smc-1103-10-Pythagoras

Measurement, STD1 M3 2025 HSC 22

An isosceles triangle is drawn inside a circle as shown. The base of the triangle is 4.8 cm long, the length of other sides is 4 cm and the height is $h$ cm.

Calculate the height, $h$, of triangle $ABC$. (2 marks)
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The area of triangle $ABC$ is 7.68 cm².
The radius of the circle is 2.5 cm.
Express the area of triangle $ABC$ as a percentage of the area of the circle, correct to 1 decimal place. (2 marks)
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a. $h=3.2\ \text{cm}$

b. $39.1\%$

Show Worked Solution

a. $\text{Since}\ \Delta ABC\ \text{is isosceles:}$

$BM\ \text{bisects}\ AC\ \ \Rightarrow\ \ AM=MC=2.4$

$\text{Using Pythagoras:}$

$h^2=4^2-2.4^2=16-5.76=10.24$

$h = \sqrt{10.24} = 3.2\ \text{cm}$

b. $\text{Area of circle}=\pi\times 2.5^2$

$\text{Percentage}$	$=\dfrac{\text{Area of triangle}}{\text{Area of circle}}\times 100\%$
	$=\dfrac{7.68}{\pi\times 2.5^2}\times 100\%$
	$=39.11…\%$
	$\approx 39.1\%\ \text{(1 decimal place)}$

Measurement, STD1 M3 2025 HSC 20

A map of a park containing a duck pond is shown.

A fence is built passing through the points $A$, $B$ and $C$ around the duck pond.

Using the scale provided on the map, calculate the length of the fence $AB$. (2 marks)
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The length of $AB$ is equal to the length of $BC$.
Use Pythagoras’ theorem to calculate the length of $AC$ in metres. Give your answer correct to 3 significant figures. (3 marks)
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What is the true bearing of point $A$ from point $C$ ? (2 marks)
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a. $75\ \text{metres}$

b. $106\ \text{metres}$

c. $225^\circ$

Show Worked Solution

a. $\text{Scale: 1 grid width = 5 metres}$

$AB = 15 \times 5 = 75\ \text{metres}$

b. $\text{Using Pythagoras:}$

$AC^2=AB^2+BC^2$

$AC^2=75^2+75^2=11250$

	$\therefore\ AC$	$=\sqrt{11250}$
		$=106.066…$
		$\approx 106\ \text{m (3 sig fig)}$

c. $\text{Since }AB=BC:$

$\angle BAC=\angle BCA=45^\circ$

$\text{Bearing of \(A$ from $C$}\ =180+45=225^\circ\)

Measurement, STD1 M3 2024 HSC 14

A hotel is located 186 m north and 50 m west of a train station.

What is the straight line distance from the hotel to the train station? Round your answer to the nearest metre. (2 marks)

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What is the bearing of the hotel from the train station? Round your answer to the nearest degree. (2 marks)

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a. $193\ \text{m}$

b. $345^\circ\ \text{(nearest degree)}$

Show Worked Solution

a. $\text{By Pythagoras:}$

	$d^2$	$=50^2+186^2$
	$d^2$	$=37\,096$
	$d$	$=\sqrt{37\,096}$
		$=192.603\dots$
		$=193\ \text{m (nearest metre)}$

♦♦ Mean mark (a) 37%.

b.	$\tan\theta$	$=\dfrac{50}{186}$
	$\theta$	$=15.046\dots^\circ$
		$\approx 15^\circ\ \text{(nearest degree)}$

$\text{Bearing}\ H\ \text{from}\ T =360-15=345^\circ$

♦♦♦ Mean mark (b) 12%.

Measurement, STD1 M3 2022 HSC 27

Shan is interested in buying a block of bushland. The price per hectare is $500. The land he wishes to purchase is in the shape of a right-angled triangle as shown.

The length of side `A B` is 7800 metres and the length of side `B C` is 3000 metres. The right angle of the triangle is angle `A C B`.

Note: 1 hectare =10 000 m²

What is the cost of the block of bushland? (4 marks)

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$540 000

Show Worked Solution

`text{Using Pythagoras:}`

`AC^2`	`=7800^2-3000^2`
	`=51\ 840\ 000`
`:.AC`	`=7200`

`text{Area}\ DeltaABC`	`=(BC xx AC)/2`
	`=3000 xx 7200`
	`=108\ 000\ 000\ text{m}^2`
	`=1080\ text{hectares (1 hectare = 10 000 m}^(2))`

`:.\ text{Cost}`	`=$500 xx 1080`
	`=$540\ 000`

♦♦ Mean mark 28%.

Measurement, STD1 M3 2019 HSC 31

Two right-angled triangles, `ABC` and `ADC`, are shown.

Calculate the size of angle `theta`, correct to the nearest minute. (3 marks)

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`41°4′\ \ text{(nearest minute)}`

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`text(Using Pythagoras in)\ DeltaACD:`

♦♦♦ Mean mark 15%.

`AC^2`	`= 2.5^2 + 6^2`
	`= 42.25`
`:.AC`	`= 6.5\ text(cm)`

`text(In)\ DeltaABC:`

`costheta`	`= 4.9/6.5`
`theta`	`= cos^(−1)\ 4.9/6.5`
	`= 41.075…`
	`= 41°4′31″`
	`= 41°5′\ \ text{(nearest minute)}`

Measurement, STD2 M6 2016 HSC 26d

The diagram shows a block of land `ABCD` that has been surveyed. All measurements are in metres.

Calculate the length of `AB`, correct to the nearest metre. (2 marks)

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`46\ text{m}`

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`text(Using Pythagoras,)`

`AB`	`= sqrt(32^2 + 33^2)`
	`= 45.967…`
	`= 46\ text{m (nearest m)}`

Measurement, STD2 M6 2005 HSC 25b

Use Pythagoras’ theorem to show that `ΔABC` is a right-angled triangle. (1 mark)

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Calculate the size of `∠ABC` to the nearest minute. (2 marks)

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`text(Proof)`
`67°23^{′}`

Show Worked Solution

i. `ΔABC\ text(is right-angled if)\ \ a^2 + b^2 = c^2`

`a^2 + b^2`	`= 5^2 + 12^2`
	`= 169`
	`= 13^2`
	`= c^2…\ text(as required.)`

MARKER’S COMMENT: Know your calculator process for producing an angle in minutes/seconds. Note >30 “seconds” rounds up to the higher “minute”.

ii. `sin ∠ABC = 12/13`

`:.∠ABC`	`= 67.38…°`
	`=67°22^{′}48^{″}`
	`= 67°23^{′}\ \ \ text{(nearest minute)}`

Measurement, STD2 M1 2004 HSC 23a

The diagram shows the shape of Carmel’s garden bed. All measurements are in
metres.

Show that the area of the garden bed is 57 square metres. (2 marks)

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Carmel decides to add a 5 cm layer of straw to the garden bed.

Calculate the volume of straw required. Give your answer in cubic metres. (2 marks)

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Each bag holds 0.25 cubic metres of straw.

How many bags does she need to buy? (2 marks)

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A straight fence is to be constructed joining point A to point B.

Find the length of this fence to the nearest metre. (2 marks)

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`text(Proof)\ \ text{(See Worked Solutions)}`
`text(2.85 m³)`
`text(She needs to buy 12 bags)`
`8\ text{m (nearest metre)}`

Show Worked Solution

i. `text(Area of)\ Delta ABC`	`= 1/2 xx b xx h`
	`= 1/2 xx 10 xx 5.1`
	`= 25.5\ text(m²)`

`text(Area of)\ Delta ACD`	`= 1/2 xx 10 xx 6.3`
	`= 31.5\ text(m²)`

`:.\ text(Total Area)`	`= 25.5 + 31.5`
	`= 57\ text(m² … as required)`

ii. `V`	`= Ah`
	`= 57 xx 0.05`
	`= 2.85\ text(m³)`

iii. `text(Bags to buy)`	`= 2.85/0.25`
	`= 11.4`

`:.\ text(She needs to buy 12 bags.)`

iv. `text(Using Pythagoras,)`

`AB^2`	`= 6.0^2 + 5.1^2`
	`= 36 + 26.01`
	`= 62.01`
`AB`	`= 7.874…`
	`=8\ text{m (nearest metre)}`

Measurement, STD2 M6 2010 HSC 3 MC

A field diagram has been drawn from an offset survey.

What is the distance from `G` to `H` correct to the nearest metre?

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`B`

Show Worked Solution

`text(Using Pythagoras:)`

`GH^2`	`=12^2+(16-11)^2`
	`=144+25`
	`=169`

`:.\ GH`	`=sqrt169 `
	`=13\ text(m)`

` => B`

\(\text{Percentage}\)	\(=\dfrac{\text{Area of triangle}}{\text{Area of circle}}\times 100\%\)
	\(=\dfrac{7.68}{\pi\times 2.5^2}\times 100\%\)
	\(=39.11…\%\)
	\(\approx 39.1\%\ \text{(1 decimal place)}\)

	\(\therefore\ AC\)	\(=\sqrt{11250}\)
		\(=106.066…\)
		\(\approx 106\ \text{m (3 sig fig)}\)

	\(d^2\)	\(=50^2+186^2\)
	\(d^2\)	\(=37\,096\)
	\(d\)	\(=\sqrt{37\,096}\)
		\(=192.603\dots\)
		\(=193\ \text{m (nearest metre)}\)

b.	\(\tan\theta\)	\(=\dfrac{50}{186}\)
	\(\theta\)	\(=15.046\dots^\circ\)
		\(\approx 15^\circ\ \text{(nearest degree)}\)