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

Answer: B

The value is 110°.

Since P lies on the line AB, angles QPA and QPB form a linear pair and add up to 180°:

\[ \begin{aligned} 7x + 5x &= 180^\circ \\ x &= 15^\circ \\ 8x - 10^\circ &= 120^\circ - 10^\circ = 110^\circ \end{aligned} \]

118° and 102° come from adding or subtracting the wrong amount, and 140° uses x = 18.75°, which would follow only if the angles added up to 225°.

Answer: B

The value is 60°.

Since PS = QS = RS, S is the midpoint of PQ and is equidistant from P, Q and R, so R lies on the circle with diameter PQ. The angle in a semicircle is a right angle, so ∠PRQ = 90°. (Equivalently, triangles PSR and QSR are isosceles, and their base angles add up to half of 180°.)

\[ \begin{aligned} \angle PQR &= 180^\circ - 90^\circ - 15^\circ = 75^\circ \\ 2\angle PQR - \angle PRQ &= 150^\circ - 90^\circ = 60^\circ \end{aligned} \]

75° is ∠PQR itself, and 50° and 45° do not follow once ∠PRQ is recognised as a right angle.

Answer: A

A triangle with all sides of different lengths is a scalene triangle.

Triangles are classified by their sides as scalene (no two sides equal, so all three angles are different too), isosceles (at least two sides equal, with the angles opposite them equal) and equilateral (all three sides equal, each angle 60°).

A right triangle is a classification by angle, one angle being 90°; it can itself be scalene or isosceles, so it is not the name for 'all sides different'. Equilateral and isosceles triangles both have equal sides, which contradicts the description.

Answer: D

A line segment has a definite length.

A line segment is the part of a line between two end points, so it can be measured, for example AB = 5 cm.

A line extends without end in both directions, so it has no definite length. A ray has one end point but extends without end in the other direction, so it cannot be measured either. A plane is a flat surface extending without end in all directions; it has no length at all. Only the segment, bounded at both ends, has a measurable length.

Answer: C

The letter H has two lines of symmetry.

H can be folded along a vertical line through its middle and along a horizontal line through its crossbar, and in each case the two halves match exactly. So it has one vertical and one horizontal line of symmetry.

E and B (in the usual printed form) have only a horizontal line of symmetry, and U has only a vertical one. Each of these has a single line, so only H has two. Other capital letters with two lines of symmetry are I, O and X.

Answer: C

The value of \((2x + 10^\circ)\) is 116°.

The interior angles of a pentagon add up to \((5 - 2) \times 180^\circ = 540^\circ\).

\[ \begin{aligned} 10x + (9 + 1 - 1 + 5 - 4)^\circ &= 540^\circ \\ 10x + 10^\circ &= 540^\circ \\ x &= 53^\circ \end{aligned} \]

So \(2x + 10^\circ = 106^\circ + 10^\circ = 116^\circ\). The other three values would each need a different x, and none of those makes the five angles add up to 540°.

Answer: C

The value of \((4x + 30^\circ)\) is 110°.

Corresponding angles formed by a transversal across parallel lines are equal:

\[ \begin{aligned} 3x - 20^\circ &= 5x - 60^\circ \\ 2x &= 40^\circ \\ x &= 20^\circ \end{aligned} \]

So \(4x + 30^\circ = 80^\circ + 30^\circ = 110^\circ\). Treating the angles as supplementary instead (adding to 180°) gives x = 32.5° and a different value.

Answer: A

\(\angle BOD\) is \(67\frac{1}{2}^\circ\).

Angles AOC, COD and DOB lie on the straight line AB, so they add up to 180°, and \(\angle COD = 90^\circ\) because OC is perpendicular to OD:

\[ \begin{aligned} x + 90^\circ + 3x &= 180^\circ \\ 4x &= 90^\circ \\ x &= 22\tfrac{1}{2}^\circ \end{aligned} \]

So \(\angle BOD = 3x = 67\frac{1}{2}^\circ\). The option \(22\frac{1}{2}^\circ\) is x itself, \(\angle AOC\), not \(\angle BOD\).

Answer: C

The value of \((3x + 2y)\) is 36.

Opposite sides of a parallelogram are equal, so PQ = RS and QR = PS:

\[ \begin{aligned} 4x + 1 &= 25 \Rightarrow x = 6 \\ 3y - 1 &= 26 \Rightarrow y = 9 \\ 3x + 2y &= 18 + 18 = 36 \end{aligned} \]

Pairing the wrong sides (PQ with QR, or PS with RS) gives fractional values and none of the other options.

Answer: C

The triangle can be constructed with AB = 8 cm, BC = 4 cm, \(\angle BCA = 90^\circ\), \(\angle ABC = 60^\circ\).

With the right angle at C, AB is the hypotenuse, and \(\cos 60^\circ = \frac{BC}{AB} = \frac{4}{8} = \frac{1}{2}\), so the four measurements agree and the triangle exists.

The others fail. In option 2, 3 + 4 = 7, so the sides cannot form a triangle. In option 4, AC = BC needs the angles at A and B to be equal, but the angle at B is 25° and the one at A is 35°. In option 1, two sides and the included 130° already fix the triangle, and its angle at A comes out near 27°, not 25°.