smc-4943-20-Parallelograms

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$

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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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$6.1\ \text{cm}$

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$\text{Area}$	$=\text{base}\times \text{height}$
$\therefore\ 75.03$	$=12.3\times h$
$h$	$=\dfrac{75.03}{12.3}$
	$=6.1$

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

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 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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$B$

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

$\Rightarrow B$

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

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

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

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

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

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