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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)

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`30\ text(cm)^2`

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`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`

Filed Under: Congruency Tagged With: num-title-ct-pathc, smc-4747-35-RHS

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)

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

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`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)}`

Filed Under: Congruency Tagged With: num-title-ct-pathc, smc-4747-35-RHS

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)

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

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

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

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

\(OA = OC\ \ \text{(radii)} \)
 

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

Filed Under: Congruency Tagged With: num-title-ct-pathc, smc-4747-35-RHS

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

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

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

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

Filed Under: Congruency Tagged With: num-title-ct-pathc, smc-4747-35-RHS

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