smc-273-20-Pythagoras

GEOMETRY, FUR1 2018 VCAA 1 MC

Henry flies a kite attached to a long string, as shown in the diagram below.

The horizontal distance of the kite to Henry’s hand is 8 m.

The vertical distance of the kite above Henry’s hand is 15 m.

The length of the string, in metres, is

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`B`

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`text(Using Pythagoras)`

`s^2`	`= 8^2 + 15^2`
	`= 289`
`:. s`	`= 17\ text(metres)`

`=> B`

GEOMETRY, FUR2 2008 VCAA 2

A shed has the shape of a prism. Its front face, `AOBCD`, is shaded in the diagram below. `ABCD` is a rectangle and `M` is the mid point of `AB`.

Show that the length of `OM` is 1.6 m. (1 mark)
Show that the area of the front face of the shed, `AOBCD`, is 18 m². (1 mark)
Find the volume of the shed in m³. (1 mark)
All inside surfaces of the shed, including the floor, will be painted.
1. Find the total area that will be painted in m². (2 marks)
  
  One litre of paint will cover an area of 16 m².
2. Determine the number of litres of paint that is required. (1 mark)

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`text(See Worked Solution)`
`text(See Worked Solution)`
`180`
1. `208`
2. `13\ text(litres)`

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`text(In)\ Delta AOM,\ text(using Pythagoras:)`

`OM`	`= sqrt (3.4^2 – 3^2)`
	`= sqrt 2.56`
	`= 1.6\ text(m … as required)`

b. `text(Area of front face of shed)`

MARKER’S COMMENT: “The “Show that” directive requires students to include all mathematical steps, as shown in the solution. Short cuts will not gain full marks!

`= text(Area)\ Delta AOB + text(Area)\ ABCD`

`= 1/2 xx 1.6 xx 6 + 2.2 xx 6`

`= 18\ text(m² … as required.)`

c.	`V`	`= Ah`
		`= 18 xx 10`
		`= 180\ text(m³)`

d.i.	`text(Roof areas)`	`= 2(1/2 xx 1.6 xx 6 + 3.4 xx 10)`
		`= 2 (4.8 + 34)`
		`= 77.6\ text(m²)`
	`text(Wall areas)`	`= 2 (2.2 xx 6 + 2.2 xx 10)`
		`= 2 (13.2 + 22)`
		`= 70.4\ text(m²)`
	`text(Floor area)`	`= 60\ text(m²)`

`:.\ text(Total Area to be painted)`

`= 77.6 + 70.4 + 60`

`= 208\ text(m²)`

ii. `text(Litres of paint required)`

`= 208/16`

`= 13\ text(litres)`

GEOMETRY, FUR1 2008 VCAA 8 MC

A regular hexagon has side length 3.0 cm and height 5.2 cm as shown in the diagram above.

The area (in cm²) of the hexagon is closest to

A. `11.7`

B. `13.5`

C. `15.6`

D. `18.0`

E. `23.4`

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`E`

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`text(Area of rectangle)`	`= 3.0 xx 5.2`
	`= 15.6\ text(cm²)`

`text(Using Pythagoras to find)\ h:`

`3.0^2`	`= 2.6^2 + h^2`
`h^2`	`= 9 – 6.76`
	`= 2.24`
`h`	`= 1.496…`

`text(Area of)\ \ Delta ABC`

`= 1/2 xx b xx h`

`= 1/2 xx 5.2 xx 1.496…`

`= 3.891…\ text(cm²)`

`:.\ text(Area of hexagon)`

`= 15.6 + (2 xx 3.891…)`

`= 23.38…\ text(cm²)`

`=> E`