Geometry (Lines, Angles, Triangles, Polygons) Questions and Answers

Answer: C

The right triangle is (iii), with sides 11 cm, 60 cm and 61 cm.

By the converse of Pythagoras' theorem, a triangle is right-angled when the square of the longest side equals the sum of the squares of the other two:

\[ \begin{aligned} 11^{2} + 60^{2} &= 121 + 3600 = 3721 \\ 61^{2} &= 3721 \end{aligned} \]

The others fail: \(20^{2} + 22^{2} = 884 \ne 576\); \(15^{2} + 32^{2} = 1249 \ne 1369\); \(19^{2} + 40^{2} = 1961 \ne 1681\). The familiar triple 9, 40, 41 has 9, not 19, so (iv) is a near miss.

Answer: D

The difference is 90°.

The angles of a quadrilateral add up to 360°. With the ratio 3 : 5 : 7 : 9:

\[ \begin{aligned} 3 + 5 + 7 + 9 &= 24\ \text{parts} \\ 1\ \text{part} &= 360^{\circ}/24 = 15^{\circ} \\ 9 \times 15 - 3 \times 15 &= 135 - 45 = 90^{\circ} \end{aligned} \]

The angles are 45°, 75°, 105° and 135°. Using 180° (the triangle sum) instead of 360° would halve every angle and give 45°, which is not offered; 60° and 72° do not follow from the ratio.

Answer: B

3 such triangles are possible.

Let the integer sides be a ≤ b ≤ c with a + b + c = 12. The triangle inequality requires a + b > c, so c < 6, while c must be at least 4 (the largest side is at least a third of the perimeter). Checking c = 4 and c = 5:

• c = 4: (4, 4, 4).
• c = 5: (2, 5, 5) and (3, 4, 5).

c = 6 fails because a + b = 6 would not exceed it (a degenerate, flat triangle). So there are 3 triangles. Counting (1, 5, 6) or (2, 4, 6) wrongly gives 4 or 5.

Answer: B

The measure of ∠ ABC is 70°.

Each given angle is an exterior angle of the triangle, and an interior angle and the exterior angle beside it on a straight line add to 180°.

\[ \begin{aligned} \angle ACB &= 180^\circ - 120^\circ = 60^\circ \\ \angle BAC &= 180^\circ - 130^\circ = 50^\circ \end{aligned} \]

The three angles of a triangle add to 180°:

\[ \angle ABC = 180^\circ - 60^\circ - 50^\circ = 70^\circ \]

Check with the exterior angle property: ∠ ACD = ∠ ABC + ∠ BAC = 70° + 50° = 120°, as given.

Answer: B

The polyhedron has 5 vertices.

Euler's formula for a polyhedron relates faces F, vertices V and edges E:

\[ \begin{aligned} F + V - E &= 2 \\ 5 + V - 8 &= 2 \\ V &= 5 \end{aligned} \]

A square pyramid fits these numbers: 5 faces (a square base and four triangles), 8 edges and 5 vertices. The option 8 repeats the number of edges. Neither 10 nor 15 satisfies the formula: 5 + 10 − 8 = 7 and 5 + 15 − 8 = 12, not 2.

Answer: B

12 angles are formed.

A transversal crossing one line makes four angles at the point where they meet. Crossing three lines, it meets each of them at a separate point, so

\[ 3 \times 4 = 12\ \text{angles} \]

A transversal of two lines makes the familiar 8 angles (corresponding, alternate and so on); adding a third line adds another four. The options 3, 6 and 9 count the points of intersection or only some of the angles at each.

Answer: C

The answer is B and C: the rectangle and the rhombus.

A rectangle (that is not a square) has exactly two lines of symmetry, joining the midpoints of opposite sides. A rhombus (that is not a square) also has exactly two, along its diagonals.

An equilateral triangle has three lines of symmetry and an isosceles triangle has only one, so any option containing A or D is wrong, and 'Only B' leaves out the rhombus.

Answer: D

The letter is Z.

Turned through half a turn (180°) about its centre, Z looks the same, so it has rotational symmetry of order 2. But no line can be drawn through it so that one half mirrors the other, so it has no line symmetry.

X has both rotational symmetry and two lines of symmetry. O has both as well. M has a vertical line of symmetry but no rotational symmetry. None of them fits 'rotational but no line symmetry'.

Answer: D

A pentagonal pyramid has six vertices and six faces.

Its faces are the pentagonal base plus five triangular side faces, 6 in all. Its vertices are the 5 corners of the base plus the apex, 6 in all. It has 10 edges: 5 round the base and 5 rising to the apex. Euler's formula checks this: \(V - E + F = 6 - 10 + 6 = 2\).

'Five faces' misses the base, 'five vertices' misses the apex, and 'six edges' is far short of the true 10, so the other options are wrong.

Answer: C

The incorrect statement is Two acute angles can be put together to make a reflex angle.

Each acute angle is less than 90°, so two together are less than 180°. A reflex angle is more than 180°, so two acute angles can never make one.

The other statements are correct. An acute and an obtuse angle can sum to exactly 180°, for example 60° and 120°, giving a straight angle. A reflex angle does lie between 180° and 360°. Two acute angles can exceed 90°, for example 50° and 60° make 110°, which is obtuse.