SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

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^@`

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`

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)

Show Answers Only

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

Show Worked Solution

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

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

\(AC = EC\ \ \text{(given)} \)
 

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

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)

Show Answers Only

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

Show Worked Solution

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

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

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

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)

Show Answers Only

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

Show Worked Solution

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

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

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

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

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)

Show Answers Only

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

Show Worked Solution

`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)}`

Plane Geometry, 2UA 2018 HSC 12c

The diagram shows the square `ABCD`. The point `E` is chosen on `BC` and the point `F` is chosen on `CD` so that  `EC = FC`.
 

  1. Prove that `Delta ADF` is congruent to `Delta ABE`.   (2 marks)

  2. The side length of the square is 14 cm and `EC` has length 4 cm. Find the area of  `AECF`.   (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `56\ text(cm)^2`
Show Worked Solution

i.    `AB = AD\ \ text{(sides of a square)}`

`DF = DC-CF`

`BE = BC-CE`

`text{Since}\ CE = CF\ \ text{(given), and}\ DC = BC\ \ text{(sides of a square)}`

`=> DF = BE`

`=> /_ ADF = /_ ABE = 90^@`

`:. Delta ADF \equiv Delta ABE\ \ text{(SAS)}`

 

ii.   `text(Area of)\ Delta ABE` `= 1/2 xx 14 xx 10`
    `= 70\ text(cm)^2`

 
`:.\ text(Area of)\ AECF`

`= text(Area of)\ ABCD-(2 xx 70)`

`= (14 xx 14)-140`

`= 56\ text(cm)^2`

Plane Geometry, 2UA 2017 HSC 15a

The triangle `ABC` is isosceles with  `AB = AC`  and the size of `/_BAC` is `x^@`.

Points `D` and `E` are chosen so that `Delta ABC, Delta ACD` and `Delta ADE` are congruent, as shown in the diagram.
 

Find the value of `x` for which `AB` is parallel to `ED`, giving reasons.  (3 marks)

Show Answers Only

`x = 36`

Show Worked Solution

`text(All base angles) = 1/2(180-x)\ \ text{(Angle sum of}\ Delta text{)}`

`text(If)\ \ AB\ text(||)\ ED,`

`/_ BAD` `= /_ ADE\ \ \ text{(alternate angles)}`
`2x` `= 1/2 (180-x)`
`4x` `= 180-x`
`5x` `= 180`
`x` `= 36^{\circ}`