smc-1121-10-Perimeter and Area

Measurement, STD1 M5 2024 HSC 29

A floor plan for a living area is shown. All measurements are in millimetres.

What is the length and width of the cupboard, in metres? (1 mark)

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The floor of the living area is to be tiled. Tiles will NOT be placed under the cupboard.
Each tile is 0.2 m × 0.5 m. The tiles are supplied in boxes of 15 at a cost of $100 per box. Only full boxes can be purchased.
What is the cost of the tiles for the living area? (4 marks)
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a. $4\ \text{m}\times 0.5\ \text{m}$

b. $$1100$

Show Worked Solution

a. $\text{1 metre = 1000 mm}$

$\text{Cupboard}\ = 4\ \text{m}\times 0.5\ \text{m}$

b. $\text{Method 1}$

$\text{Total area to be tiled}\ =6\times 3-4\times 0.5=16\ \text{m}^2$

$\text{Area of each tile}\ =0.2\times0.5=0.1\ \text{m}^2$

$\text{Tiles needed}\ =\dfrac{16}{0.1}=160\ \text{tiles}$

$\text{Number of boxes}\ =\dfrac{160}{15}=10.66\dots=11\ \text{boxes}$

$\therefore\ \text{Total cost}\ =100\times 11=$1100$

$\text{Method 2}$

$\text{Tiles to fit 6 m width}\ =\dfrac{6}{0.2}=30 $

$\text{Tiles to fit 2 m width}\ =\dfrac{2}{0.2}=10 $

$\text{Total tiles}\ =30\times 5\ \text{rows}+10\times 1\ \text{rows}\ =160$

$\text{Number of boxes}\ =\dfrac{160}{15}=10.66\dots=11\ \text{boxes}$

$\therefore\ \text{Total cost}\ =100\times 11=$1100$

Measurement, STD1 M1 2024 HSC 9 MC

An equilateral triangle and an isosceles triangle are shown.

The triangles have the same perimeter.

What is the value of $x$?

$8$
$9$
$11$
$12$

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$D$

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$\text{Perimeter triangle 1 }=3\times 11=33\ \text{m}$

$\text{Perimeter triangle 2 }=2\times x+9=(2x+9)\ \text{m}$

$\therefore\ 2x+9$	$=33$
$2x$	$=24$
$x$	$=12$

$\Rightarrow D$

Measurement, STD1 M5 2023 HSC 12

A floor plan is shown.

What are the dimensions, in metres, of the living room? (2 marks)

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The shaded area of the kitchen floor is to be tiled. Each tile is 40 cm by 40 cm. How many tiles are needed to cover the kitchen floor? (2 marks)

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The tiles are supplied in boxes of 10 . Only full boxes can be purchased. How many boxes need to be purchased to tile the kitchen floor? (1 mark)

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a. $5.2\times 5.94$

b. $72\text{ tiles}$

c. $8\text{ boxes}$

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a. $\text{Dimensions}=5.2\times (2.15+0.16+3.6)=5.2\times5.94$

♦ Mean mark (a) 51%.

b. $\text{Method 1}$

$\text{Kitchen Area}$	$=3.2\times 3.6$
	$=11.52\text{ m}^2$
$\text{Tile area}$	$=0.4\times 0.4$
	$=0.16\text{ m}^2$
$\text{Tiles needed}$	$=\dfrac{11.52}{0.16}$
	$=72\text{ tiles}$

$\text{Method 2}$

$\text{Tiles to fit width}$	$=\dfrac{3.6}{0.4}$
	$=9$
$\text{Tiles to fit length}$	$=\dfrac{3.2}{0.4}$
	$=8$
$\text{Tiles needed}$	$=9\times 8$
	$=72\text{ tiles}$

♦ Mean mark (b) 43%.

c. $\text{Boxes}=\dfrac{72}{10}=7.2$

$\therefore\ \text{Boxes }=8$

Measurement, STD2 M1 2021 HSC 1 MC

Which of the following shapes has the largest perimeter?

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`A`

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`\text{Consider each option:}`

`\text{Option A:} \ 4 \times 8 = 32 \ \text{cm}`

`\text{Option B:} \ 2 \times (3 + 11) = 28 \ \text{cm}`

`\text{Option C:} \ 3 \times 10 = 30 \ \text{cm}`

`\text{Option D:} \ 4 \times 2 + 3 + 9 = 20 \ \text{cm}`

`=> A`

Measurement, STD1 M1 2019 HSC 36

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.

The path is to be covered with triangular pavers with side lengths of 15 cm and 20 cm as shown.

The pavers are to be laid to cover the path with no gaps or overlaps.

How many pavers are needed? (4 marks)

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`4176`

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`text(Shaded Area)`	`=\ text(Large rectangle − garden area)`
	`=(1.8+8.4+1.8) xx (1.8+5.4+1.8) – (8.4 xx 5.4)`
	`= 12 xx 9 – 8.4 xx 5.4`
	`= 62.64\ text(m²)`

♦♦ Mean mark 18%.
COMMENT: Convert all measurements to the same units to minimise errors (either m² or cm²).

`text(Area of 1 paver (in m²))`	`= 1/2 xx 0.15 xx 0.20`
	`= 0.015\ text(m²)`

`:.\ text(Number of pavers needed)`

`= 62.64/0.015`

`= 4176`

Measurement, STD2 M1 2009 HSC 23c

The diagram shows the shape and dimensions of a terrace which is to be tiled.

Find the area of the terrace. (2 marks)

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Tiles are sold in boxes. Each box holds one square metre of tiles and costs $55. When buying the tiles, 10% more tiles are needed, due to cutting and wastage.

Find the total cost of the boxes of tiles required for the terrace. (2 marks)

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i. `13.77\ text(m²)`

ii. `$880`

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`text(Area)`	`=\ text(Area of big square – Area of 2 cut-out squares`
	`= (2.7 + 1.8) xx (2.7 + 1.8)\-2 xx (1.8 xx 1.8)`
	`= 20.25\-6.48`
	`= 13.77\ text(m²)`

ii.	`text(Tiles required)`	`= (13.77 +10 text{%}) xx 13.77`
		`= 15.147\ text(m²)`

`=>\ text(16 boxes are needed)`

`:.\ text(Total cost of boxes)`	`=16 xx $55`
	`= $880`

\(\therefore\ 2x+9\)	\(=33\)
\(2x\)	\(=24\)
\(x\)	\(=12\)

\(\text{Kitchen Area}\)	\(=3.2\times 3.6\)
	\(=11.52\text{ m}^2\)
\(\text{Tile area}\)	\(=0.4\times 0.4\)
	\(=0.16\text{ m}^2\)
\(\text{Tiles needed}\)	\(=\dfrac{11.52}{0.16}\)
	\(=72\text{ tiles}\)

\(\text{Tiles to fit width}\)	\(=\dfrac{3.6}{0.4}\)
	\(=9\)
\(\text{Tiles to fit length}\)	\(=\dfrac{3.2}{0.4}\)
	\(=8\)
\(\text{Tiles needed}\)	\(=9\times 8\)
	\(=72\text{ tiles}\)