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

Q: A triangular pyramid has :

A triangular pyramid has 4 faces and 6 edges.A triangular pyramid (tetrahedron) has a triangular base and three triangular side faces, so 4 faces. The base has 3 edges and three more run from the base corners to the apex, so 6 edges. It has 4 vertices.Euler's formula checks it: \(F + V - E = 4 + 4 - 6 = 2\). '3 faces' leaves out the base, and '4 edges' counts too few, so the other options are wrong.

Q: In a parallelogram PQRS, &ang; P = 2x + 20°, &ang; Q = 2y - 40°, &ang; R = 5x - 100° and &ang; S = 3y - 100°. Then, the value of (x + y) is equal to :

The value of \(x + y\) is 100.In a parallelogram opposite angles are equal, so \(\angle P = \angle R\) and \(\angle Q = \angle S\):\[ \begin{aligned} 2x + 20 &= 5x - 100 \Rightarrow x = 40 \\ 2y - 40 &= 3y - 100 \Rightarrow y = 60 \end{aligned} \]So \(x + y = 100\). Check: the angles are \(P = R = 100^{\circ}\) and \(Q = S = 80^{\circ}\); adjacent angles add to \(180^{\circ}\) and all four add to \(360^{\circ}\), as they must. The other options come from using the wrong pairs of angles.

Q: Two alternate interior angles formed by the intersection of two parallel lines by a transversal are (5y - 10°) and (7y - 90°). Then, the value of (3y - 20°) is :

The value of \(3y - 20\) is 100°.Alternate interior angles formed by a transversal across parallel lines are equal, so set the two expressions equal:\[ \begin{aligned} 5y - 10 &= 7y - 90 \\ 2y &= 80 \Rightarrow y = 40 \\ 3y - 20 &= 120 - 20 = 100 \end{aligned} \]The question tests only that equality. Note that with \(y = 40\) each given angle works out to \(190^{\circ}\), which a real interior angle cannot be, so the numbers do not describe an actual figure; the keyed answer is what the equality gives. 120° is \(3y\) without subtracting 20.

Q: Sides AB and AC of a triangle ABC are produced to points D and E respectively. Bisectors of angles CBD and BCE intersect each other at the point O. If &ang; BAC = 50°, then &ang; BOC is equal to :

\(\angle BOC\) is 65°.O is where the bisectors of the two exterior angles at B and C meet. With \(\angle B\) and \(\angle C\) the interior angles of the triangle:\[ \begin{aligned} \angle OBC &= \tfrac{1}{2}(180^{\circ} - B) \\ \angle OCB &= \tfrac{1}{2}(180^{\circ} - C) \end{aligned} \]\[ \begin{aligned} \angle BOC &= 180^{\circ} - \angle OBC - \angle OCB \\ &= \tfrac{1}{2}(B + C) \\ &= \tfrac{1}{2}(180^{\circ} - 50^{\circ}) = 65^{\circ} \end{aligned} \]This is the standard result \(\angle BOC = 90^{\circ} - \frac{A}{2}\) for exterior bisectors. 115° is \(90^{\circ} + \frac{A}{2}\), the result for the bisectors of the interior angles, which is the usual mix-up.

Q: Which of the following figures has rotational symmetry, but no line symmetry ?

The figure is a parallelogram.A general parallelogram looks the same after a half-turn (180°) about the point where its diagonals meet, so it has rotational symmetry of order 2. But no line through it folds one half exactly onto the other, so it has no line of symmetry.A kite has one line of symmetry (along its main diagonal) and no rotational symmetry. A rectangle and a rhombus have both: two lines of symmetry each, as well as rotational symmetry of order 2.

Q: ABC is a right triangle, right-angled at B. D and E are points on side AB such that AD = DE = EB. Then, \(\dfrac{DC^{2} - BC^{2}}{AC^{2} - EC^{2}}\) is equal to :

The ratio is \(\dfrac{1}{2}\).Let \(AD = DE = EB = a\), so \(AB = 3a\), \(DB = 2a\) and \(EB = a\). Because the angle at B is a right angle, Pythagoras' theorem applies to triangles DBC, ABC and EBC:\[ \begin{aligned} DC^{2} - BC^{2} &= DB^{2} = 4a^{2} \\ AC^{2} - EC^{2} &= AB^{2} - EB^{2} \\ &= 9a^{2} - a^{2} = 8a^{2} \end{aligned} \]\[ \frac{4a^{2}}{8a^{2}} = \frac{1}{2} \]BC cancels in both differences, so the answer does not depend on the length of BC. \(\frac{4}{9}\) is \(DB^{2} : AB^{2}\), which forgets to subtract \(EB^{2}\).

Q: If (3x − 30°) and (2x + 20°) are two supplementary angles, then the value of (2x − 4°) is :

The value of \((2x - 4^\circ)\) is 72°.Supplementary angles add up to 180°.\[ \begin{aligned} (3x - 30^\circ) + (2x + 20^\circ) &= 180^\circ \\ 5x - 10^\circ &= 180^\circ \\ x &= 38^\circ \\ 2x - 4^\circ &= 76^\circ - 4^\circ = 72^\circ \end{aligned} \]76° is \(2x\) before subtracting 4°. Treating the angles as complementary (sum 90°) gives \(x = 20^\circ\) and a value of 36°, which is not offered.

Q: Angles P and Q of a triangle PQR are respectively 30° and 50°. If the bisectors of angles P and Q meet at a point O, then the value of (2&ang;POQ − 10°) is :

The value is 270°.The bisectors split the angles at P and Q in half, so in triangle POQ the angles at P and Q are 15° and 25°.\[ \begin{aligned} \angle POQ &= 180^\circ - 15^\circ - 25^\circ = 140^\circ \\ 2\angle POQ - 10^\circ &= 280^\circ - 10^\circ = 270^\circ \end{aligned} \]The same result follows from the rule \(\angle POQ = 90^\circ + \frac{1}{2}\angle R\), with \(\angle R = 100^\circ\). The option 150° comes from using \(\angle P + \angle Q = 80^\circ\) in place of \(\angle POQ\): \(2 \times 80^\circ - 10^\circ = 150^\circ\).

Q: Which of the following measurements (in cm) cannot form the sides of a right triangle?

The set that cannot form a right triangle is 8, 16, 12.Three lengths form a right triangle only when the square of the longest equals the sum of the squares of the other two (the converse of Pythagoras' theorem).\[ \begin{aligned} 8^{2} + 12^{2} &= 64 + 144 = 208 \\ 16^{2} &= 256 \ne 208 \end{aligned} \]The other sets pass: \(2.5^{2} + 6^{2} = 6.25 + 36 = 42.25 = 6.5^{2}\); \(12^{2} + 16^{2} = 400 = 20^{2}\); and \(4.5^{2} + 6^{2} = 20.25 + 36 = 56.25 = 7.5^{2}\).

Q: Which of the following letters does not have rotational symmetry?

The letter without rotational symmetry is E.A figure has rotational symmetry if it looks the same after a rotation of less than a full turn. E only looks the same after a complete 360° turn, so it has no rotational symmetry (it has only a horizontal line of symmetry).H and N each look the same after a half turn of 180°, so both have rotational symmetry of order 2. O, drawn as a circle or an oval, looks the same after a half turn as well. So H, O and N are all wrong answers.

Important Instructions

31.

A triangular pyramid has :

A.4 faces and 6 edges.
B.3 faces and 6 edges.
C.3 faces and 4 edges.
D.4 faces and 4 edges.

Exam Year(s)

CTET - 2026,

Answer: A

A triangular pyramid has 4 faces and 6 edges.

A triangular pyramid (tetrahedron) has a triangular base and three triangular side faces, so 4 faces. The base has 3 edges and three more run from the base corners to the apex, so 6 edges. It has 4 vertices.

Euler's formula checks it: \(F + V - E = 4 + 4 - 6 = 2\). '3 faces' leaves out the base, and '4 edges' counts too few, so the other options are wrong.

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32.

In a parallelogram PQRS, ∠ P = 2x + 20°, ∠ Q = 2y - 40°, ∠ R = 5x - 100° and ∠ S = 3y - 100°. Then, the value of (x + y) is equal to :

A.100°
B.120°
C.140°
D.160°

Exam Year(s)

CTET - 2026,

Answer: A

The value of \(x + y\) is 100.

In a parallelogram opposite angles are equal, so \(\angle P = \angle R\) and \(\angle Q = \angle S\):

\[ \begin{aligned} 2x + 20 &= 5x - 100 \Rightarrow x = 40 \\ 2y - 40 &= 3y - 100 \Rightarrow y = 60 \end{aligned} \]

So \(x + y = 100\). Check: the angles are \(P = R = 100^{\circ}\) and \(Q = S = 80^{\circ}\); adjacent angles add to \(180^{\circ}\) and all four add to \(360^{\circ}\), as they must. The other options come from using the wrong pairs of angles.

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33.

Two alternate interior angles formed by the intersection of two parallel lines by a transversal are (5y - 10°) and (7y - 90°). Then, the value of (3y - 20°) is :

A.90°
B.110°
C.120°
D.100°

Exam Year(s)

CTET - 2026,

Answer: D

The value of \(3y - 20\) is 100°.

Alternate interior angles formed by a transversal across parallel lines are equal, so set the two expressions equal:

\[ \begin{aligned} 5y - 10 &= 7y - 90 \\ 2y &= 80 \Rightarrow y = 40 \\ 3y - 20 &= 120 - 20 = 100 \end{aligned} \]

The question tests only that equality. Note that with \(y = 40\) each given angle works out to \(190^{\circ}\), which a real interior angle cannot be, so the numbers do not describe an actual figure; the keyed answer is what the equality gives. 120° is \(3y\) without subtracting 20.

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34.

Sides AB and AC of a triangle ABC are produced to points D and E respectively. Bisectors of angles CBD and BCE intersect each other at the point O. If ∠ BAC = 50°, then ∠ BOC is equal to :

A.115°
B.105°
C.75°
D.65°

Exam Year(s)

CTET - 2026,

Answer: D

\(\angle BOC\) is 65°.

O is where the bisectors of the two exterior angles at B and C meet. With \(\angle B\) and \(\angle C\) the interior angles of the triangle:

\[ \begin{aligned} \angle OBC &= \tfrac{1}{2}(180^{\circ} - B) \\ \angle OCB &= \tfrac{1}{2}(180^{\circ} - C) \end{aligned} \]

\[ \begin{aligned} \angle BOC &= 180^{\circ} - \angle OBC - \angle OCB \\ &= \tfrac{1}{2}(B + C) \\ &= \tfrac{1}{2}(180^{\circ} - 50^{\circ}) = 65^{\circ} \end{aligned} \]

This is the standard result \(\angle BOC = 90^{\circ} - \frac{A}{2}\) for exterior bisectors. 115° is \(90^{\circ} + \frac{A}{2}\), the result for the bisectors of the interior angles, which is the usual mix-up.

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35.

Which of the following figures has rotational symmetry, but no line symmetry ?

A.Kite
B.Parallelogram
C.Rectangle
D.Rhombus

Exam Year(s)

CTET - 2026,

Answer: B

The figure is a parallelogram.

A general parallelogram looks the same after a half-turn (180°) about the point where its diagonals meet, so it has rotational symmetry of order 2. But no line through it folds one half exactly onto the other, so it has no line of symmetry.

A kite has one line of symmetry (along its main diagonal) and no rotational symmetry. A rectangle and a rhombus have both: two lines of symmetry each, as well as rotational symmetry of order 2.

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36.

ABC is a right triangle, right-angled at B. D and E are points on side AB such that AD = DE = EB. Then, \(\dfrac{DC^{2} - BC^{2}}{AC^{2} - EC^{2}}\) is equal to :

A.\(\dfrac{2}{5}\)
B.\(\dfrac{4}{9}\)
C.\(\dfrac{1}{3}\)
D.\(\dfrac{1}{2}\)

Exam Year(s)

CTET - 2026,

Answer: D

The ratio is \(\dfrac{1}{2}\).

Let \(AD = DE = EB = a\), so \(AB = 3a\), \(DB = 2a\) and \(EB = a\). Because the angle at B is a right angle, Pythagoras' theorem applies to triangles DBC, ABC and EBC:

\[ \begin{aligned} DC^{2} - BC^{2} &= DB^{2} = 4a^{2} \\ AC^{2} - EC^{2} &= AB^{2} - EB^{2} \\ &= 9a^{2} - a^{2} = 8a^{2} \end{aligned} \]

\[ \frac{4a^{2}}{8a^{2}} = \frac{1}{2} \]

BC cancels in both differences, so the answer does not depend on the length of BC. \(\frac{4}{9}\) is \(DB^{2} : AB^{2}\), which forgets to subtract \(EB^{2}\).

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37.

If (3x − 30°) and (2x + 20°) are two supplementary angles, then the value of (2x − 4°) is :

A.72°
B.76°
C.80°
D.84°

Exam Year(s)

CTET - 2026,

Answer: A

The value of \((2x - 4^\circ)\) is 72°.

Supplementary angles add up to 180°.

\[ \begin{aligned} (3x - 30^\circ) + (2x + 20^\circ) &= 180^\circ \\ 5x - 10^\circ &= 180^\circ \\ x &= 38^\circ \\ 2x - 4^\circ &= 76^\circ - 4^\circ = 72^\circ \end{aligned} \]

76° is \(2x\) before subtracting 4°. Treating the angles as complementary (sum 90°) gives \(x = 20^\circ\) and a value of 36°, which is not offered.

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38.

Angles P and Q of a triangle PQR are respectively 30° and 50°. If the bisectors of angles P and Q meet at a point O, then the value of (2∠POQ − 10°) is :

A.150°
B.190°
C.270°
D.250°

Exam Year(s)

CTET - 2026,

Answer: C

The value is 270°.

The bisectors split the angles at P and Q in half, so in triangle POQ the angles at P and Q are 15° and 25°.

\[ \begin{aligned} \angle POQ &= 180^\circ - 15^\circ - 25^\circ = 140^\circ \\ 2\angle POQ - 10^\circ &= 280^\circ - 10^\circ = 270^\circ \end{aligned} \]

The same result follows from the rule \(\angle POQ = 90^\circ + \frac{1}{2}\angle R\), with \(\angle R = 100^\circ\). The option 150° comes from using \(\angle P + \angle Q = 80^\circ\) in place of \(\angle POQ\): \(2 \times 80^\circ - 10^\circ = 150^\circ\).

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39.

Which of the following measurements (in cm) cannot form the sides of a right triangle?

A.2.5, 6.5, 6
B.12, 16, 20
C.4.5, 6, 7.5
D.8, 16, 12

Exam Year(s)

CTET - 2026,

Answer: D

The set that cannot form a right triangle is 8, 16, 12.

Three lengths form a right triangle only when the square of the longest equals the sum of the squares of the other two (the converse of Pythagoras' theorem).

\[ \begin{aligned} 8^{2} + 12^{2} &= 64 + 144 = 208 \\ 16^{2} &= 256 \ne 208 \end{aligned} \]

The other sets pass: \(2.5^{2} + 6^{2} = 6.25 + 36 = 42.25 = 6.5^{2}\); \(12^{2} + 16^{2} = 400 = 20^{2}\); and \(4.5^{2} + 6^{2} = 20.25 + 36 = 56.25 = 7.5^{2}\).

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40.

Which of the following letters does not have rotational symmetry?

Exam Year(s)

CTET - 2026,

Answer: C

The letter without rotational symmetry is E.

A figure has rotational symmetry if it looks the same after a rotation of less than a full turn. E only looks the same after a complete 360° turn, so it has no rotational symmetry (it has only a horizontal line of symmetry).

H and N each look the same after a half turn of 180°, so both have rotational symmetry of order 2. O, drawn as a circle or an oval, looks the same after a half turn as well. So H, O and N are all wrong answers.

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Tags: Geometry Lines, Angles, Triangles Questions || Polygons MCQ Questions and Answers || Geometry Quantitative Aptitude || Angles and Triangles GK Questions || Quantitative Aptitude Geometry GK

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

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Other Subjects

Tags: Geometry Lines, Angles, Triangles Questions || Polygons MCQ Questions and Answers || Geometry Quantitative Aptitude || Angles and Triangles GK Questions || Quantitative Aptitude Geometry GK

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