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GEOMETRY, FUR2 2021 VCAA 2

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.

  1. Calculate the area, in square metres, of the shaded region in the diagram above.
  2. Round your answer to two decimal places.   (1 mark)
  3. Calculate the perimeter, in metres, of the shaded region in the diagram.
  4. Round your answer to one decimal place.   (1 mark)
  5. Two squash players, Wei-Yi and Bao, are playing a match.
  6. The diagram below shows the lines on the court floor.

 
             

  1. Wei-Yi is serving from point `A` and and Bao is standing at point `B`.
  2. Point `A` is 2.7 m from Point `C`.
  3. Point `B` is 3.1 m from Point `C`.
  4. The angle `ACB` is 119°.
  5. Show that the distance between Wei-Yi and Bao is 5 m, rounded to the nearest metre.   (1 mark)
Show Answers Only
  1. `32.66 \ text{m}^2`
  2. `26.5 \ text{m}`
  3. `5 \ text{m}`
Show Worked Solution
a.   `text{Shaded Area}` `= 1/2 xx 9.75 xx (4.57 + 2.13)`
    `= 32.6625`
    `= 32.66 \ text{m}^2 \ text{(to 2 d.p.)}`

 

b.

`x = 4.57 – 2.13 = 2.44 \ text{m}`
  

`text{Using Pythagoras}`

`y` `= sqrt{2.44^2 + 9.75^2}`
  `= 10.050 …\ text{m}`

 

`:. \ text{Perimeter}` `= 4.57 + 9.75 + 2.13 + 10.05`
  `= 26.5 \ text{m}`

 
c. 

`text{Using cosine rule:}`

`AB^2` `=  AC^2 + CB^2 – 2 xx AC xx CB xx cos 119^@`
  `= 2.7^2 + 3.1^2 – 2 xx 2.7 xx 3.1 xx cos119^@`
  `= 25.0157`
`:.AB` `= 5 \ text{m (nearest metre)}`

GEOMETRY, FUR2 2013 VCAA 1

A spectator, `S`, in the grandstand of an athletics ground is 13 m vertically above point `G`.

Competitor `X`, on the athletics ground, is at a horizontal distance of 40 m from `G`.
  

Geometry and Trig, FUR2 2013 VCAA 1_1
 

  1. Find the distance, `SX`, correct to the nearest metre.  (1 mark)

Competitor `X` is 40 m from `G` and competitor `Y` is 52 m from `G`.

The angle `XGY` is 113°.
 

Geometry and Trig, FUR2 2013 VCAA 1_2
 

    1. Calculate the distance, `XY`, correct to the nearest metre.  (1 mark)
    2. Find the area of triangle `XGY`, correct to the nearest square metre.  (1 mark)
  2. Determine the angle of elevation of spectator `S` from competitor `Y`, correct to the nearest degree.

    Note that `X`, `G` and `Y` are on the same horizontal level.  (1 mark)
     

Geometry and Trig, FUR2 2013 VCAA 1_3

Show Answers Only
  1. `42\ text{m  (nearest m)}`
    1. `77\ text{m  (nearest m)}`
    2. `957\ text{m²  (nearest m²)}`
  3. `14^@\ \ text{(nearest degree)}`
Show Worked Solution

a.   `text(Using Pythagoras:)`

`SX^2` `= 40^2 + 13^2`
  `= 1769`
`:. SX` `= 42.05…`
  `= 42\ text{m  (nearest m)}`

 

b.i.   `text(Using the cosine rule:)`

`XY^2` `= 40^2 + 52^2 – 52^2 xx 40 xx 52cos113^@`
  `= 5929.44…`
`:. XY` `= 77.00…`
  `= 77\ text{m  (nearest m)}`

 

b.ii.   `text(Using the sine rule:)`

`text(Area)` `= 1/2ab sinC`
  `= 1/2 xx 40 xx 52 xx sin113^@`
  `= 957.32…`
  `= 957\ text{m²  (nearest m²)}`

 

c.    Geometry-FUR2-2013-VCAA-1 Answer
`theta` `= text(angle of elevation of)\ S\ text(from)\ Y`
`tantheta` `= 13/52`
`:. theta` `= tan^(−1)\ 1/4`
  `= 14.036…`
  `= 14^@\ \ text{(nearest degree)}`

GEOMETRY, FUR2 2015 VCAA 4

Wires support the communications tower, as shown in the diagram below.
 

Geometry and Trig, FUR2 2015 VCAA 4
 

The shortest wire is 31 m long.

The shortest wire makes an angle of 38° with the communications tower.

The longest wire is 37 m long.

The longest wire is attached to the communications tower `x` metres above the shortest wire.

What is the value of `x`?

Write your answer in metres, correct to one decimal place.  (2 marks)

Show Answers Only

`7.3\ text(m)`

Show Worked Solution

VCAA 2015 fur2 Q4bi

`text(Using the sine rule:)`

♦ Mean mark sub 50% (exact data not available).
`(sin theta)/31` `= sin142^@/37`
`sin theta` `= (31 xx sin142^@)/37`
`:. theta` `=31.05…^@`

 

`:.phi` `=180 – 142 – 31.05…`
  `= 6.94…^@`

 

`text(Using the cosine rule:)`

`x^2` `=37^2+31^2 – 2 xx 31 xx 37 xx cos 6.95^@`
  `=52.85…`
`:. x`  `=7.27…`
  `=7.3\ text(m)`

GEOMETRY, FUR1 2015 VCAA 3 MC

The lengths of the sides of a triangle are 3 cm, 6 cm and 5 cm, as shown below.
 

GEO & TRIG, FUR1 2015 VCAA 3 MC
 

The angle, `x`, can be found using

A.   `cos(x) = (3^2 + 6^2 - 5^2)/(2 xx 3 xx 6)`

B.   `cos(x) = (3^2 + 6^2 + 5^2)/(2 xx 3 xx 6)`

C.   `cos(x) = (3^2 + 5^2 - 6^2)/(2 xx 3 xx 6)`

D.   `cos(x) = 3/5`

E.   `cos(x) = 3/6`

Show Answers Only

`A`

Show Worked Solution

`text(Using the cosine rule:)`

`cosx = (3^2 + 6^2 – 5^2)/(2 xx 3 xx 6)`

`=> A`

GEOMETRY, FUR1 2013 VCAA 2 MC

The distances from a kiosk to points `A` and `B` on opposite sides of a pond are found to be 12.6 m and 19.2 m respectively.

The angle between the lines joining these points to the kiosk is 63°.
 

 The distance, in m, across the pond between points `A` and `B` can be found by evaluating

A.  `1/2 xx 12.6 xx 19.2 xx sin(63°)`

B.  `{19.2 xx sin(63°)}/12.6`

C.  `sqrt(12.6^2 + 19.2^2)`

D.  `sqrt(12.6^2 + 19.2^2 - 2 xx 12.6 xx 19.2 xx cos(63°)`

E.  `sqrt{s(s - 12.6)(s - 19.2)(s - 63)} , text(where)\ s = 1/2 (12.6 + 10.2 + 63)` 

Show Answers Only

`D`

Show Worked Solution

`text(Using the cosine rule:)`

`(AB)^2` `= 12.6^2 xx 19.2^2 – 2 xx 12.6 xx 19.2 xx cos(63°)`
`:. AB` `= sqrt{12.6^2 xx 19.2^2 – 2 xx 12.6 xx 19.2 xx cos(63°)}`

 
`=>D`