SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

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`

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-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-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`