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Similarity, SMB-027

 


 

  1. What scale factor is used to convert Circle A into Circle B.   (1 mark)

  2. Complete this equation:
  3.      Area of Circle A = _____ × Area of Circle B.  (1 mark)

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  1. \(\dfrac{1}{3}\)
  2. \(9\)
Show Worked Solution

i.     \(\text{Scale factor}\ = \dfrac{\text{Diameter B}}{\text{Diameter A}} = \dfrac{0.8}{2.4} = \dfrac{1}{3} \)
 

ii.     \(\text{Scale factor (B to A)} = \dfrac{\text{Diameter A}}{\text{Diameter B}} = \dfrac{2.4}{0.8} = 3 \)

\(\text{Scale factor (Area)} = 3^2 = 9 \)

\(\therefore\ \text{Area of Circle A = 9 × Area of Circle B} \)

Similarity, SMB-025

A triangular prism is pictured below.
 

By what factor will its volume change if

  1. Each dimension is doubled?   (1 mark)

  2. Each dimension is decreased by two-thirds?   (2 marks)
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  1. \(\text{Increases by a factor of 8}\)
  2. \(\text{Decreases by a factor of}\ \ \dfrac{1}{27} \)
Show Worked Solution

i.    \(\text{Dimensions increase by a factor of 2}\)

\(\Rightarrow\ \text{Volume increases by a factor of}\ 2^3 = 8\)
 

ii.    \(\text{Dimensions decrease by two-thirds}\)

\(\Rightarrow\ \text{i.e. adjust dimensions by a factor of}\ \ \dfrac{1}{3} \)

\(\Rightarrow\ \text{Volume decreases by a factor of}\ \Big{(} \dfrac{1}{3} \Big{)}^3 = \dfrac{1}{27} \)

Similarity, SMB-016

Triangle I and Triangle II are similar. Pairs of equal angles are shown.
 

Find the area of Triangle II?  (3 marks)

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`24\ text{cm}^2`

Show Worked Solution

`text(In Triangle I, using Pythagoras:)`

`text{Base}` `= sqrt(5^2-3^2)`
  `= 4`

 
`text(Triangle I ||| Triangle II (given))`

♦♦ Mean mark 29%.

`=>\ text(corresponding sides are in the same ratio)`

`text{Scale factor}\ = 6/2=2`

`text{Scale factor (Area)}\ = 2^2=4`

`:. text(Area (Triangle II))` `= 4 xx text{Area of triangle I}`
  `= 4 xx 1/2 xx 3 xx 4`
  `=24\ text{cm}^2`

Similarity, SMB-012

Poppy uses a photocopier to enlarge this picture.
 

   

The enlarged picture is 3 times as high and 3 times as wide as the original.

By what factor is the area of the picture increased?   (2 marks)

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`text(9 times the area of the original)`

Show Worked Solution

`text{Method 1}`

`text{Dimensions increased by a factor of 3}`

`:.\ text{Area increased by a factor of}\ 3^2 = 9`
 

`text{Method 2}`

`text(Area of original picture)\ = 3 xx 5 = 15\ text(cm)^2`

`text(Area of enlarged picture)\ = 9 xx 15 = 135\ text(cm)^2`

`:.\ text(Factor)\ = 135/15 = 9\ text(times)`

Measurement, STD1 M5 2021 HSC 26

The diagrams show two similar shapes. The dimensions of the small shape are enlarged by a scale factor of 1.5 to produce the large shape.
 


 

Calculate the area of the large shape.  (3 marks)

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`279\ text(cm)^2`

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`text(Dimension of larger shape:)`

♦♦ Mean mark 32%.

`text(Width) = 16 xx 1.5 = 24\ text(cm)`

`text(Height) = 9 xx 1.5 = 13.5\ text(cm)`

`text(Triangle height) = 2.5 xx 1.5 = 3.75\ text(cm)`

`:.\ text(Area)` `= 24 xx (13.5-3.75) + 1/2 xx 24 xx 3.75`
  `= 279\ text(cm)^2`

Measurement, STD1 M5 2020 HSC 28

Two similar right-angled triangles are shown.
 


 

The length of side `AB` is 8 cm and the length of side `EF` is 4 cm.

The area of triangle `ABC` is 20 cm2.

Calculate the length in centimetres of side `DF` in Triangle II, correct to two decimal places.   (4 marks)

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`7.55\ \text{cm}`

Show Worked Solution

`text{Consider} \ Δ ABC :`

`text{Area}` `= frac{1}{2} xx AB xx BC`
`20` `= frac{1}{2} xx 8 xx BC`
`therefore \ BC` `= 5`

 

`text{Using Pythagoras in} \ Δ ABC :`

♦♦♦ Mean mark 11%.

`AC = sqrt(8^2 + 5^2) = sqrt89`

 

`text{S} text{ince} \ Δ ABC\ text{|||}\ Δ DEF,`

`frac{AC}{BC}` `= frac{DF}{EF}`
`frac{sqrt89}{5}` `= frac{DF}{4}`
`therefore \ DF` `= frac{4 sqrt89}{5}`
  `= 7.547 …`
  `= 7.55 \ text{cm (to 2 d.p.)}`