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

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°.

Answer: D

The value of (2y + 15°) is 65°.

Complementary angles add up to 90°. Let the larger be a and the smaller y:

\[ \begin{aligned} a + y &= 90^{\circ},\ a - y = 40^{\circ} \\ y &= 25^{\circ} \\ 2y + 15^{\circ} &= 65^{\circ} \end{aligned} \]

The angles are 65° and 25°, and they do add to 90° and differ by 40°. Note that 2y + 15° happens to equal the larger angle, a useful check. Any other value of y breaks one of the two conditions, so 45°, 50° and 55° are wrong.

Answer: B

The pair that does not have an equal number of lines of symmetry is Kite and parallelogram.

• Isosceles triangle: 1 line; kite: 1 line, so pair A is equal.
• Kite: 1 line; a general parallelogram: 0 lines, so pair B is unequal.
• Rectangle: 2 lines; rhombus: 2 lines (its diagonals), so pair C is equal.
• Kite: 1 line; isosceles trapezium: 1 line, so pair D is equal.

A parallelogram has rotational symmetry of order 2 but, unless it is a rectangle or rhombus, no line of symmetry.

Answer: B

The true statement is ΔPQS ≅ ΔPRS, by ASA.

In triangles PQS and PRS:

• ∠QPS = ∠RPS, because PS bisects ∠QPR (angle);
• PS = PS, the common side (side);
• ∠PSQ = ∠PSR = 90°, because PS is perpendicular to QR (angle).

So the triangles are congruent by ASA, with P matching P, Q matching R and S matching S, which is written ΔPQS ≅ ΔPRS. RHS cannot be used because it needs the hypotenuses PQ and PR to be known equal, which is not given. Writing ΔPSR matches Q with S, the wrong correspondence.

Answer: C

The third side can be 33 cm.

By the triangle inequality, each side must be less than the sum and greater than the difference of the other two:

\[ 19 - 15 \lt c \lt 19 + 15,\ \text{so}\ 4 \lt c \lt 34 \]

Only 33 cm lies strictly between 4 cm and 34 cm. A side of 4 cm or 34 cm would make the three points lie on one straight line, so no triangle forms, and 3 cm is too short to close the triangle.

Answer: B

The statement that is not correct is 'One of them is not a parallelogram'.

The four shapes are a slanting parallelogram, a square, a rectangle and a rhombus. A parallelogram is any quadrilateral with both pairs of opposite sides parallel, and squares, rectangles and rhombuses all qualify. So all four are parallelograms, and the claim that one is not is false.

The other statements are correct: all are parallelograms; two are rhombuses (the square and the rhombus, since a square has four equal sides); and two are rectangles (the square and the rectangle, since a square has four right angles).

Answer: D

The shape without a vertex is the sphere.

A vertex is a corner point where edges meet. A sphere has one continuous curved surface, with no edges and no corners at all.

A pyramid has an apex and the corners of its base, a prism has the corners of its two bases (a triangular prism has 6), and a cone has one vertex at its tip, even though its base is a curved edge.

Answer: C

The true statement is ΔABC ≅ ΔBAD.

Since DA ⊥ AB and CB ⊥ AB, both triangles are right-angled: ΔABC at B and ΔBAD at A. Their hypotenuses are equal (AC = BD) and they share the side AB. By the RHS congruence rule the triangles are congruent.

The correspondence must match equal parts: the right angle B of ΔABC goes to the right angle A, the vertex A goes to B, and C goes to D. So A ↔ B, B ↔ A, C ↔ D, written ΔABC ≅ ΔBAD. The other orderings pair the right angles or the hypotenuses wrongly, even though they name the same two triangles.