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GEOMETRY, FUR1-NHT 2019 VCAA 7 MC

Two similar right-angled triangles are shown in the diagram below.
 

     

The total perimeter of the two triangles, in metres, is

  1. 21
  2. 30
  3. 36
  4. 42
  5. 45
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`C`

Show Worked Solution

     

`text(Ratio of similar sides)\ = 3 : 6 = 1 : 2`

`text(Base of 12 m in ratio)\ 1 : 2 = 4 : 8`

 
 

`text(Length of hypotenuses:)`

`h_1 = sqrt(3^2 + 4^2) = 5`

`h_2 = sqrt(6^2 + 8^2) = 10`
 

`:. \ text(Perimeter) ` `= 3 + 4 + 5 + 6 + 8 + 10`
  `= 36 \ text(m)`

 
`=> \ C`

GEOMETRY, FUR1 2019 VCAA 4 MC

Triangle `M`, shown below, has side lengths of 3 cm, 4 cm and 5 cm.
 


 

Four other triangles have the following side lengths:

    • Triangle `N` has side lengths of 3 cm, 6 cm and 8 cm.
    • Triangle `O` has side lengths of 4 cm, 8 cm and 12 cm.
    • Triangle `P` has side lengths of 6 cm, 8 cm and 10 cm.
    • Triangle `Q` has side lengths of 9 cm, 12 cm and 15 cm.

The triangles that are similar to triangle `M` are

  1. triangle `N` and triangle `O`.
  2. triangle `N`, triangle `O` and triangle `P`.
  3. triangle `O` and triangle `P`.
  4. triangle `O` and triangle `Q`.
  5.  triangle `P` and triangle `Q`.
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`E`

Show Worked Solution

`text(Similarity test: all side pairs are in the same ratio.)`

`text(Triangle)\ P:`

`3/6 = 4/8 = 5/10 =>\ text(similar)`
 

`text(Triangle)\ Q:`

`3/9 = 4/12 = 5/15 =>\ text(similar)`
 

`text(Triangles)\ N, O \ => \ text(ratios not equal)`

`=>  E`

GEOMETRY, FUR1 2016 VCAA 2 MC

Triangle ABC is similar to triangle DEF.
 


 

The length of DF, in centimetres, is

  1. `0.9`
  2. `1.2`
  3. `1.8`
  4. `2.7`
  5. `3.6`
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`D`

Show Worked Solution

`text(Using similar ratios,)`

`(DF)/(DE)` `= (AC)/(AB)`
`(DF)/1.8` `= 3.6/2.4`
`:. DF` `= ((3.6 xx 1.8))/2.4`
  `= 2.7\ text(cm)`

`=> D`

GEOMETRY, FUR2 2010 VCAA 4

A flying fox suspension wire begins at `V`, 12.5 metres above `T` as shown in the diagram below. It ends at `S`, 4.5 metres above `K`.
 

GEOMETRY, FUR2 2010 VCAA 4
 

At `P`, the flying fox wire passes over `H`.

The horizontal distances `TH` and `HK` are 95 metres and 65 metres respectively.

Calculate the vertical distance, `PH`, in metres.  (2 marks)

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`7.75\ text(m)`

Show Worked Solution

GEOMETRY, FUR2 2010 VCAA 4 Answer 
 

`Delta VAS\ text(|||)\ Delta PBS\ \ (A A A)`

♦♦♦ Mean mark 11%.
`(PB)/(BS)` `= (VA)/(AS)`
`(PB)/(65)` `= 8/160`
`:. PB` `= (8 xx 65)/160`
  `= 3.25\ text(m)`

 

`:. PH` `= PB + BH`
  `= 3.25 + 4.5`
  `= 7.75\ text(m)`

 

`text(Alternative solution)`

`text(In)\ Delta VAS,`

`tantheta` `= 8/160`
`:. theta` `= 2.862…^@`

 

`text(In)\ DeltaPBS,`

`tantheta` `= (PB)/65`
`PB` `= 65 xx tan2.862…^@`
  `= 3.25\ text(m)`

 

`:. PH = 3.25 + 4.5 = 7.75\ text(m)`

GEOMETRY, FUR1 2011 VCAA 8 MC

The diagram below shows a cross-section, ` PQRS`, of a swimming pool.

The swimming pool is 11 metres long and the depth increases uniformly from 1 metre at the shallow end to 1.8 metres at the deep end.

The depth of the water at a point 8 metres from the shallow end, represented by `TU` on the diagram is closest to

A.  `1.25\ text(metres)`

B.  `1.31\ text(metres)`

C.  `1.34\ text(metres)`

D.  `1.58\ text(metres)`

E.  `1.62\ text(metres)`

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`D`

Show Worked Solution

`/_ PXU = /_ PYS = 90^@`

♦ Mean mark 40%.
MARKERS’ COMMENT: The most common mistake was to apply the properties of similar triangles to trapeziums, leading to an incorrect answer of B.

`/_ XPU\ text(is common)`

`:. Delta PXU\ text(|||)\ Delta PYS\ text{(equiangular)}`

`(XU)/8` `= (YS)/11` `text{(corresponding sides of}`
`\ text{similar triangles)}`
`:. XU` `= (8 xx 0.8)/11`  
  `= 0.581…\ \ text(m)`  

 

`:. text(Depth)\ TU = 1.581…\ \ text(m)`

`=> D`

GEOMETRY, FUR1 2006 VCAA 7 MC

GEOMETRY, FUR1 2006 VCAA 7 MC
 

In the diagram, `AD = 9\ text(cm),  AC = 24\ text(cm) and DB = 27\ text(cm.)`

Line segments `AC` and `DE` are parallel.

The length of `DE` is

A.     `6\ text(cm)`

B.     `8\ text(cm)`

C.   `12\ text(cm)`

D.   `16\ text(cm)`

E.   `18\ text(cm)`

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`E`

Show Worked Solution

GEOMETRY, FUR1 2006 VCAA 7 MC Answer 

 
`/_CAD = /_EDB\ \ \ text{(} text(corresponding angles,)\ AC\ text(||)\ DE text{)}`

`/_DBE\ text(is common)`

`:. Delta ACB\ text(|||)\ Delta DEB\ \ text{(equiangular)}`

 

`:.\ (DE)/(DB)` `= (AC)/(AB)` `\ \ \ text{(corresponding sides}`
`\ \ \ \ text{of similar triangles)}`
`(DE)/27` `= 24/36`  
`DE` `= (24 xx 27)/36`  
  `= 18\ text(cm)`  

 
`=>  E`

GEOMETRY, FUR1 2013 VCAA 6 MC

In triangle `MNR`, point `P` lies on side `MR` and point `Q` lies on side `NR`.

The lines `PQ` and `MN` are parallel.

The length of `RQ` is 4 cm, the length of `QN` is 6 cm and the length of `PQ` is 5 cm.
 

The length of `MN`, in cm, is equal to

A.    `7.5`

B.    `8.3`

C.  `12.0`

D.  `12.5`

E.  `15.0`

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`D`

Show Worked Solution

`/_ PRQ\ \ text(is common)`

`/_ RPQ = /_ RMN\ \ (text{corresponding}\ PQ\ text{||}\ MN)`

`:. Delta RPQ\ text(|||)\ Delta RMN\ text{(equiangular)}`

`∴ (MN)/(RN)` `= (PQ)/(RQ)` `\ \ \ \ text{(corresponding sides of}`
`\ \ \ \ \ text{similar triangles)}`
`(MN)/10` `= 5/4`  
`:.MN` `= 12.5\ text(cm)`  

`=> D`

GEOMETRY, FUR1 2009 VCAA 1 MC

GEOMETRY, FUR1 2009 VCAA 1 MC
 

The two triangles, `ABC` and `FGH`, are similar.

The length `GH` is

A.   `14\ text(cm)`

B.   `24\ text(cm)`

C.   `26\ text(cm)`

D.   `28\ text(cm)`

E.   `32\ text(cm)`

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`D`

Show Worked Solution

`Delta ABC\ text(|||)\ Delta FGH`

`:. (GH)/(BC)` `=(GF)/(BA)`
`(GH)/(14)` `=(24)/(12)`
`:.GH` `= 28 \ text(cm)`

 
`=>  D`