"Mathematics — Geometry" — Practice

Question 1

Two angles of a triangle measure 55° and 65°. The third angle is:
(a) 60°  (b) 120°  (c) 55°  (d) 70°

Accepted Answer

The three angles of a triangle sum to 180°. Third angle = 180° − 55° − 65° = 60°.

Answer: (a) 60°. Trap (b) 120° comes from forgetting to subtract from 180° and instead just adding the two given angles (55+65=120). Trap (c) 55° comes from wrongly assuming the triangle must be isosceles with the third angle repeating one of the given ones. Trap (d) 70° comes from a small subtraction slip (180−55−65 miscalculated as 70 instead of 60).

Question 2

An exterior angle of a triangle measures 120°, and one of the interior opposite (remote) angles is 50°. The other interior opposite angle is:
(a) 70°  (b) 60°  (c) 130°  (d) 50°

Accepted Answer

By the exterior angle theorem, the exterior angle equals the sum of the two interior opposite angles: 120° = 50° + x ⇒ x = 70°.

Answer: (a) 70°. Trap (b) 60° comes from wrongly halving the exterior angle instead of subtracting the known remote angle. Trap (c) 130° comes from adding instead of subtracting (120+50 misapplied, then misread). Trap (d) 50° comes from wrongly assuming the two remote angles must be equal to each other.

Question 3

Three angles of a quadrilateral measure 80°, 95°, and 100°. The fourth angle is:
(a) 85°  (b) 95°  (c) 105°  (d) 75°

Accepted Answer

The interior angles of a quadrilateral sum to 360°. Fourth angle = 360° − 80° − 95° − 100° = 85°.

Answer: (a) 85°. Trap (b) 95° comes from wrongly assuming the quadrilateral must have a repeated angle. Trap (c) 105° comes from using 180° (the triangle angle sum) as a starting point by mistake instead of 360°, then compensating incorrectly. Trap (d) 75° comes from a subtraction slip while adding the three given angles before subtracting from 360°.

Question 4

The sum of the interior angles of a regular octagon (8 sides) is:
(a) 1080°  (b) 1440°  (c) 900°  (d) 720°

Accepted Answer

Sum of interior angles = (n−2) × 180°. For n=8: (8−2) × 180° = 6 × 180° = 1080°.

Answer: (a) 1080°. Trap (b) 1440° comes from using n×180° instead of (n−2)×180°, i.e. 8×180=1440 — the single most common slip in this formula. Trap (c) 900° comes from using (n−3)×180° instead of (n−2)×180°, an off-by-one in the wrong direction. Trap (d) 720° comes from applying the hexagon's angle sum (n=6) by mistake instead of the octagon's (n=8).

Question 5

Each exterior angle of a regular polygon measures 24°. The number of sides of this polygon is:
(a) 15  (b) 12  (c) 18  (d) 24

Accepted Answer

For a regular polygon, each exterior angle = 360°/n, so n = 360°/24° = 15.

Answer: (a) 15. Trap (b) 12 comes from dividing 360 by 30 instead of 24 (a misread digit). Trap (c) 18 comes from dividing 360 by 20 instead of 24 (another misread). Trap (d) 24 comes from simply repeating the given exterior angle value as if it were the answer, without performing the division at all.

Question 6

In triangles ABC and PQR, AB = PQ, BC = QR, and the angle included between them, ∠B = ∠Q. Which congruence criterion proves △ABC ≅ △PQR?
(a) SAS  (b) SSA  (c) AAA  (d) SSS

Accepted Answer

Two sides (AB, BC) and the angle INCLUDED between them (∠B) are given, matching (PQ, QR) and the included ∠Q — this is exactly the SAS criterion.

Answer: (a) SAS. Trap (b) SSA is a plausible-sounding but invalid label for this exact data pattern — it would apply only if the given angle were NOT the included one, and even then SSA is not a valid criterion at all. Trap (c) AAA is wrong because no angle-only data is given here, and AAA proves similarity, not congruence, in any case. Trap (d) SSS is wrong because only two sides are given, not three — the third piece of information given is an angle, not a side.

Question 7

A convex polyhedron has 6 faces and 8 vertices. Using Euler's formula, the number of edges is:
(a) 12  (b) 14  (c) 10  (d) 2

Accepted Answer

Euler's formula: F + V − E = 2. Substituting: 6 + 8 − E = 2 ⇒ E = 12 (this matches a cube).

Answer: (a) 12. Trap (b) 14 comes from adding F and V and forgetting to subtract 2 at all (6+8=14). Trap (c) 10 comes from a sign error, using F+V+E=2 or a similar misapplied rearrangement instead of F+V−E=2. Trap (d) 2 comes from writing down the right-hand side of Euler's formula itself instead of solving for E.

Question 8

To construct the perpendicular bisector of a line segment AB using a ruler and compasses, the radius used to draw arcs from A and from B must be:
(a) more than half of AB  (b) exactly half of AB  (c) less than half of AB  (d) any length — it doesn't matter

Accepted Answer

For the two arcs (equal radius, centred at A and at B) to intersect at two distinct points — which is what defines the perpendicular bisector line — the radius must be greater than half the length of AB.

Answer: (a) more than half of AB. Trap (b) exactly half of AB is wrong because the two arcs would then just touch at a single point (the midpoint of AB) rather than crossing at two points off the segment. Trap (c) less than half of AB is wrong because the arcs would be too far apart to meet at all. Trap (d) is wrong because the construction fails outside the 'more than half' condition — the radius is not arbitrary.

Question 9

A regular hexagon has:
(a) 6 lines of symmetry and rotational symmetry of order 6  (b) 3 lines of symmetry and rotational symmetry of order 6  (c) 6 lines of symmetry and rotational symmetry of order 3  (d) 12 lines of symmetry and rotational symmetry of order 6

Accepted Answer

A regular polygon with n sides always has exactly n lines of symmetry and rotational symmetry of order n. For a hexagon, n=6, so it has 6 lines of symmetry and rotational symmetry of order 6.

Answer: (a) 6 lines of symmetry and rotational symmetry of order 6. Trap (b) halves the line-symmetry count, confusing '3 pairs of opposite sides' with the actual line count (a hexagon has 6 lines: 3 through opposite vertices and 3 through opposite edge midpoints). Trap (c) halves the rotational order instead, by the same kind of pairing error. Trap (d) doubles the true line-symmetry count, perhaps by double-counting each axis.

Question 10

A Class VI student is shown a square tilted 45° (resembling a 'diamond') and insists it is not a square, even after the teacher points out that all four sides are equal and all angles are right angles. This response is most characteristic of which Van Hiele level?
(a) Level 0 — Visualization  (b) Level 1 — Analysis  (c) Level 2 — Informal deduction  (d) Level 3 — Deduction

Accepted Answer

The student is judging the shape by its overall visual appearance/orientation rather than by its stated properties, even when those properties are pointed out directly — this appearance-over-properties dependency is the defining feature of Van Hiele Level 0 (Visualization).

Answer: (a) Level 0 — Visualization. Trap (b) Level 1 — Analysis is tempting because 'equal sides, right angles' was mentioned in the scenario, but it was the teacher, not the student, who used the properties — the student is failing to apply exactly the property-based reasoning that defines Level 1. Trap (c) Level 2 — Informal deduction concerns relationships BETWEEN shape classes (like square-rectangle inclusion), a more advanced skill than this basic recognition failure. Trap (d) Level 3 — Deduction requires formal, axiomatic proof reasoning, far beyond a Class VI child's classification error.

Question 11

Which of the following correctly reflects the inclusive (hierarchical) definition of quadrilaterals used in the NCERT curriculum?
(a) Every square is a rectangle, because it satisfies all the defining properties of a rectangle (four right angles, opposite sides equal and parallel) in addition to having all sides equal
(b) A square and a rectangle are entirely separate categories of quadrilaterals with no overlap
(c) A rectangle is a type of square, since a rectangle is the more general shape
(d) Only shapes with unequal adjacent sides can correctly be called rectangles

Accepted Answer

NCERT's inclusive (hierarchical) definition treats a square as a special case of a rectangle: it satisfies every defining property of a rectangle and additionally has all sides equal, so a square is always also a rectangle.

Answer: (a). Trap (b) is exactly the common student (and sometimes teacher) misconception this chapter names — mutually exclusive categorisation instead of inclusive hierarchy. Trap (c) reverses the direction of the hierarchy (a square is a special case of a rectangle, not the other way around) — a plausible trap for someone who has heard 'special case' language but applied it backwards. Trap (d) smuggles in a false exclusionary condition that would incorrectly disqualify squares from ever being called rectangles — the opposite of the correct inclusive definition.

Question 12

NCERT defers formal, axiomatic proof-based deductive reasoning in geometry (Van Hiele Level 3) to which stage, and why is this sequencing pedagogically appropriate for Classes VI-VIII?
(a) Class IX-X, because most VI-VIII learners are still consolidating Van Hiele Levels 0-2 through hands-on activities before formal proof becomes meaningful
(b) Class VI itself, because CTET expects two-column proofs to be taught from the very first geometry chapter
(c) It is never introduced in school and is reserved for undergraduate mathematics
(d) Class III-V, because younger children find formal proofs easier to accept without questioning

Accepted Answer

NCERT deliberately reserves formal, axiomatic proof-based deduction for Class IX-X CBSE geometry (Euclid's Geometry, Lines and Angles, Triangles), after Classes VI-VIII have built Van Hiele Levels 0-2 through hands-on, activity-based instruction — introducing formal proof earlier would outrun most learners' developmental readiness.

Answer: (a). Trap (b) reverses the entire premise of Van Hiele sequencing and contradicts NCF 2005's activity-based approach — a confident-sounding trap because it asserts the opposite of correct practice. Trap (c) is factually wrong — Class IX-X CBSE geometry explicitly does introduce formal proofs; this option over-corrects by pushing formal proof entirely out of school. Trap (d) inverts the developmental logic completely — younger children are the LEAST ready for axiomatic reasoning, not the most ready.

Practice — "Mathematics — Geometry"

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