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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 137

The sector shown has a radius of 13 cm and an angle of 230°. 
 

 

 What is the area of the sector to the nearest square centimetre?    (2 marks) 

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

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\(339\ \text{cm}^2\ (\text{nearest cm}^2)\)

Show Worked Solution
\(\text{Sector area}\) \(=\dfrac{\theta}{360}\times \pi r^2\)
  \(=\dfrac{230}{360}\times \pi\times 13^2\)
  \(=339.204\dots\)
  \(\approx 339\ \text{cm}^2\ (\text{nearest cm}^2)\)

Area, SM-Bank 136

The diagram shows an annulus.
 

 
Calculate the area of the shaded region (annulus), correct to 2 decimal places.  (2 marks)

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\(\approx 50.27\ \text{cm}^2\ (2\text{ d.p.})\)

Show Worked Solution

\(\text{Method 1:}\)

\(\text{Radius small circle}\ (r)=3\)

\(\text{Radius large circle}\ (R)=\dfrac{10}{2}=5\)

\(\text{Shaded region}\) \(=\text{Area large circle}-\text{Area small circle}\)
  \(=\pi\times 5^2-\pi \times 3^2\)
  \(=25\pi-9\pi\)
  \(=16\pi\)
  \(=50.27\ \text{cm}^2\ (2\text{ d.p.})\)

 
\(\text{Method 2:}\)

\(\text{Area of annulus}\)

\(=\pi(R^2 − r^2)\)

\(=\pi(5^2 − 3^2)\)

\(=\pi(25 − 9)\)

\(=16\pi\)

\(=50.27\ \text{cm}^2\ (2\text{ d.p.})\)

Area, SM-Bank 135

A shape consisting of a quadrant and a right-angled triangle is shown.
 

  1. Use Pythagoras' Theorem to calculate the radius of the quadrant.  (2 marks)

  2. What is the area of this shape, correct to one decimal place?  (2 marks)
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a.    \(8\ \text{cm}\)

b.    \(74.3\ \text{cm}^2\ (1\text{d.p.})\)

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a.    \(\text{Using Pythagoras to find radius}\ (r):\)

\(a^2+b^2\) \(=c^2\)
\(r^2+6^2\) \(=10^2\)
\(r^2\) \(=10^2-6^2\)
\(r\) \(=\sqrt{64}\)
  \(=8\ \text{cm}\)

 

b.    \(\text{Total area}\) \(=\text{Area of triangle}+\text{Area of quadrant}\)
    \(=\dfrac{1}{2}\times 8\times 6+\dfrac{1}{4}\times \pi\times 8^2\)
    \(=24+50.265\dots\)
    \(=74.265\dots\)
    \(\approx 74.3\ \text{cm}^2\ (1\text{d.p.})\)

Area, SM-Bank 134

A kite has an area of \(32\ 240\) square centimetres. Given that one of the diagonals has a length of 124 centimetres, calculate the length of the other diagonal.  (2 marks)

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\(520\ \text{cm}\)

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\(\text{Let the unknown diagonal}=x\)

\(A\) \(=\dfrac{1}{2}xy\)
\(32\ 240\) \(=\dfrac{1}{2}\times 124\times x\)
\(62x\) \(=32\ 240\)
\(x\) \(=\dfrac{32\ 240}{62}\)
  \(=520\ \text{cm}\)

Area, SM-Bank 133

A kite has an area of 52 square metres. Given that one of the diagonals has a length of 8 metres, calculate the length of the other diagonal.  (2 marks)

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\(13\ \text{m}\)

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\(\text{Let the unknown diagonal}=x\)

\(A\) \(=\dfrac{1}{2}xy\)
\(52\) \(=\dfrac{1}{2}\times 8\times x\)
\(4x\) \(=52\)
\(x\) \(=\dfrac{52}{4}\)
  \(=13\ \text{m}\)

Area, SM-Bank 132

Johan builds a kite with diagonals of 0.7 metres and 1.2 metres as shown below.

Calculate the area of Johan's kite (not including the tail) in square metres.   (2 marks)
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\(0.42\ \text{m}^2\)

Show Worked Solution
\(A\) \(=\dfrac{1}{2}xy\)
  \(=\dfrac{1}{2}\times 0.7\times 1.2\)
  \(=0.42\ \text{m}^2\)

Area, SM-Bank 128

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

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

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\(27\pi\ \text{mm}^2\)

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\(\theta=30^{\circ} \ \ r=18\ \text{mm}\)

\(A\) \(=\dfrac{\theta}{360}\times \pi r^2\)
  \(=\dfrac{30}{360}\times \pi \times 18^2\)
  \(=\dfrac{1}{12}\times 324\pi\)
  \(=27\pi\ \text{mm}^2\)

Area, SM-Bank 127

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

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

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\(\dfrac{100\pi}{3}\ \text{m}^2\)

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\(\theta=120^{\circ} \ \ r=10\ \text{m}\)

\(A\) \(=\dfrac{\theta}{360}\times \pi r^2\)
  \(=\dfrac{120}{360}\times \pi\times 10^2\)
  \(=\dfrac{1}{3}\times 100\pi\)
  \(=\dfrac{100\pi}{3}\ \text{m}^2\)

Area, SM-Bank 126

Calculate the area of the shaded region in the following composite shape, giving your answer correct to one decimal place.   (2 marks)
 

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\(2789.0\ \text{m}^2\ (1 \text{ d.p.})\)

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\(\text{Diameter semi-cirle}=114\ \text{cm}\)

\(\text{Radius semi-cirle}(r)=57\ \text{cm}\)

\(\text{Total area}=\text{Area square}-\text{Area semi-cirle}\)

\(A\) \(=s^2-\pi r^2\)
  \(=114^2-\pi\times 57^2\)
  \(=12\ 996-10\ 207.034\dots\)
  \(=2788.965\dots\approx 2789.0\ \text{m}^2\ (1 \text{ d.p.})\)

Area, SM-Bank 125

Calculate the area of the shaded region in the following composite shape, giving your answer correct to one decimal place.   (2 marks)
 

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\(22.0\ \text{m}^2\ (1 \text{ d.p.})\)

Show Worked Solution

\(\text{Radius small cirle}(r)=3\ \text{m}\)

\(\text{Radius large cirle}(R)=4\ \text{m}\)

\(\text{Total area}=\text{Area large cirle}-\text{Area small cirle}\)

\(A\) \(=\pi R^2-\pi r^2\)
  \(=\pi\times 4^2-\pi\times 3^2\)
  \(=50.265\dots-28.274\dots\)
  \(=21.991\dots\approx 22.0\ \text{m}^2\ (1 \text{ d.p.})\)

Area, SM-Bank 124

Calculate the area of the following composite shape, giving your answer correct to one decimal place.   (2 marks)
 

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\(321.0\ \text{cm}^2\ (1 \text{ d.p.})\)

Show Worked Solution

\(\text{Radius small semi-cirle}(r)=4.5\ \text{mm}\)

\(\text{Radius large semi-cirle}(R)=9\ \text{mm}\)

\(\text{Total area}=\text{Area small semi-cirle}+\text{Area large semi-cirle}+\text{Area rectangle}\)

\(A\) \(=\dfrac{1}{2}\times \pi r^2+\dfrac{1}{2}\times \pi R^2+lb\)
  \(=\dfrac{1}{2}\times \pi\times 4.5^2+\dfrac{1}{2}\times \pi\times 9^2+18\times 9\)
  \(=31.808\dots+127.234\dots+162\)
  \(=321.043\dots\approx 321.0\ \text{cm}^2\ (1 \text{ d.p.})\)

Area, SM-Bank 123

Calculate the area of the following composite shape, giving your answer correct to one decimal place.   (2 marks)
 

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\(178.3\ \text{cm}^2\ (1 \text{ d.p.})\)

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\(\text{Radius}=8\ \text{cm}\)

\(\text{Rectangle length}=24-8=16\ \text{cm}\)

\(\text{Total area}=\text{Area Quadrant}+\text{Area rectangle}\)

\(A\) \(=\dfrac{1}{4}\times \pi r^2+lb\)
  \(=\dfrac{1}{4}\times \pi\times 8^2+16\times 8\)
  \(=50.265\dots+128\)
  \(=178.265\dots\approx 178.3\ \text{cm}^2\ (1 \text{ d.p.})\)

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

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

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

Angle Properties, SM-Bank 007

The diagram below shows two parallel lines intersected by transversal \(CG\).
 

  1. Name two angles that are complementary to \(\angle DBF\).   (2 marks)

  2. Name the angle that is corresponding to \(\angle HFG\).   (1 mark)

  3. Name the angle that is alternate to \(\angle CFH\).   (1 mark)
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i.     \(\text{Correct answers include two of:}\)

\(\angle DBC,\ \angle EFB,\ \angle GFH,\ \text{or}\ \angle GBA.\)

ii.    \(\angle ABF\)

iii.  \(\angle DBF\)

Show Worked Solution

i.     \(\text{Complementary angles sum to 180°.}\)

\(\text{Correct answers include two of:}\)

\(\angle DBC,\ \angle EFB,\ \angle GFH,\ \text{or}\ \angle GBA.\)
 

ii.    \(\angle ABF\)

 

 
iii.  \(\angle DBF\)

Area, SM-Bank 121

Milan cuts a sector from a circle so that  \(\dfrac{3}{8}\)  of the area of the circle remains.
 


 

If the circle's radius is 4 cm, what is the area of the shape, to the nearest square centimetre?  (2 marks)

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\(19\ \text{cm}^2\ (\text{nearest cm}^2)\)

Show Worked Solution
\(\text{Area}\) \(=\dfrac{3}{8}\times \pi r^2\)
  \(=\dfrac{3}{8}\times \pi\times 4^2\)
  \(=18.849\dots\)
  \(=19\ \text{cm}^2\ (\text{nearest cm}^2)\)

Area, SM-Bank 120

A one-on-one basketball court is a composite shape made up of a rectangle and a semicircle, as shown below.
 

Calculate the area of the court, giving your answer correct to one decimal place.   (2 marks)

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\(104.5\ \text{m}^2\ (1 \text{ d.p.})\)

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\(\text{Diameter}=12\ \text{m}\)

\(\therefore\ \text{Radius}=6\ \text{m}\)

\(\text{Total area}=\text{Area semi-circle}+\text{Area rectangle}\)

\(A\) \(=\dfrac{1}{2}\times \pi r^2+lb\)
  \(=\dfrac{1}{2}\times \pi\times 6^2+12\times 4\)
  \(=104.548\dots\)
  \(=104.5\ \text{m}^2\ (1 \text{ d.p.})\)

Area, SM-Bank 119

Calculate the area of a semi-circle with a diameter of 60 centimetres. Give your answer correct to one decimal place.   (2 marks)

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\(1413.7\ \text{cm}^2\ (1 \text{ d.p.})\)

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\(\text{Diameter}=60\ \text{cm}\)

\(\therefore\ \text{Radius}=30\ \text{cm}\)

\(\text{Area semi-circle}\) \(=\dfrac{1}{2}\times \pi r^2\)
  \(=\dfrac{1}{2}\times \pi\times 30^2\)
  \(=1413.716\dots\)
  \(=1413.7\ \text{cm}^2\ (1 \text{ d.p.})\)

Area, SM-Bank 118

Calculate the area of a circle with a diameter of 37.4 millimetres. Give your answer correct to one decimal place.   (2 marks)

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\(1098.6\ \text{mm}^2\ (1 \text{ d.p.})\)

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\(\text{Diameter}=37.4\ \text{mm}\)

\(\therefore\ \text{Radius}=18.7\ \text{mm}\)

\(\text{Area}\) \(=\pi r^2\)
  \(=\pi\times 18.7^2\)
  \(=1098.583\dots\)
  \(=1098.6\ \text{mm}^2\ (\text{1 d.p.})\)

Area, SM-Bank 117

Calculate the area of a circle with a radius of 72.3 centimetres. Give your answer correct to one decimal place.   (2 marks)

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\(16\ 422.0\ \text{cm}^2\ (1\ \text{d.p.})\)

Show Worked Solution
\(\text{Area}\) \(=\pi r^2\)
  \(=\pi\times 72.3^2\)
  \(=16\ 422.015\dots\)
  \(=16\ 422.0\ \text{cm}^2\ (1 \text{ d.p.})\)

Area, SM-Bank 116

Calculate the area of a circle with a radius of 20 metres. Give your answer as an exact value in term of \(\pi\).   (2 marks)

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\(400\pi\ \text{m}^2\)

Show Worked Solution
\(\text{Area}\) \(=\pi r^2\)
  \(=\pi\times 20^2\)
  \(=400\pi\ \text{m}^2\)

Area, SM-Bank 115

Calculate the area of the following shape, giving your answer correct to 1 decimal place.  (2 marks)
 

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\(855.3\ \text{m}^2\ (1\ \text{d.p.})\)

Show Worked Solution
\(\text{Area quadrant}\) \(=\dfrac{1}{4}\times\pi r^2\)
  \(=\dfrac{1}{4}\times\pi\times 33^2\)
  \(=855.2985\dots\)
  \(\approx 855.3\ \text{m}^2\ (1\ \text{d.p.})\)

Area, SM-Bank 114

Calculate the area of the following shape, giving your answer correct to 1 decimal place.  (2 marks)

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\(84.8\ \text{m}^2\ (1\ \text{d.p.})\)

Show Worked Solution

\(\text{Diameter}=12\ \text{m}\)

\(\therefore\ \text{Radius}=6\ \text{m}\)

\(\text{Area}\) \(=\dfrac{3}{4}\times\pi r^2\)
  \(=\dfrac{3}{4}\times\pi\times 6^2\)
  \(=84.8230\dots\)
  \(\approx 84.8\ \text{m}^2\ (1\ \text{d.p.})\)

Area, SM-Bank 113

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

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\(12\pi \ \text{m}^2\)

Show Worked Solution
\(\text{Area}\) \(=\dfrac{3}{4}\times\pi r^2\)
  \(=\dfrac{3}{4}\times\pi\times 4^2\)
  \(=12\pi \ \text{m}^2\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Area, SM-Bank 107

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

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

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

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

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

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.    \(60\ \text{m}^2\)

Show Worked Solution

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)

Show Answers Only

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)

Show Answers Only

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

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. \(19\ \text{cm}^2\)
Show Answers Only

\(C\)

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)

Show Answers Only

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

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)

Show Answers Only

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

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)

Show Answers Only

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

Show Worked Solution

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

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

Area, SM-Bank 089

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

Show Answers Only

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

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)

Show Answers Only

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

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)

Show Answers Only

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

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. \(25\ 000\ \text{cm}^2\)
Show Answers Only

\(C\)

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)

Show Answers Only

\(5\ \text{litres}\)

Show Worked Solution

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

\(=6^2\)

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

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

\(=6\times 10\)

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

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

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

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