Quadrilaterals

Area, SM-Bank 149

A rhombus has an area of 250 square centimetres. If one diagonal measures 10 centimetres, find the length of the other diagonal. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$50\ \text{cm}$

Show Worked Solution

$\text{Area of rhombus}=250\ \text{cm}^2$

$\text{Length of diagonal:}\ x=10\ \text{m}$

$A$	$=\dfrac{1}{2}xy$
$250$	$=\dfrac{1}{2}\times 10\times y$
$5y$	$=250$
$y$	$=\dfrac{250}{5}$
	$=50\ \text{cm}$

$\therefore\ \text{The other diagonal is }50\ \text{cm long.}$

Area, SM-Bank 148

A rhombus has an area of 140.8 square metres. If one diagonal measures 16 metres, find the length of the other diagonal. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$17.6\ \text{m}$

Show Worked Solution

$\text{Area of rhombus}=140.8\ \text{m}^2$

$\text{Length of diagonal:}\ x=16\ \text{m}$

$A$	$=\dfrac{1}{2}xy$
$140.8$	$=\dfrac{1}{2}\times 16\times y$
$8y$	$=140.8$
$y$	$=\dfrac{140.8}{8}$
	$=17.6\ \text{m}$

$\therefore\ \text{The other diagonal is }17.6\ \text{m long.}$

Area, SM-Bank 147

Calculate the area of the following rhombus in millimetres squared. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$96\ \text{mm}^2$

Show Worked Solution

$\text{Length of diagonal 1:}\ x=16\ \text{mm}$

$\text{Length of diagonal 2:}\ y=12\ \text{mm}$

$\text{Area of rhombus}$	$=\dfrac{1}{2}xy$
	$=\dfrac{1}{2}\times 16\times 12$
	$=96\ \text{mm}^2$

Area, SM-Bank 146

Calculate the area of the following rhombus. All measurements are in metres. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$90.24\ \text{m}^2$

Show Worked Solution

$\text{Length of diagonal 1:}\ x=14.1\ \text{m}$

$\text{Length of diagonal 2:}\ y=12.8\ \text{m}$

$\text{Area of rhombus}$	$=\dfrac{1}{2}xy$
	$=\dfrac{1}{2}\times 14.1\times 12.8$
	$=90.24\ \text{m}^2$

Area, SM-Bank 145

Calculate the area of the following rhombus. All measurements are in metres. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$76.44\ \text{m}^2$

Show Worked Solution

$\text{Length of diagonal 1:}\ x=2\times 4.2=8.4\ \text{m}$

$\text{Length of diagonal 2:}\ y=2\times 9.1=18.2\ \text{m}$

$\text{Area of rhombus}$	$=\dfrac{1}{2}xy$
	$=\dfrac{1}{2}\times 8.4\times 18.2$
	$=76.44\ \text{m}^2$

Area, SM-Bank 144

Calculate the area of the following rhombus. All measurements are in centimetres. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$484\ \text{cm}^2$

Show Worked Solution

$\text{Length of diagonal 1:}\ x=2\times 11=22\ \text{cm}$

$\text{Length of diagonal 2:}\ y=2\times 22=44\ \text{cm}$

$\text{Area of rhombus}$	$=\dfrac{1}{2}xy$
	$=\dfrac{1}{2}\times 22\times 44$
	$=484\ \text{cm}^2$

Area, SM-Bank 143

Calculate the area of the following rhombus. All measurements are in millimetres. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$24\ \text{mm}^2$

Show Worked Solution

$\text{Length of diagonal 1:}\ x=2\times 3=6\ \text{mm}$

$\text{Length of diagonal 2:}\ y=2\times 4=8\ \text{mm}$

$\text{Area of rhombus}$	$=\dfrac{1}{2}xy$
	$=\dfrac{1}{2}\times 6\times 8$
	$=24\ \text{mm}^2$

Area, SM-Bank 142

Calculate the area of the following kite. All measurements are in millimetres. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$180\ \text{mm}^2$

Show Worked Solution

$\text{Length of diagonal 1:}\ x=4+11=15\ \text{mm}$

$\text{Length of diagonal 2:}\ y=2\times 12=24\ \text{mm}$

$\text{Area of kite}$	$=\dfrac{1}{2}xy$
	$=\dfrac{1}{2}\times 34\times 20$
	$=180\ \text{mm}^2$

Area, SM-Bank 141

Calculate the area of the following kite. All measurements are in metres. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$340\ \text{m}^2$

Show Worked Solution

$\text{Length of diagonal 1:}\ x=26+8=34\ \text{m}$

$\text{Length of diagonal 2:}\ y=2\times 10=20\ \text{m}$

$\text{Area of kite}$	$=\dfrac{1}{2}xy$
	$=\dfrac{1}{2}\times 34\times 20$
	$=340\ \text{m}^2$

Area, SM-Bank 140

Calculate the area of the following kite. All measurements are in centimetres. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$156\ \text{cm}^2$

Show Worked Solution

$\text{Length of diagonal 1:}\ x=8+18=26\ \text{cm}$

$\text{Length of diagonal 2:}\ y=2\times 6=12\ \text{cm}$

$\text{Area of kite}$	$=\dfrac{1}{2}xy$
	$=\dfrac{1}{2}\times 26\times 12$
	$=156\ \text{cm}^2$

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)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$62.64\ \text{m}^2$

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 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)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$520\ \text{cm}$

Show Worked Solution

$\text{Let the unknown diagonal}=x$

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

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)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$13\ \text{m}$

Show Worked Solution

$\text{Let the unknown diagonal}=x$

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

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)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$0.42\ \text{m}^2$

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 131

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

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$15\ 200\ \text{cm}^2$

Show Worked Solution

$A$	$=\dfrac{1}{2}xy$
	$=\dfrac{1}{2}\times 152\times 200$
	$=15\ 200\ \text{cm}^2$

Area, SM-Bank 130

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

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$288\ \text{m}^2$

Show Worked Solution

$A$	$=\dfrac{1}{2}xy$
	$=\dfrac{1}{2}\times 18\times 32$
	$=288\ \text{m}^2$

Area, SM-Bank 129

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

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$5.115\ \text{cm}^2$

Show Worked Solution

$A$	$=\dfrac{1}{2}xy$
	$=\dfrac{1}{2}\times 3.3\times 3.1$
	$=5.115\ \text{cm}^2$

Area, SM-Bank 122

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

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

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$487\ \text{m}^2$

Show Worked Solution

$\text{Total Area}=\text{Area Rectangle}+\text{Area trapezium}$

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

Area, SM-Bank 104

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

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$15.5\ \text{cm}^2$

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)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$1116\ \text{m}^2$

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)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$182\ \text{cm}^2$

Show Worked Solution

$\text{Area}$	$=\text{Area triangle 1}+\text{Area triangle 2}$
	$=\dfrac{1}{2}\times 14\times 12+\dfrac{1}{2}\times 14\times 14$
	$=84+98$
	$=182\ \text{cm}^2$

Area, SM-Bank 096

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

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

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

--- 4 WORK AREA LINES (style=lined) ---
He is creating the path using pavers at a cost of $92 per square metre. Calculate the cost of the pavers. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---

Show Answers Only

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.

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

--- 4 WORK AREA LINES (style=lined) ---
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)

--- 2 WORK AREA LINES (style=lined) ---

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 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?

$13.5\ \text{cm}^2$
$14.5\ \text{cm}^2$
$17.5\ \text{cm}^2$
$19\ \text{cm}^2$

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)

--- 4 WORK AREA LINES (style=lined) ---

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 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)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$5\ \text{litres}$

Show Worked Solution

$\text{Area of square floor}$

$=6^2$

$=36\ \text{m}^2$

$\text{Area of rectangular floor}$

$=6\times 10$

$=60\ \text{m}^2$

$\text{Paint needed for rectangular wall}$

$=\dfrac{60}{36}\times 3$

$=5\ \text{litres}$

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)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$15\ \text{litres}$

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)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$15\ \text{litres}$

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 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)

--- 4 WORK AREA LINES (style=lined) ---

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)

--- 4 WORK AREA LINES (style=lined) ---

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)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$13.5\ \text{m}^2$

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)

--- 4 WORK AREA LINES (style=lined) ---
What is the cost of tiling Bob's outdoor area, if tiles cost $42.50 per square metre? (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

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?

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

Show Answers Only

$B$

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?

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

Show Answers Only

$D$

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?

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

Show Answers Only

$A$

Show Worked Solution

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

$\Rightarrow A$

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)

--- 4 WORK AREA LINES (style=lined) ---
Calculate the cost of tiling the shaded area if the tiles cost $85 per square metre. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

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)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$306\ \text{mm}^2$

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)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$24\ \text{m}$

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)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$6.1\ \text{cm}$

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)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$84\ \text{m}^2$

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)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$115.73\ \text{mm}^2$

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)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$48.96\ \text{cm}^2$

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)

--- 4 WORK AREA LINES (style=lined) ---
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)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

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)

--- 4 WORK AREA LINES (style=lined) ---
Calculate the are of the square correct to one decimal place. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

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)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$2304\ \text{cm}^2$

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)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$9.61\ \text{cm}^2$

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)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$81\ \text{m}^2$

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)

--- 4 WORK AREA LINES (style=lined) ---
(2 marks)

--- 4 WORK AREA LINES (style=lined) ---

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.

(2 marks)

--- 4 WORK AREA LINES (style=lined) ---
(2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

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)

--- 4 WORK AREA LINES (style=lined) ---
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)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

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)

--- 4 WORK AREA LINES (style=lined) ---
Using your answer from (a) calculate the area of the quilt in square metres, correct to 1 decimal place? (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

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)

--- 4 WORK AREA LINES (style=lined) ---
What is the cost of tiling the living area if tiles cost $45 per square metre? (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

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)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$1.68\ \text{km}^2$

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)

--- 4 WORK AREA LINES (style=lined) ---
(2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

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)

--- 4 WORK AREA LINES (style=lined) ---
(2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

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)

--- 3 WORK AREA LINES (style=lined) ---
(1 mark)

--- 3 WORK AREA LINES (style=lined) ---

Show Answers Only

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 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)

--- 4 WORK AREA LINES (style=lined) ---
Calculate the total area of the front of the cabin. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

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)

--- 4 WORK AREA LINES (style=lined) ---

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)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$6700\ \text{m}^2$

Show Worked Solution

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

Area, SM-Bank 035

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

The distance $ST$ is 1 m.

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

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$10\ \text{m}^2$

Show Worked Solution

$\text{Method 1:}$

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

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

$\text{Method 2:}$

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

\(A\)	\(=\dfrac{1}{2}xy\)
\(250\)	\(=\dfrac{1}{2}\times 10\times y\)
\(5y\)	\(=250\)
\(y\)	\(=\dfrac{250}{5}\)
	\(=50\ \text{cm}\)

\(A\)	\(=\dfrac{1}{2}xy\)
\(140.8\)	\(=\dfrac{1}{2}\times 16\times y\)
\(8y\)	\(=140.8\)
\(y\)	\(=\dfrac{140.8}{8}\)
	\(=17.6\ \text{m}\)

\(\text{Area of rhombus}\)	\(=\dfrac{1}{2}xy\)
	\(=\dfrac{1}{2}\times 16\times 12\)
	\(=96\ \text{mm}^2\)

\(\text{Area of rhombus}\)	\(=\dfrac{1}{2}xy\)
	\(=\dfrac{1}{2}\times 14.1\times 12.8\)
	\(=90.24\ \text{m}^2\)

\(\text{Area of rhombus}\)	\(=\dfrac{1}{2}xy\)
	\(=\dfrac{1}{2}\times 8.4\times 18.2\)
	\(=76.44\ \text{m}^2\)

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

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

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

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