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v1 Measurement, STD2 M7 2008 HSC 20 MC

A point `P` lies between a lamp post, 1.5 metres high, and a building, 7 metres high. `P` is 2.5 metres away from the base of the post.

From `P`, the angles of elevation to the top of the lamp post and to the top of the building are equal.
 

What is the distance, `x`, from `P` to the top of the tower?

  1. 10.60
  2. 12.50
  3. 13.60
  4. 14.55
Show Answers Only

`C`

Show Worked Solution

`text(Triangles are similar)\ \ text{(equiangular)}`

`text(In smaller triangle:)`

`h^2` `= 1.5^2 + 2.5^2`
  `= 8.5`
`h` `= sqrt 8.5`
   
`x/sqrt8.5` `= 7/1.5 \ \ \ text{(sides of similar Δs in same ratio)}`
`x` `= (7 sqrt 8.5)/1.5`
  `= 13.60…`

 
`=>  C`

v1 Measurement, STD2 M7 2016 HSC 16 MC

The width (`W`) of a road can be calculated using two similar triangles, as shown in the diagram.
  
  

What is the approximate width of the road?

  1. `12.8\ text(m)`
  2. `13.3\ text(m)`
  3. `14.6\ text(m)`
  4. `17.8\ text(m)`
Show Answers Only

`=> C`

Show Worked Solution

`text{Triangles are similar (equiangular)}`

`text(Using similar ratios:)`

`W/(6.5)` `= 18/8`
`:. W` `= (18 xx 6.5)/8`
  `= 14.62…`

 
`=> C`

v1 Algebra, STD2 A4 2014 HSC 29a

A golf club hires an entire course for a charity event at a total cost of `$40\ 000`. The cost will be shared equally among the players, so that `C` (in dollars) is the cost per player when `n` players attend.

  1. Complete the table below by filling in the three missing values.   (1 mark)
    \begin{array} {|l|c|c|c|c|c|c|}
    \hline
    \rule{0pt}{2.5ex}\text{Number of players} (n) \rule[-1ex]{0pt}{0pt} & \ 50\ & \ 100 \ & 200 \ & 250 \ & 400\ & 500 \ \\
    \hline
    \rule{0pt}{2.5ex}\text{Cost per person} (C)\rule[-1ex]{0pt}{0pt} &  &  &  & 160 & 100\ & 80 \ \\
    \hline
    \end{array}
  2. Using the values from the table, draw the graph showing the relationship between `n` and `C`.   (2 marks)
     
  3. What equation represents the relationship between `n` and `C`?   (1 mark)

  4. Give ONE limitation of this equation in relation to this context.   (1 mark)

  5. Is it possible for the cost per person to be $94? Support your answer with appropriate calculations.   (1 mark)

Show Answers Only

a.   

\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\text{Number of players} (n) \rule[-1ex]{0pt}{0pt} & \ 500\ & \ 1000 \ & 1500 \ & 2000 \ & 2500\ & 3000 \ \\
\hline
\rule{0pt}{2.5ex}\text{Cost per person} (C)\rule[-1ex]{0pt}{0pt} & 240 & 120 & 80 & 60 & 48\ & 40 \ \\
\hline
\end{array}
 

b. 

c.   `C = (40\ 000)/n`

`n\ text(must be a whole number)`
 

d.    `text(Limitations can include:)`

  `•\ n\ text(must be a whole number)`

  `•\ C > 0`
 

e.   `text(If)\ C = 94:`

`94` `= (120\ 000)/n`
`94n` `= 120\ 000`
`n` `= (120\ 000)/94`
  `= 1276.595…`

 
`:.\ text(C)text(ost cannot be $94 per person, because)\ n\ text(isn’t a whole number.)`

Show Worked Solution

a.   

\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\text{Number of players} (n) \rule[-1ex]{0pt}{0pt} & \ 50\ & \ 100 \ & 200 \ & 250 \ & 400\ & 500 \ \\
\hline
\rule{0pt}{2.5ex}\text{Cost per person} (C)\rule[-1ex]{0pt}{0pt} & 800 & 400 & 200 & 160 & 100\ & 80 \ \\
\hline
\end{array}

b.   
       
      

c.   `C = (40\ 000)/n`

 

TIP: Limitations require looking at possible restrictions of both the domain and range.

d.   `text(Limitations can include:)`

  `•\ n\ text(must be a whole number)`

  `•\ C > 0`
 

e.   `text(If)\ C = 120`

`120` `= (40\ 000)/n`
`120n` `= 40\ 000`
`n` `= (40\ 000)/120`
  `= 333.33..`

  
`:.\ text{Cost cannot be $120 per person because the required}\ n\ \text{is not a whole number.}`

v1 Algebra, STD2 A4 2022 HSC 24

A chef believes that the time it takes to defrost a turkey (`D` hours) varies inversely with the room temperature (`T^\circ \text{C}`). The chef observes that at a room temperature of `20^\circ \text{C}`, it takes 15 hours for the turkey to fully defrost.

  1. Find the equation relating `D` and `T`.   (2 marks)

  2. By first completing this table of values, graph the relationship between temperature and time.   (2 marks)

\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \ \ T\ \  \rule[-1ex]{0pt}{0pt} & \ \ \ 10\ \ \  & \ \ 15\ \ \  & \ \ \ 20\ \ \  & \ \ \ 25\ \ \ & \ \ \ 30\ \ \ \\
\hline
\rule{0pt}{2.5ex} \ \ D\ \  \rule[-1ex]{0pt}{0pt} & \ \ \ \  & \ \ \  & \ \ \ \  & \ \ \ \ & \ \ \ \\
\hline
\end{array}

 
           

Show Answers Only

a.   `D \prop 1/T\ \ =>\ \ D=k/T`

  `15` `=k/20`
  `k` `=15 xx 20=300`

 
`:.D=300/T`

b.   

\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \ \ T\ \  \rule[-1ex]{0pt}{0pt} & \ \ \ 10\ \ \  & \ \ 15\ \ \  & \ \ \ 20\ \ \  & \ \ \ 25\ \ \ & \ \ \ 30\ \ \ \\
\hline
\rule{0pt}{2.5ex} \ \ D\ \  \rule[-1ex]{0pt}{0pt} & \ \ \ 30\ \ \  & \ \ 20\ \ \  & \ \ \ 15\ \ \  & \ \ \ 12\ \ \ & \ \ \ 10\ \ \ \\
\hline
\end{array}

 

     

Show Worked Solution

a.   `D \prop 1/T\ \ =>\ \ D=k/T`

  `15` `=k/20`
  `k` `=15 xx 20=300`

 
`:.D=300/T`

b.   

\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \ \ D\ \  \rule[-1ex]{0pt}{0pt} & \ \ \ 10\ \ \  & \ \ 15\ \ \  & \ \ \ 20\ \ \  & \ \ \ 25\ \ \ & \ \ \ 30\ \ \ \\
\hline
\rule{0pt}{2.5ex} \ \ T\ \ \rule[-1ex]{0pt}{0pt} & 30 & 20 & 15 & 12 & 10  \\
\hline
\end{array}

 

     

v2 Functions, 2ADV F1 2017 HSC 1 MC

What is the gradient of the line  \(6x+7y-1 = 0\)?

  1. \(-\dfrac{6}{7}\)
  2. \(\dfrac{6}{7}\)
  3. \(-\dfrac{7}{6}\)
  4. \(\dfrac{7}{6}\)
Show Answers Only

\(A\)

Show Worked Solution
\(6x+7y-1\) \(=0\)  
\(7y\) \(=-6x+1\)  
\(y\) \(=-\dfrac{6}{7}x+\dfrac{1}{7}\)  

 
\(\Rightarrow A\)

v1 Functions, 2ADV F1 2009 HSC 1a

Sketch the graph of  \(y+\dfrac{x}{3} = 2\), showing the intercepts on both axes.   (2 marks)

Show Answers Only

Show Worked Solution

\(y+\dfrac{x}{3} = 2\ \ \Rightarrow\ \ y=-\dfrac{1}{3}x+2\)

\(y\text{-intercept}\ = 2\)

\(\text{Find}\ x\ \text{when}\ y=0:\)

\(\dfrac{x}{3}=2\ \ \Rightarrow\ \ x=6\)
 

v1 Functions, 2ADV F1 2017 HSC 1 MC

What is the gradient of the line \(4x-5y-2 = 0\)?

  1. \(-\dfrac{4}{5}\)
  2. \(\dfrac{4}{5}\)
  3. \(\dfrac{5}{4}\)
  4. \(-\dfrac{5}{4}\)
Show Answers Only

\(B\)

Show Worked Solution
\(4x-5y-2\) \(=0\)  
\(-5y\) \(=-4x + 2\)  
\(y\) \(=\dfrac{4}{5}x-\dfrac{2}{5}\)  

 
\(\Rightarrow B\)

Algebra, STD2 A1 2010 HSC 18 MC v1

Which of the following correctly express  \(h\)  as the subject of  \(A=\dfrac{bh}{2}\) ?

  1. \(h=\dfrac{A-2}{b}\)
  2. \(h=2A-b\)
  3. \(h=\dfrac{2A}{b}\)
  4. \(h=\dfrac{Ab}{2}\)
Show Answers Only

\(C\)

Show Worked Solution
\(A\) \(=\dfrac{bh}{2}\)
\(bh\) \(=2A\)
\(\therefore\ h\) \(=\dfrac{2A}{b}\)

 
\(\Rightarrow C\)

Special Properties, SMB-031

In the diagram, `ABCDE` is a regular pentagon. The diagonals `AC` and `BD` intersect at `F`.

  1. Show that the size of `/_ABC` is 108°.  (1 mark)

  2. Find the size of `/_BAC`. Give reasons for your answer.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `36°`
Show Worked Solution
i.    

`text(Sum of all internal angles)`

`= (n-2) xx 180°`

`= (5-2) xx 180°`

`= 540°`
 

`:. /_ABC= 540/5= 108°`
 

ii.  `BA = BC\ \ text{(equal sides of a regular pentagon)}`

`:. Delta BAC\ text(is isosceles)`

`/_BAC= 1/2 (180-108)=36^{\circ} \ \ \ text{(base angle of}\ Delta BAC text{)}`

Congruency, SMB-016

Igor was designing a shield using 10 congruent (isosceles) triangles, as shown in the diagram below.
 

How many degrees in the angle marked `x`?   (3 marks)

Show Answers Only

`72^@`

Show Worked Solution

`text(Angles at centre of circle)\ = 360/10 = 36^@`

`text{Since triangles are isosceles:}`

`180` `= 36 + 2x`
`2x` `= 180-36`
  `= 144 `
`:. x` `= 72^@`

Special Properties, SMB-020 MC

`AB` is the diameter of a circle, centre `O`.

There are 3 triangles drawn in the lower semi-circle and the angles at the centre are all equal to `x^@`.

The three triangles are best described as:

  1. isosceles
  2. scalene
  3. right-angled
  4. equilateral
Show Answers Only

`D`

Show Worked Solution

`3x=180^{\circ}\ \=>\ x=60^{\circ} `

`AO=OC=OD=OB\ \ text{(radii of circle)}`

`=>\ text{Since angles opposite equal sides of a triangle are}`

`text(equal, all triangle angles can be found to equal 60°.)`

`:.\ text(The three triangles are equilateral.)`

`=>D`

Special Properties, SMB-019

A regular pentagon, a square and an equilateral triangle meet at a point.
 

 
What is the size of the angle `x°`?   (3 marks)

Show Answers Only

`102°`

Show Worked Solution

`text(Sum of internal angles of pentagon)`

`= (n-2) xx 180`

`= 3 xx 180`

`= 540^@`
 

`text(Internal angle in pentagon)`

`= 540/5`

`= 108^@`
 

`:. x` `= 360-(108 + 90 + 60)`
  `= 102^@`

Special Properties, SMB-018

A six sided figure is drawn below.
  

What is the sum of the six interior angles?   (2 marks)

Show Answers Only

`720^@`

Show Worked Solution

`\text{Method 1}`

`text(Reflex angle) = 360-90 = 270^@`

`:.\ text(Sum of interior angles)`

`= (270 xx 2) + (30 xx 2) + (60 xx 2)`

`= 720^@`
 

`\text{Method 2}`

`text{Sum of interior angles (formula)}`

`= (n-2) xx 180`

`=4 xx 180`

`= 720^@`

Special Properties, SMB-016

Two identical quadrilaterals fit together to make this regular pentagon.
 

What is the value of `x`?   (2 marks)

Show Answers Only

`108^@`

Show Worked Solution

`text(Consider regular pentagon:)`

`text{Sum of internal angles (formula)}`

`= (n-2) xx 180`

`= 3 xx 180`

`= 540^@`
 

`:. x` `= 540/5`
  `= 108^@`
TIP: Two quadrilaterals joining does not make the internal angle sum = 2 × 360°!

Special Properties, SMB-015

A star is drawn on the inside of a regular pentagon, as shown below.
 

What is the size of the angle marked `x`?   (3 marks)

Show Answers Only

`36^@`

Show Worked Solution

`text(Consider the triangle)\ ABC\ \ text(in the pentagon:)`

STRATEGY: The internal angle sum is  3 × 180 = 540 (since 3 triangles can be drawn internally from one point).

`text(Total degrees in a pentagon)`

`= 3 xx 180`

`= 540^@`

 
`text{Internal angle}\ =540/5 = 108^@\ \ \text{(regular pentagon)}`
  
`DeltaABC\ text(is isosceles)`

`:. x + x + 108` `= 180`
`2x` `= 72`
`x` `= 36^@`

Special Properties, SMB-014

The sum of the interior angles of a 6 sided polygon can be found by first dividing it into triangles from one vertex.
 

 

What is the sum of the interior angles of this polygon?   (2 marks)

Show Answers Only

`720\ text(degrees)`

Show Worked Solution

`text{Since the polygon can be divided into 4 separate triangles:}`

`text(Sum of interior angles)`

`= 4 xx 180`

`= 720\ text(degrees)`

Special Properties, SMB-013 MC

Which statement is always true?

  1. Scalene triangles have two angles that are equal.
  2. All angles in a parallelogram are equal.
  3. The opposite sides of a trapezium are equal in length.
  4. The diagonals of a rhombus are perpendicular to each other.
Show Answers Only

`D`

Show Worked Solution

`text{Consider each option:}`

`A:\ \text{Isosceles (not scalene) have two equal angles.}`

`B:\ \text{Only opposite angles in a parallelogram are equal.}`

`C:\ \text{At least one pair of opposite sides of a trapezium are not equal.}`

`D:\ \text{Rhombuses have perpendicular diagonals.}`

`=>D`

Special Properties, SMB-010 MC

Which of these are always equal in length?

  1. the diagonals of a rhombus
  2. the diagonals of a parallelogram
  3. the opposite sides of a parallelogram
  4. the opposite sides of a trapezium
Show Answers Only

`C`

Show Worked Solution

`text{Consider each option:}`

`A:\ \text{rhombus diagonals are perpendicular but not always equal}`

`B:\ \text{parallelogram diagonals not always equal (see below)}`

`C:\ \text{always true (see above)}`

`D:\ \text{at least 1 pair of opposite sides of a trapezium are not equal}`
\(\Rightarrow C\)

Special Properties, SMB-009 MC

`PQRS` is a parallelogram.

Which of these must be a property of `PQRS`?

  1. Line `PS` is perpendicular to line `PQ`.
  2. Line `PQ` is parallel to line `PS`.
  3. Diagonals `PR` and `SQ` are perpendicular.
  4. Line `PS` is parallel to line `QR`.
Show Answers Only

`D`

Show Worked Solution

`text{By elimination:}`

`A\ \text{and}\ B\ \text{clearly incorrect.}`

`C\ \text{true if all sides are equal (rhombus) but not true for all parallelograms.}`

`text(Line)\ PS\ text(must be parallel to line)\ QR.`

`=>D`

Special Properties, SMB-008

The sum of the internal angles of a polygon can be calculated by drawing triangles from any given vertex as shown below. 
 

What is the size of the angle marked `x` in the diagram below?   (2 marks)

Show Answers Only

`107°`

Show Worked Solution

`text{Since the quadrilateral was divided into two triangles}`

`=>\ \text{Sum of internal angles}\ = 2 xx 180 = 360^{\circ}`

`:. x` `= 360-(103 + 88 + 62)`
  `= 360-253`
  `= 107°`

Congruency, SMB-013

Which two of the triangles below are congruent?   (2 marks)

Show Answers Only

\(\text{Triangle B and Triangle C}\)

Show Worked Solution

\(\text{Unknown side}\ (x)\ \text{in Triangle B (Pythagoras):}\)

\(x=\sqrt{11^2-(\sqrt{96})^2} = \sqrt{25} = 5\)

\(\Rightarrow\ \text{Triangle B and Triangle C are congruent (SSS)} \)

\(\text{Triangle A can be shown to have different dimensions but this is not necessary.}\)

Congruency, SMB-012

Which two of the triangles below are congruent?   (2 marks)

Show Answers Only

\(\text{Triangle A and Triangle C}\)

Show Worked Solution

\(\text{Unknown side}\ (x)\ \text{in Triangle A (Pythagoras):}\)

\(x=\sqrt{8^2-(\sqrt{48})^2} = \sqrt{16} = 4\)

\(\Rightarrow\ \text{Triangle A and Triangle C are congruent (SSS)} \)

\(\text{Triangle B can be shown to have different dimensions but this is not necessary.}\)

Congruency, SMB-015

The two triangles below are congruent.

  1. Which congruency test would be used to prove the two triangles above are congruent?   (1 mark)

  2. Find the values of  `a` and `b`.  (2 marks)

Show Answers Only
  1. `text{(SAS)}`
  2. `a=12.1, \ b=7.4`
Show Worked Solution

i.    `text{Congruency test}\ => \ text{SAS} `
 

ii.   `text{Matching corresponding sides:}`

`a=12.1, \ b=7.4`

Congruence, SMB-014

The area of the rectangle in the diagram below is 15 cm2.
 

Giving reasons, find the area of the trapezium `ABCD`.   (4 marks)

Show Answers Only

`30\ text(cm)^2`

Show Worked Solution

`text{Opposite sides of rectangles are equal and parallel}`

`\Delta AEI ≡ \Delta DHJ\ \ text{(RHS)}`

`/_ EAI = /_ BEF\ \ text{(corresponding,}\ AD\ text{||}\ EH )`

`\Delta AEI ≡ \Delta EBF\ \ text{(AAS)}`

`text{Similarly,}\ \Delta DHJ ≡ \Delta HCG\ \ text{(AAS)}`

`=>\ \text{All triangles in diagram are congruent.}`

`text(Rearranging the diagram:)`

`:.\ text(Area of trapezium)`

`= 2 xx 15`

`= 30\ text(cm)^2`

Special Properties, SMB-004 MC

Which one of the following triangles is impossible to draw?

  1. a right angled triangle with two acute angles
  2. an isosceles triangle with one right angle
  3. a scalene triangle with three acute angles
  4. a right angled triangle with one obtuse angle
Show Answers Only

`D`

Show Worked Solution

`text(A right angle = 90°.)`

`text{Since an obtuse angle is greater than 90°, it is impossible for}`

`text(a triangle, with an angle sum less than 180°, to have both.)`

`=>D`

Special Properties, SMB-003 MC

Select the statement that is true about triangle `ABC`.

  1. Triangle `ABC` is a scalene triangle.
  2. Triangle `ABC` has exactly 2 equal sides.
  3. Triangle `ABC` is an equilateral triangle.
  4. Triangle `ABC` is an obtuse triangle.
Show Answers Only

`C`

Show Worked Solution

`angleA = 180-(60 + 60) = 60^@`

`:. text(All angles are)\ 60^@.`

`:. text(Triangle)\ ABC\ text(is an equilateral.)`

`=>C`

Special Properties, SMB-002 MC

A triangle has two acute angles.

What type of angle couldn't the third angle be?

  1. an acute angle
  2. an obtuse angle
  3. a right-angle
  4. a reflex angle
Show Answers Only

`D`

Show Worked Solution

`text(A triangle’s angles add up to 180°, and a reflex angle is)`

`text(greater than 180°.)`

`:.\ text(The third angle cannot be reflex.)`

`=>D`

Special Properties, SMB-001 MC

Which of the following triangle types is impossible to draw?

  1. a right-angled, scalene triangle
  2. a right-angled, equilateral triangle
  3. an obtuse-angled, isosceles triangle
  4. an acute-angled, scalene triangle
Show Answers Only

`B`

Show Worked Solution

`text(An equilateral triangle has all angles = 60°.)`

`:.\ text(A right-angled, equilateral triangle is impossible.)`

`=>B`

Congruency, SMB-011

The diagram shows a circle with centre `O` and radius 5 cm.

The length of the arc `PQ` is 9 cm. Lines drawn perpendicular to `OP` and `OQ` at `P` and `Q` respectively meet at  `T`.
 

Prove that `Delta OPT` is congruent to `Delta OQT`.   (2 marks)

Show Answers Only

`text(Proof)  text{(See Worked Solutions)}`

Show Worked Solution

`text(Prove)\ Delta OPT ≡ Delta OQT`

`OT\ text(is common)`

COMMENT: Know the difference between the congruency proof of `RHS` and `SAS`. Incorrect identification will lose a mark.

`/_OPT = /_OQT = 90°\ \ text{(given)}`

`OP = OQ\ \ \ text{(radii)}`

`:.\ Delta OPT ≡ Delta OQT\ \ \ text{(RHS)}`

Congruency, SMB-010

In the diagram, `AD` is parallel to `BC`, `AC` bisects `/_BAD` and `BD` bisects `/_ABC`. The lines `AC` and `BD` intersect at `P`.

  1. Prove that `/_BAC = /_BCA`.  (1 mark)

  2. Prove that `Delta ABP ≡ Delta CBP`.  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
i.  

`text(Prove)\ /_BAC = /_BCA`

`/_BCA` `= /_CAD\ \ \ text{(alternate angles,}\ BC\ text(||)\ AD text{)}`
`/_CAD` `= /_BAC\ \ \ text{(}AC\ text(bisects)\ /_BAD text{)}`
`:. /_BAC` `= /_BCA\ …\ text(as required)`

 

ii.  `text(Prove)\ Delta ABP ≡ Delta CBP`

`/_BAC` `= /_BCA\ \ \ text{(from part (i))}`
`/_ABP` `= /_CBP\ text{(}BD\ text(bisects)\ /_ABC text{)}`
`BP\ text(is common)`

 

`:. Delta ABP ≡ Delta CBP\ \ text{(AAS)}`

Congruency, SMB-008

In the figure below, \(ABCD\) is a parallelogram where opposite sides of the quadrilateral are equal.
 

Prove that a diagonal of the parallelogram produces two triangles that are congruent.  (2 marks)

Show Answers Only

\(\text{Proof (See Worked Solutions)}\)

Show Worked Solution

\(\text{One of multiple solutions:}\)

\( AB=CD\ \ \text{and}\ \ AC=BD\ \ \text{(given)} \)

\(BC\ \text{is common} \) 

\(\therefore\ \Delta ABC \equiv \Delta DCB\ \ \text{(SSS)}\)

Congruency, SMB-007

In the figure below, \(AB \parallel DE, \ AC = CE\)  and the line \(AE\) intersects \(DB\) at \(C\).
 

Prove that this pair of triangles are congruent.  (2 marks)

Show Answers Only

\(\text{Proof (See Worked Solutions)}\)

Show Worked Solution

\( \angle ACB = \angle DCE\ \ \text{(vertically opposite)} \)

\(AC = CE\ \ \text{(given)} \)

\(\angle BAC = \angle DEC\ \ \text{(alternate,}\ AB \parallel DE \text{)} \)
 

\(\therefore\ \Delta ABC \equiv \Delta EDC\ \ \text{(AAS)}\)

Congruency, SMB-006

In the quadrilateral \(ABCD\), \(AB \parallel CD, \angle BAD = \angle BCD\)  and  \(\angle DBC = \angle BDA = 90^{\circ} \).
 

Prove that this pair of triangles are congruent.  (2 marks)

Show Answers Only

\(\text{Proof (See Worked Solutions)}\)

Show Worked Solution

\(\angle BAD = \angle BCD\ \ \text{(given)} \) 

\(\angle DBC = \angle BDA = 90^{\circ} \ \ \text{(given)} \)

\(BD\ \text{is common} \) 
 

\(\therefore\ \Delta BAD \equiv \Delta DCB\ \ \text{(AAS)}\)

Congruency, SMB-005

In the figure below, \(BE = BC\), \(AB = BD\) and the line \(AD\) intersects \(CE\) at \(B\).
 

Prove that this pair of triangles are congruent.  (2 marks)

Show Answers Only

\(\text{Proof (See Worked Solutions)}\)

Show Worked Solution

\( \angle ABE = \angle CBD\ \ \text{(vertically opposite)} \)

\(AB = BD\ \ \text{(given)} \)

\(EB = BC\ \ \text{(given)} \)
 

\(\therefore\ \Delta ABE \equiv \Delta DBC\ \ \text{(SAS)}\)

Congruency, SMB-004

In the circle below, centre \(O\), \(OB\) is perpendicular to chord \(AC\).
 

Prove that a pair of triangles in this figure are congruent.  (2 marks)

Show Answers Only

\(\text{Proof (See Worked Solutions)}\)

Show Worked Solution

\(\text{One of multiple proofs:}\)

\(OB\ \ \text{common} \)

\( \angle OBA = \angle OBC = 90^{\circ}\ \ \text{(given)} \)

\(OA = OC\ \ \text{(radii)} \)
 

\(\therefore\ \Delta AOB \equiv \Delta COB\ \ \text{(RHS)}\)

Congruency, SMB-003

In the figure below, the line \(AD\) intersects \(BE\) at \(C\), \(BC = CD\) and \(AC = EC\).
 

Prove that this pair of triangles are congruent.  (2 marks)

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\(\text{Proof (See Worked Solutions)}\)

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\( \angle BCE = \angle DCE\ \ \text{(vertically opposite)} \)

\(BC = CD\ \ \text{(given)} \)

\(AC = EC\ \ \text{(given)} \)
 

\(\therefore\ \Delta ABC \equiv \Delta EDC\ \ \text{(SAS)}\)

Congruency, SMB-002

The diagram shows two triangles that touch at the middle of a circle.

Prove that this pair of triangles are congruent.  (2 marks)

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\(\text{Proof (See Worked Solutions)}\)

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\(\text{Base of each triangle (chords) are equal} \)

\(\text{All other sides are equal radii of the circle} \)

\(\therefore\ \text{Two given triangles are congruent}\ \ \text{(SSS)}\)

Congruency, SMB-001

The diagram shows two right-angled triangles where \(\angle BAC = \angle BDC =  90^{\circ}\), and  \(AB = BD\). 

Prove that this pair of triangles are congruent.  (2 marks)

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\(\text{Proof (See Worked Solutions)}\)

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\(BC\ \text{(hypotenuse) is common} \)

\(BA = BD\ \text{(given)} \)

\(\angle BAC = \angle BDC =  90^{\circ}\ \ \text{(given)} \)

\(\therefore \Delta ABC \equiv \Delta DBC\ \ \text{(RHS)}\)

Similarity, SMB-026

The two triangles below are similar.
 

Find the length of \(ED\).   (3 marks)

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\(ED = 15\)

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\(\text{Using Pythagoras in}\ \Delta ABC: \)

\( AB=\sqrt{13^2-12^2}=\sqrt{25}=5 \)

\(\text{Scale factor}\ = \dfrac{FD}{AC} = \dfrac{39}{13} = 3 \)

\(\therefore ED = 3 \times AB = 3 \times 5 = 15 \)

Similarity, SMB-024

Prove that the two triangles in the right cone pictured below are similar.   (2 marks)
 

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\(\text{Proof (See Worked Solution)}\)

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\(\angle ACD \ \text{is common to}\ \Delta ACE\ \text{and}\ \Delta BCD \)

\(\angle CAE=\angle CBD=90^\circ \ \text{(Right cone)} \)

\(\therefore \Delta ACE\ \text{|||}\ \Delta BCD\ \ \text{(equiangular)}\)

Similarity, SMB-023

Show that \(\Delta ACD\) and \(\Delta DCB\) in the figure below are similar.   (2 marks)
 

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\(\text{Proof (See Worked Solution)}\)

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\(\angle BDC = 180-(88+44) = 48^{\circ} \ \text{(angle sum of Δ)} \)

\(\angle CAD = \angle CDB = 48^{\circ} \)

\(\angle ACD=44^{\circ} \ \text{is common to}\ \Delta ACD\ \text{and}\ \Delta BCD \)

\(\therefore \Delta ACD\ \text{|||}\ \Delta DCB\ \ \text{(equiangular)}\)

Similarity, SMB-022

Prove that this pair of triangles are similar.   (2 marks)

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\(\text{Proof (See Worked Solution)}\)

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\(\angle PQT = \angle SQR \ \text{(vertically opposite)} \)

\(\dfrac{PQ}{QS} = \dfrac{2.5}{5} = \dfrac{1}{2} \)

\(\dfrac{TQ}{QR} = \dfrac{2}{4} = \dfrac{1}{2} \)

\(\therefore \Delta PQT\ \text{|||}\ \Delta SQR\ \ \text{(sides adjacent to equal angles in proportion)}\)

Similarity, SMB-021

Prove that this pair of triangles are similar.   (2 marks)
 

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\(\text{Proof (See Worked Solution)}\)

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\(\angle ACD = \angle BEA \ \text{(given)} \)

\(\angle BAE\ \text{is common to}\ \Delta ACD\ \text{and}\ \Delta BEA \)

\(\therefore \Delta ACD\ \text{|||}\ \Delta BEA\ \ \text{(equiangular)}\)

Similarity, SMB-020

Prove that this pair of triangles are similar.   (3 marks)
 

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\(\text{Proof (See Worked Solution)}\)

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\(\text{Find unknown side}\ (x)\ \text{of smaller triangle}\)

\(\text{Using Pythagoras:}\)

\(x=\sqrt{5^2-4^2} = \sqrt{9} = 3\)
 

\(\dfrac{AC}{DE} = \dfrac{5}{15} = \dfrac{1}{3} \)

\(\dfrac{BC}{EF} = \dfrac{3}{9} = \dfrac{1}{3} \)

\(\angle ABC = \angle DEF = 90^{\circ} \ \ \text{(given)} \)

\(\therefore \Delta ABC\ \text{|||}\ \Delta DEF\ \ \text{(hypotenuse and second side of right-angled triangle in proportion)}\)

Similarity, SMB-019

Prove that this pair of triangles are similar.   (2 marks)
 

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\(\text{Proof (See Worked Solution)}\)

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\(\dfrac{AB}{GH} = \dfrac{11}{22} = \dfrac{1}{2} \)

\(\dfrac{BC}{FG} = \dfrac{3.5}{7} = \dfrac{1}{2} \)

\(\angle ABC = \angle FGH = 95^{\circ} \ \ \text{(given)} \)
 

\(\therefore \Delta ABC\ \text{|||}\ \Delta HGF\ \ \text{(sides adjacent to equal angles in proportion)}\)

Similarity, SMB-018

Prove that this pair of triangles are similar.   (2 marks)
 

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\(\text{Proof (See Worked Solution)}\)

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\(\angle ABE = \angle CBD\ \ \text{(vertically opposite)} \)

\(\angle AEB = \angle BCD\ \ \text{(alternate,}\ AE \parallel CD \text{)} \)

\(\therefore \Delta ABE\ \text{|||}\ \Delta DBC\ \ \text{(equiangular)}\)

Similarity, SMB-017

Prove that this pair of triangles are similar.   (2 marks)
 

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\(\text{Proof (See Worked Solution)}\)

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\(\dfrac{RP}{AC} = \dfrac{18}{12} = \dfrac{3}{2} \)

\(\dfrac{QR}{BA} = \dfrac{12}{8} = \dfrac{3}{2} \)

\(\dfrac{PQ}{CB} = \dfrac{9}{6} = \dfrac{3}{2} \)
 

\(\therefore \Delta PQR\ \text{|||}\ \Delta CBA\ \ \text{(three pairs of sides in proportion)}\)

Congruency, SMB-009

 

The diagram shows a right-angled triangle `ABC` with `∠ABC = 90^@`. The point `M` is the midpoint of `AC`, and `Y` is the point where the perpendicular to `AC` at `M` meets `BC`.

Show that `\Delta AYM \equiv \Delta CYM`.  (2 marks)

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`text(Proof)\ \ text{(See Worked Solutions)}`

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`text(In)\ ΔAYM\ text(and)\ ΔCYM`

`∠AMY` `= ∠CMY = 90^@\ \ \ (MY ⊥ AC)`
`AM` `=CM\ \ \ text{(given)}`
`YM\ text(is common)`

 
`:. \Delta AYM \equiv \Delta CYM\ \ text{(SAS)}`

Trigonometry, SMB-067

Find the value of \(\theta\), correct to the nearest minute.   (3 marks)

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\(\theta=112^{\circ} 12^{′}\)

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\(\cos A\) \(= \dfrac{b^2+c^2-a^2}{2bc} \)  
\(\cos \theta\) \(= \dfrac{7.8^2+10.2^2-15.0^2}{2 \times 7.8 \times 10.2}\)  
  \(= -0.3778…\)  
\(\therefore \theta\) \(= \cos^{-1} (-0.3778…) \)  
  \(=112.199…^{\circ} \)  
  \(=112^{\circ} 12^{′}\ \ \text{(nearest minute)} \)  

Trigonometry, SMB-066

Find the value of \(\theta\), correct to the nearest degree.   (2 marks)

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\(\theta=35^{\circ}\)

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\(\cos A\) \(= \dfrac{b^2+c^2-a^2}{2bc} \)  
\(\cos \theta\) \(= \dfrac{16.2^2+18.1^2-10.5^2}{2 \times 16.2 \times 18.1}\)  
  \(= 0.818…\)  
\(\therefore \theta\) \(= \cos^{-1} (0.818…) \)  
  \(=35.09…^{\circ} \)  
  \(=35^{\circ}\ \ \text{(nearest degree)} \)  

Trigonometry, SMB-065

Find the value of \(\theta\), correct to 1 decimal place.   (2 marks)

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\(\theta=38.9^{\circ}\)

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\(\cos A\) \(= \dfrac{b^2+c^2-a^2}{2bc} \)  
\(\cos \theta\) \(= \dfrac{9^2+5^2-6^2}{2 \times 9 \times 5}\)  
  \(= 0.777…\)  
\(\therefore \theta\) \(= \cos^{-1} (0.777…) \)  
  \(=38.94…^{\circ} \)  
  \(=38.9^{\circ}\ \ \text{(1 d.p.)} \)  

Trigonometry, SMB-064

Find the value of \(d\), correct to 2 decimal places.   (3 marks)

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\(d=12.9\ \text{km}\)

Show Worked Solution
\(a^2\) \(=b^2+c^2-2bc\ \cos A \)  
\(d^2\) \(= 8.9^2+5.2^2-2 \times 8.9 \times 5.2 \times \cos 130^{\circ}\)  
  \(= 165.746…\)  
\(\therefore d\) \(=12.87…\)  
  \(=12.9\ \text{km (1 d.p.)} \)