Year 9 Geometry: free practice test with answers

Sample geometry questions with answers

Try each one before you open the answer.

1. Angles and triangles

   In triangle $ABC$, $\angle BAC = 48^\circ$. Side $BC$ is extended past $C$ to a point $D$, and the exterior angle $\angle ACD = 125^\circ$. Find $\angle ABC$ and $\angle ACB$, giving reasons.

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   Key idea: An exterior angle of a triangle equals the sum of the two opposite interior angles. Angles on a straight line add to $180^\circ$.

   1. Exterior angle: $\angle ACD = \angle BAC + \angle ABC$, so $125 = 48 + \angle ABC$ and $\angle ABC = 77^\circ$
   2. Angles on a straight line: $\angle ACB = 180 - 125 = 55^\circ$
   3. Check the angle sum: $48 + 77 + 55 = 180$ ✓

   Answer: $\angle ABC = 77^\circ$, $\angle ACB = 55^\circ$

2. Parallel lines

   Lines $AB$ and $CD$ are parallel and horizontal, with $AB$ above $CD$, and $A$ and $C$ on the left. A transversal $EF$ crosses $AB$ at $G$ and $CD$ at $H$, with $E$ above $AB$ and $F$ below $CD$. The angle $\angle EGB = 65^\circ$. Find each angle, giving a reason.

   (a) $\angle GHD$ (b) $\angle AGH$ (c) $\angle BGH$ (d) $\angle CHG$

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   Key idea: When a transversal crosses parallel lines, corresponding angles are equal, alternate angles are equal and co-interior angles add to $180^\circ$.

   1. (a) $\angle GHD$ and $\angle EGB$ are corresponding angles (both above their line and to the right of the transversal), so $\angle GHD = 65^\circ$
   2. (b) $\angle AGH$ and $\angle EGB$ are vertically opposite, so $\angle AGH = 65^\circ$
   3. (c) $\angle BGH$ and $\angle EGB$ are on a straight line, so $\angle BGH = 180 - 65 = 115^\circ$
   4. (d) $\angle CHG$ and $\angle BGH$ are alternate angles, so $\angle CHG = 115^\circ$ (check: $\angle CHG$ and $\angle GHD$ make a straight line, $115 + 65 = 180$)

   Answer: (a) $65^\circ$ (b) $65^\circ$ (c) $115^\circ$ (d) $115^\circ$

3. Quadrilaterals and other polygons

   Find the angle sum of a hexagon.

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   Key idea: Angle sum of an $n$-sided polygon $= (n - 2) \times 180^\circ$.

   1. A hexagon has $6$ sides
   2. $(6 - 2) \times 180 = 720^\circ$

   Answer: $720^\circ$

4. Congruent triangles

   State the congruence test (SSS, SAS, AAS or RHS) that proves each pair of triangles congruent.

   (a) All three sides of one triangle are equal to the three sides of the other.

   (b) Two sides and the angle between them are equal.

   (c) Two angles and a side in the same position are equal.

   (d) Both triangles have a right angle, and the hypotenuse and one other side are equal.

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   Key idea: The four tests for congruent triangles are SSS, SAS, AAS and RHS. In SAS the angle must be the one between the two sides.

   1. (a) Three pairs of equal sides: SSS
   2. (b) Two sides and the included angle: SAS
   3. (c) Two angles and a matching side: AAS
   4. (d) Right angle, hypotenuse, side: RHS

   Answer: (a) SSS (b) SAS (c) AAS (d) RHS

5. Using congruence in proof

   $ABCD$ is a parallelogram, so $AB \parallel DC$ and $AD \parallel BC$. The diagonal $AC$ is drawn. Prove that $\triangle ABC \equiv \triangle CDA$.

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   Key idea: Parallel sides give pairs of equal alternate angles, and the diagonal is a side shared by both triangles.

   1. $\angle BAC = \angle DCA$ (alternate angles, $AB \parallel DC$)
   2. $\angle BCA = \angle DAC$ (alternate angles, $AD \parallel BC$)
   3. $AC$ is common to both triangles
   4. So $\triangle ABC \equiv \triangle CDA$ (AAS)

   Answer: $AC$ is common, $\angle BAC = \angle DCA$ and $\angle BCA = \angle DAC$ (alternate angles), so $\triangle ABC \equiv \triangle CDA$ (AAS)

6. Enlargement and similar figures

   Triangle $ABC$ is enlarged by a scale factor of $3$ to produce triangle $A'B'C'$. Triangle $ABC$ has side lengths $4$ cm, $6$ cm and $9$ cm. Find the side lengths of triangle $A'B'C'$.

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   Key idea: Under an enlargement with scale factor $k$, every side length of the image is $k$ times the corresponding side length of the original shape.

   1. Multiply each side length by the scale factor $k = 3$:\

   $$\begin{gathered}4 \times 3 = 12 \text{ cm} \\ 6 \times 3 = 18 \text{ cm} \\ 9 \times 3 = 27 \text{ cm}\end{gathered}$$

   2. The side lengths of triangle $A'B'C'$ are $12$ cm, $18$ cm and $27$ cm.

   Answer: $12$ cm, $18$ cm and $27$ cm.

7. Similar triangles

   Two right-angled triangles are similar. In the smaller triangle, the two shorter sides are $6$ cm and $8$ cm. The longer of those two sides in the larger triangle is $20$ cm. Find the hypotenuse of the larger triangle.

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   Key idea: Use Pythagoras to find the hypotenuse of the smaller triangle, then use the similarity ratio to scale it up.

   1. Find the hypotenuse of the smaller triangle:\

   $$c^2 = 6^2 + 8^2 = 36 + 64 = 100 \implies c = 10 \text{ cm}$$

   2. The side of length $8$ cm in the smaller triangle corresponds to the side of length $20$ cm in the larger triangle. Find the scale factor:\

   $$k = \frac{20}{8} = 2.5$$

   3. Multiply the smaller hypotenuse by the scale factor:\

   $$\text{Larger hypotenuse} = 10 \times 2.5 = 25 \text{ cm}$$

   Answer: The hypotenuse of the larger triangle is $25$ cm.

8. Proving and applying similar triangles

   In $\triangle ABC$, $\angle A = 50^\circ$ and $\angle B = 60^\circ$. In $\triangle DEF$, $\angle D = 50^\circ$ and $\angle E = 60^\circ$. Are the triangles similar? State the test and write a similarity statement.

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   Key idea: If two angles of one triangle equal two angles of another, the third angles are also equal, so the triangles are similar (AAA).

   1. $\angle C = 180 - 50 - 60 = 70^\circ$ and $\angle F = 180 - 50 - 60 = 70^\circ$
   2. All three pairs of angles are equal, so $\triangle ABC \sim \triangle DEF$ (AAA)

   Answer: Yes, by AAA; $\triangle ABC \sim \triangle DEF$

Free worksheets for this chapter

Worked examples and a free PDF with answers for each topic.

• Year 9 pythagoras (applied) worksheet
• Year 9 direct proportion, rates and scale worksheet
• Year 9 enlargements and scale factors worksheet