked Solution

\(\text{Area of square wall}\)

\(=4^2\)

\(=16\ \text{m}^2\)

\(\text{Area of rectangular wall}\)

\(=3\times 10\)

\(=30\ \text{m}^2\)

\(\text{Paint needed for rectangular wall}\)

\(=\dfrac{30}{16}\times 8\)

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

Area, SM-Bank 077

Vlad has a giant chess board that has an area of 4 square metres.

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

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\(40\ 000\ \text{square centimetres}\)

Show Worked Solution

\(\text{Strategy 1 (convert square metres):}\)

\(\text{1 m}^{2} = 100\ \text{cm}\ \times 100\ \text{cm}\ = 10\ 000\ \text{cm}^{2}\)

\(\text{Side length}\ =\sqrt{4} =2\ \text{m}\)

\(\text{Area}\ =2^2 = 4\ \text{m}\ = 4 \times 10\ 000 = 40\ 000\ \text{cm}^{2}\)
 

\(\text{Strategy 2 (convert sides to centimetres):}\)

\(\text{Side length}\ =\sqrt{4} =2\ \text{m}\ = 200\ \text{cm}\)

\(\text{Area}\ =200^2 = 40\ 000\ \text{square centimetres}\)

Area, SM-Bank 076

Jeremy's back deck is square in shape and has an area of 16 square metres.

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

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\(160\ 000\ \text{square centimetres}\)

Show Worked Solution

\(\text{Strategy 1 (convert square metres):}\)

\(\text{1 m}^{2} = 100\ \text{cm}\ \times 100\ \text{cm}\ = 10\ 000\ \text{cm}^2 \)

\(\text{Side length of deck}\ =\sqrt{16} =4\ \text{m}\)

\(\text{Area} =4^2 = 16\ \text{m}^{2} = 16 \times 10\ 000 = 160\ 000\ \text{cm}^{2}\)
 

\(\text{Strategy 2 (convert side lengths):}\)

\(\text{Side length of deck}\ =\sqrt{16} =4\ \text{m}= 4 \times 100 = 400\ \text{cm}\)

\(\text{Area} =400^2 =160\ 000\ \text{cm}^{2}\)

Area, SM-Bank 075

A square fridge magnet has an area of 900 square millimetres.

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

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\(9\ \text{cm}^2\)

Show Worked Solution

\(\text{Strategy 1 (convert square millimetres):}\)

\(\text{1 cm}^{2}\ = 10\ \text{mm}\ \times 10\ \text{mm}\ = 100\ \text{mm}^{2} \)

\(\text{Side length}\ =\sqrt{900}\ = 30\ \text{mm}\)

\(\text{Area} =30^2 =900\ \text{mm}^2 = \dfrac{900}{100} = 9\ \text{cm}^2 \)
 

\(\text{Strategy 2 (convert side lengths):}\)

\(\text{Side length}\ =\sqrt{900}\ =30\ \text{mm}= \dfrac{30}{10} = 3\ \text{cm}\)

\(\text{Area} =3^2 =9\ \text{cm}^2\)

Area, SM-Bank 074 MC

A square table top has an area of 4225 square centimetres.

What is the area of the table top in square metres?

  1. \(0.4225\ \text{square metres}\)
  2. \(4.225\ \text{square metres}\)
  3. \(42.25\ \text{square metres}\)
  4. \(42\ 250\ \text{square metres}\)
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Side length}\ =\sqrt{4225} =65\ \text{cm}= \dfrac{65}{10\ 000} = 0.65\ \text{m}\)

\(\text{Area}\ =0.65^2 =0.4225\ \text{square metres}\)

\(\Rightarrow A\)

Area, SM-Bank 073 MC

A large square feature tile has an area of 15 625 square centimetres.

What is the area of the feature tile in square millimetres?

  1. \(156.25\ \text{square millimetres}\)
  2. \(1562.5\ \text{square millimetres}\)
  3. \(156\ 250\ \text{square millimetres}\)
  4. \(1\ 562\ 500\ \text{square millimetres}\)
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Side length}\ =\sqrt{15\ 625}\ =125\ \text{cm} = 125 \times 10 = 1250\ \text{mm}\)

\(\text{Area}\ =1250^2 =1\ 562\ 500\ \text{square millimetres}\)

\(\Rightarrow D\)

Area, SM-Bank 072 MC

A square table top has an area of 9025 square centimetres.

What is the area of the table top in square millimetres?

  1. \(9.025\ \text{square millimetres}\)
  2. \(90.25\ \text{square millimetres}\)
  3. \(90\ 250\ \text{square millimetres}\)
  4. \(902\ 500\ \text{square millimetres}\)
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Side length} =\sqrt{9025} =95\ \text{cm}= 95 \times 10 = 950\ \text{mm}\)

\(\text{Area}\ =950^2 =902\ 500\ \text{square millimetres}\)

\(\Rightarrow D\)

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)

Show Answers Only

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

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)

Show Answers Only

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

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

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

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

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

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

 

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

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.    \($7650\)

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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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\(B\)

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

\(\Rightarrow B\)

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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\(D\)

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

\(\Rightarrow D\)

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 064 MC

A circular pool is located in a square lawn, as shown below.
 

The sides of the square lawn are 10 m in length.

The pool has a radius of 3 m.

The area of the lawn surrounding the pool, in square metres, is closest to

  1. \(59\)
  2. \(72\)
  3. \(81\)
  4. \(128\)
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\(B\)

Show Worked Solution
\(\text{Area of square}\) \(=\text{side}^2\)
  \(=10^2\)
  \(=100\ \text{m}^2\)

   

\(\text{Area of pool}\) \(=\pi r^2\)
  \(=\pi \times 3^2\)
  \(= 28.27\dots\ \text{m}^2\)

 

\(\therefore\ \text{Area of lawn}\) \(=100-28.27\dots\)
  \(=71.72\dots\ \text{m}^2\)

 
\(\Rightarrow B\)

Area, SM-Bank 063

A large mosaic tile artwork has been created inside a rectangle in the shape of a parallelogram as shown below.

  1. Calculate the shaded area outside the parallelogram in square metres.  (2 marks)

  2. Calculate the cost of tiling the shaded area if the tiles cost $85 per square metre.  (2 marks)

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a.    \(42\ \text{m}^2\)

b.    \($3570\)

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a.    \(\text{Area to be tiled}\) \(=\text{Area of rectangle}-\text{Area of parallelogram}\)
    \(=11\times 6-8\times 3\)
    \(=42\ \text{m}^2\)

   

b.    \(\text{Cost of tiling}\) \(=\text{Shaded area}\times $85\)
    \(=42\times $85\)
    \(=$3570\)

Area, SM-Bank 062

Rorke is designing a new logo that is made up of two identical parallelograms as shown below.

Calculate the area of the logo in square millimetres.  (2 marks)

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\(306\ \text{mm}^2\)

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\(\text{Area}\) \(=2\times\text{base}\times \text{height}\)
  \(=2\times 17\times 9\)
  \(=306\ \text{mm}^2\)

 
\(\therefore\ \text{The area of the logo is }306\ \text{mm}^2.\)

Area, SM-Bank 061

A parallelogram has an area of 1872 square metres and a perpendicular height of 78 metres.

Calculate the base length of the parallelogram.  (2 marks)

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\(24\ \text{m}\)

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\(\text{Area}\) \(=\text{base}\times \text{height}\)
\(\therefore\ 1872\) \(=b\times 78\)
\(b\) \(=\dfrac{1872}{78}\)
  \(=24\)

 
\(\therefore\ \text{The base length of the parallelogram is }24\ \text{m.}\)

Area, SM-Bank 060

The parallelogram below has an area of 75.03 square centimetres and a base length of 12.3 centimetres.

Calculate the perpendicular height of the parallelogram.  (2 marks)

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\(6.1\ \text{cm}\)

Show Worked Solution
\(\text{Area}\) \(=\text{base}\times \text{height}\)
\(\therefore\ 75.03\) \(=12.3\times h\)
\(h\) \(=\dfrac{75.03}{12.3}\)
  \(=6.1\)

 
\(\therefore\ \text{The perpendicular height of the parallelogram is }6.1\ \text{cm.}\)

Area, SM-Bank 059

Calculate the area of the parallelogram below, in metres squared.  (2 marks)

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\(84\ \text{m}^2\)

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\(\text{Area}\) \(=\text{base}\times \text{height}\)
  \(=6\times 14\)
  \(=84\ \text{m}^2\)

Area, SM-Bank 058

Calculate the area of the parallelogram below, in millimetres squared.  (2 marks)

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\(115.73\ \text{mm}^2\)

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\(\text{Area}\) \(=\text{base}\times \text{height}\)
  \(=7.1\times 16.3\)
  \(=115.73\ \text{mm}^2\)

Area, SM-Bank 057

Calculate the area of the parallelogram below, in centimetres squared.  (2 marks)

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\(48.96\ \text{cm}^2\)

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\(\text{Area}\) \(=\text{base}\times \text{height}\)
  \(=10.2\times 4.8\)
  \(=48.96\ \text{cm}^2\)

Area, SM-Bank 056

A sporting field in the shape of a square has a side length of 110 metres.

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

  2. During the off-season, the sporting field is to be covered in fertiliser. If fertiliser costs $6.50 per 100 square metres, calculate the cost of fertilising the field.  (2 marks)

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a.    \(12\ 100\ \text{m}^2\)

b.    \($786.50\)

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a.    \(\text{Area}\) \(=s^2\)
    \(=110^2\)
    \(=12\ 100\ \text{m}^2\)

 

b.    \(\text{Cost}\) \(=\dfrac{12\ 100}{100}\times 6.50\)
    \(=121\times 6.50\)
    \(=$786.50\)

Area, SM-Bank 055

The square below has a diagonal of 12 metres.

  1. Use Pythagoras' Theorem to calculate the side length of the square. Give your answer in exact surd form.  (2 marks)


  2. Calculate the are of the square correct to one decimal place.  (2 marks)

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a.    \(\sqrt{72}\ \text{m}\)

b.    \(72\ \text{m}^2\)

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a.    \(\text{Using Pythagoras to find the side length of the square:}\)

\(a^2+b^2\) \(=c^2\)
\(a^2+a^2\) \(=12^2\)
 \(2a^2\) \(=144\)
\(a^2\) \(=\dfrac{144}{2}=72\)
 \(a\) \(=\sqrt{72}\ \text{m}\)

b.   

\(\text{Area}\) \(=s^2\)
  \(=(\sqrt{72})^2\)
  \(=72\ \text{m}^2\)

Area, SM-Bank 054

Calculate the area of a square with a perimeter of 192 centimetres.  (2 marks)

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\(2304\ \text{cm}^2\)

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\(\text{Perimeter}\) \(=192\ \text{cm}\)
\(\therefore\ \text{Side}\) \(=\dfrac{192}{4}\)
  \(=48\ \text{cm}\)

 

\(\text{Area}\) \(=s^2\)
  \(=48^2\)
  \(=2304\ \text{cm}^2\)

Area, SM-Bank 053

The following shape has a perimeter of 12.4 centimetres. Calculate its' area.  (2 marks)

 

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\(9.61\ \text{cm}^2\)

Show Worked Solution
\(\text{Perimeter}\) \(=12.4\ \text{cm}\)
\(\therefore\ \text{Side}\) \(=\dfrac{12.4}{4}\)
  \(=3.1\ \text{cm}\)

 

\(\text{Area}\) \(=s^2\)
  \(=3.1^2\)
  \(=9.61\ \text{cm}^2\)