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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?

$0.4225\ \text{square metres}$
$4.225\ \text{square metres}$
$42.25\ \text{square metres}$
$42\ 250\ \text{square metres}$

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$A$

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?

$156.25\ \text{square millimetres}$
$1562.5\ \text{square millimetres}$
$156\ 250\ \text{square millimetres}$
$1\ 562\ 500\ \text{square millimetres}$

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$D$

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?

$9.025\ \text{square millimetres}$
$90.25\ \text{square millimetres}$
$90\ 250\ \text{square millimetres}$
$902\ 500\ \text{square millimetres}$

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)

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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.

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

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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?

$(c+d)-(a+b)$
$(c\times d)-(a\times b)$
$(c\times d)\times (a\times b)$
$(c\times d)+(a\times b)$

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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?

$(e\times f)\times (a\times b)\times (c\times d)$
$(e\times f)+(a\times b)+(c\times d)$
$(e+f)+(a+b)+(c+d)$
$(e\times f)+(a\times (b+d))+(c\times d)$

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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?

$(a\times b)+(c\times d)$
$(a\times b)\times (c\times d)$
$(a+b)+(c+d)$
$(a+b)\times (c+d)$

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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

$59$
$72$
$81$
$128$

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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.

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

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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$

Show Worked Solution

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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$306\ \text{mm}^2$

Show Worked Solution

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

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

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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$24\ \text{m}$

Show Worked Solution

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

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

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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$84\ \text{m}^2$

Show Worked Solution

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

Area, SM-Bank 058

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

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$115.73\ \text{mm}^2$

Show Worked Solution

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

Area, SM-Bank 057

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

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$48.96\ \text{cm}^2$

Show Worked Solution

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

Area, SM-Bank 056

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

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

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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$

Show Worked Solution

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.

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

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

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a. $\sqrt{72}\ \text{m}$

b. $72\ \text{m}^2$

Show Worked Solution

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

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

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

Area, SM-Bank 054

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

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$2304\ \text{cm}^2$

Show Worked Solution

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

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

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

$\text{Perimeter}$	$=12.4\ \text{cm}$
$\therefore\ \text{Side}$	$=\dfrac{12.4}{4}$
	$=3.1\ \text{cm}$

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

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.

(2 marks)

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(2 marks)

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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.

(2 marks)

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(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.

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)

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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.

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

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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.

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

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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.

(2 marks)

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(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.

(2 marks)

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(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 mark)

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(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.

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

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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)

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$32.66 \ \text{m}^2$

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

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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$6700\ \text{m}^2$

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$\text{Area}$	$= 160\times 40+12\times 25$
	$=6700\ \text{m}^2$

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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$643\ \text{mm}^2$

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$\text{Agree represents }80\text{% of the circle}$

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

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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$1382.3\ \text{cm}^2$

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

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

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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$10\ \text{m}^2$

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$\text{Method 1:}$

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

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

$\text{Method 2:}$

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

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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$11.5\ \text{cm}$

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$\text{Area of circle}$	$=\pi r^2$
	$=\pi\times 6.5^2$
	$= 132.73\dots\ \text{cm}^2$

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

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

Area, SM-Bank 033 MC

A piece of cardboard is shown in the diagram below.

The area of the cardboard, in square centimetres, is

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$B$

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

$\Rightarrow B$

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

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$A$

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

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

$\Rightarrow A$

Area, SM-Bank 031 MC

Consider the diagram below.

The shaded area, in square centimetres, is

$35$
$45$
$60$
$95$

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$A$

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

$\Rightarrow A$

Area, SM-Bank 042

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

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$37\ \text{cm}^2$

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$\text{Area}$	$=\text{Area of upper rectangle}+\text{Area of lower rectangle}$
	$=9\times 3+5\times 2$
	$=37\ \text{cm}^2$

Area, SM-Bank 030

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

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$95\ \text{cm}^2$

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$\text{Area}$	$=\text{Area of large rectangle}-\text{Area of small rectangle}$
	$=16\times 11-9\times 9$
	$=95\ \text{cm}^2$

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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$41.9\ \text{m}^2$

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

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

$\therefore\ \text{Total area of gym}$

$=40.7+1.2$

$=41.9\ \text{m}^2$

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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$368.56\ \text{m}^2$

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

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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$23.4\ \text{cm}^2$

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

$\text{Using Pythagoras to find}\ h:$

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

$\text{Area of}\ \Delta ABC$

$=\dfrac{1}{2}\times bh$

$=\dfrac{1}{2}\times 5.2\times 1.496\dots$

$= 3.891\dots\ \text{cm}^2$

$\therefore\ \text{Area of hexagon}$

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

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

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.

$25\ \text{m}^2$
$100\ \text{m}^2$
$200\ \text{m}^2$
$300\ \text{m}^2$

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$C$

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

$\text{Area} =x\times y$

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

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

$\Rightarrow C$

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?

$3$
$4$
$6$
$9$

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$D$

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

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

$\Rightarrow D$

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)

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$506.25\ \text{cm}^2$

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

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

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

Area, SM-Bank 023

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

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$44\ \text{cm}^2$

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$\text{Method 1: Subtraction}$

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

$\text{Method 2: Addition}$

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

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?

$3\ \text{cm}\times 6\ \text{cm}$
$6\ \text{cm}\times 9\ \text{cm}$
$12\ \text{cm}\times 6\ \text{cm}$
$21\ \text{cm}\times 3\ \text{cm}$

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$B$

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)

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$3\ \text{cm}$

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

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

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

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

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)

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$588\ \text{m}^2$

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$\text{Longer side}$	$=21\times \dfrac{4}{3}$
	$=28\ \text{m}$

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

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

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

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

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

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

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

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

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

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