Year 10 and 10A Geometry and networks: free practice test with answers

Sample geometry and networks questions with answers

Try each one before you open the answer.

1. Review of geometry

   Find the size of each exterior angle of a regular octagon.

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   Key idea: Exterior angles of a polygon add to $360^\circ$.

   1. The exterior angles of any polygon add to $360^\circ$.
   2. A regular octagon has 8 equal exterior angles: $360^\circ \div 8 = 45^\circ$.

   Answer: $45^\circ$

2. Congruent triangles

   Triangle $ABC$ has $AB = 5$ cm, $BC = 7$ cm and $\angle ABC = 48°$. Triangle $DEF$ has $DE = 5$ cm, $EF = 7$ cm and $\angle DEF = 48°$. Prove that $\triangle ABC \cong \triangle DEF$.

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   Key idea: Two triangles are congruent by SAS if two pairs of corresponding sides and the included angle are equal.

   1. $AB = DE = 5$ cm (given)
   2. $BC = EF = 7$ cm (given)
   3. $\angle ABC = \angle DEF = 48°$ (given), and this angle is included between the two known sides in each triangle.
   4. Therefore $\triangle ABC \cong \triangle DEF$ (SAS).

   Answer: $\triangle ABC \cong \triangle DEF$ by SAS.

3. Similar figures

   A garden shed is modelled by a rectangular prism with length 3 m, width 2 m and height 2.5 m. A scale model of the shed is built using a scale factor of $\dfrac{1}{20}$ (that is, every length on the model is $\dfrac{1}{20}$ of the real length). Find the surface area and volume of the model, and compare them with the real shed's surface area and volume by stating the scale factors for area and volume.

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   Key idea: When lengths scale by a factor of $s$, areas scale by $s^2$ and volumes scale by $s^3$.

   1. Find the real shed's surface area. The six faces of the rectangular prism have areas: two $3 \times 2.5 = 7.5 \text{ m}^2$ faces, two $2 \times 2.5 = 5 \text{ m}^2$ faces, and two $3 \times 2 = 6 \text{ m}^2$ faces.\

   $$SA_{\text{real}} = 2(7.5) + 2(5) + 2(6) = 15 + 10 + 12 = 37 \text{ m}^2$$

   2. Find the real shed's volume:\

   $$V_{\text{real}} = 3 \times 2 \times 2.5 = 15 \text{ m}^3$$

   3. The length scale factor is $s = \dfrac{1}{20}$, so the area scale factor is $s^2 = \dfrac{1}{400}$ and the volume scale factor is $s^3 = \dfrac{1}{8000}$.
   4. Model surface area:\

   $$SA_{\text{model}} = \frac{37}{400} = 0.0925 \text{ m}^2$$

   5. Model volume:\

   $$V_{\text{model}} = \frac{15}{8000} = 0.001875 \text{ m}^3$$

   Answer: The model's surface area is $0.0925 \text{ m}^2$ (area scale factor $\tfrac{1}{400}$) and its volume is $0.001875 \text{ m}^3$ (volume scale factor $\tfrac{1}{8000}$).

4. Circle terminology and chord properties

   A chord of length 16 cm is drawn in a circle of radius 10 cm. Find the perpendicular distance from the centre of the circle to the chord.

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   Key idea: The perpendicular from the centre to a chord bisects the chord. This creates a right-angled triangle with the radius as the hypotenuse.

   1. Half the chord: $16 \div 2 = 8$ cm.
   2. The radius (10 cm) is the hypotenuse, half the chord (8 cm) is one side, and the distance $d$ is the other side.
   3. $d = \sqrt{10^2 - 8^2} = \sqrt{100 - 64} = \sqrt{36} = 6$ cm

   Answer: 6 cm

5. Angle properties of circles: Theorems 3 and 4

   Points $A$, $B$, $C$ and $D$ lie on a circle in order, forming cyclic quadrilateral $ABCD$. Given that $\angle DAB = 73°$ and $\angle ABC = 105°$, find $\angle BCD$ and $\angle CDA$.

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   Key idea: Opposite angles of a cyclic quadrilateral sum to $180°$.

   1. $\angle DAB$ and $\angle BCD$ are opposite angles, so:\

   $$\angle BCD = 180° - 73° = 107°$$

   2. $\angle ABC$ and $\angle CDA$ are opposite angles, so:\

   $$\angle CDA = 180° - 105° = 75°$$

   3. Check: $73° + 107° = 180°$ ✓ and $105° + 75° = 180°$ ✓

   Answer: $\angle BCD = 107°$, $\angle CDA = 75°$

6. Theorems involving circles and tangents

   A straight fence is built so that it just touches a circular pond at a single point $P$. The centre of the pond is $O$ and the radius is $8$ metres. A post is placed at point $T$ on the fence, $15$ metres from $P$. Find the distance from the post at $T$ to the centre $O$.

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   Key idea: A tangent to a circle is perpendicular to the radius at the point of contact, so $OP \perp TP$ and $\angle OPT = 90°$.

   1. Triangle $OPT$ is right-angled at $P$.
   2. $OP = 8$ m (radius), $TP = 15$ m.
   3. By Pythagoras' theorem:\

   $$OT^2 = OP^2 + TP^2 = 8^2 + 15^2 = 64 + 225 = 289$$

   4. $OT = \sqrt{289} = 17 \text{ m}$

   Answer: The distance from the post to the centre is $17$ metres.

7. Introduction to networks

   A network has five vertices: A, B, C, D and E. The edges are: A, B, A, C, B, C, B, D, and D, E. State the number of vertices, the number of edges, and the degree of each vertex.

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   Key idea: The degree of a vertex is the number of edges connected to it. The sum of all degrees equals twice the number of edges.

   1. Count the vertices: A, B, C, D, E gives 5 vertices.
   2. Count the edges: A, B, A, C, B, C, B, D, D, E gives 5 edges.
   3. Find the degree of each vertex by counting how many edges meet it:

   • A appears in A, B and A, C: $\deg(A) = 2$
   • B appears in A, B, B, C and B, D: $\deg(B) = 3$
   • C appears in A, C and B, C: $\deg(C) = 2$
   • D appears in B, D and D, E: $\deg(D) = 2$
   • E appears in D, E: $\deg(E) = 1$

   4. Check: sum of degrees $= 2 + 3 + 2 + 2 + 1 = 10 = 2 \times 5$ ✓

   Answer: 5 vertices, 5 edges; $\deg(A)=2,\; \deg(B)=3,\; \deg(C)=2,\; \deg(D)=2,\; \deg(E)=1$.

8. Trails, paths and Eulerian circuits

   A network has six vertices: P, Q, R, S, T and U. The edges are: P, Q, P, R, Q, R, Q, S, R, T, S, T, S, U and T, U. (a) State the degree of each vertex and determine whether the network is connected. (b) Determine whether an Eulerian path exists. If it does, state a valid starting and finishing vertex.

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   Key idea: An Eulerian path (which travels every edge exactly once) exists if and only if the network is connected and has exactly 0 or 2 vertices of odd degree. With 0 odd-degree vertices, the path is a circuit (starts and ends at the same vertex); with exactly 2, it starts at one odd-degree vertex and ends at the other.

   1. Find the degree of each vertex:\

   $$\begin{aligned} \deg(P) &= 2 (\text{P, Q, P, R})\\ \deg(Q) &= 3 (\text{P, Q, Q, R, Q, S})\\ \deg(R) &= 3 (\text{P, R, Q, R, R, T})\\ \deg(S) &= 3 (\text{Q, S, S, T, S, U})\\ \deg(T) &= 3 (\text{R, T, S, T, T, U})\\ \deg(U) &= 2 (\text{S, U, T, U}) \end{aligned}$$

   2. Check connectivity: every vertex can be reached from every other vertex by tracing the edges, so the network is connected.
   3. Identify odd-degree vertices: Q, R, S and T each have degree 3, that is 4 odd-degree vertices.
   4. Because there are more than 2 odd-degree vertices, no Eulerian path exists.

   Answer: The network is connected. The odd-degree vertices are Q, R, S and T (four in total), so no Eulerian path exists.

Free worksheets for this chapter

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

• Year 10 geometric reasoning and proofs worksheet
• Year 10 direct and inverse proportion worksheet
• Year 10A circle theorems worksheet
• Year 10 networks worksheet