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Samacheer Kalvi 7th Maths (Term 1) Geometry Exercise Solutions

Complete Samacheer Kalvi 7th Maths Term 1 Chapter 5 Geometry exercise solutions — angles, lines, triangles, symmetry, congruence solved step-by-step.

Updated 12 June 202618 min read

Exercise Solutions Available

Step-by-step answers for all textbook questions in this chapter

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How to Use This Page

These are complete solutions for Samacheer Kalvi 7th Maths (Term 1) Chapter 5 — Geometry. For every problem, show all steps: Given → Property → Reasoning → Answer. Solve each exercise on your own first, then check the method below.

Key Concepts (Quick Recap)

  • Complementary angles: Two angles whose sum is 90°.
  • Supplementary angles: Two angles whose sum is 180°.
  • Linear pair: Two adjacent angles on a straight line that add up to 180°.
  • Vertically opposite angles: When two lines intersect, the angles opposite each other are equal.
  • Parallel lines and transversal: Corresponding angles are equal; alternate interior angles are equal; co-interior (same-side interior) angles are supplementary.
  • Angle sum property of a triangle: The three interior angles of a triangle always add up to 180°.
  • Isosceles triangle: Angles opposite equal sides are equal. If vertex angle is given, each base angle = (180° − vertex angle) / 2.
  • Equilateral triangle: All sides equal, all angles equal to 60°.
  • Congruence criteria: SSS, SAS, ASA, RHS.

Exercise Solutions

Exercise 5.1 — Angles

Q1. Can two adjacent angles be supplementary?
Answer

Yes. Two adjacent angles can be supplementary if they form a linear pair — that is, if their non-common arms form a straight line and the sum of the two angles is 180°.

Example: 110° and 70° are adjacent supplementary angles because 110° + 70° = 180°.

Q2. Can two obtuse angles form a linear pair?
Answer

No. A linear pair must add up to 180°. Each obtuse angle is greater than 90°. If both angles are obtuse, their sum would be greater than 90° + 90° = 180°, which exceeds 180°. Therefore, two obtuse angles cannot form a linear pair.

Q3. Can two right angles form a linear pair?
Answer

Yes. Two right angles each measure 90°.

90° + 90° = 180°

Since their sum is 180°, two right angles can form a linear pair.

Q4. Find the values of x, y and z from the given figure (intersecting lines forming angles).
GivenTwo lines intersect. One angle = 55°.
Vertically opposite angles are equal: x = 55° (vertically opposite to the given angle) Linear pair at the same point: y + 55° = 180° y = 180° − 55° y = 125° Vertically opposite to y: z = 125° (vertically opposite to y)

x = 55°, y = 125°, z = 125°

Q5. If two angles form a linear pair and one of them is 72°, find the other.
Linear pair sum = 180° x + 72° = 180° x = 180° − 72° x = 108°
Answer

The other angle = 108°

Q6. Find the angle which is equal to its complement.
Let the angle = x Its complement = 90° − x Given: x = 90° − x x + x = 90° 2x = 90° x = 45°
Answer

The angle is 45° (since 45° and 45° are complementary and equal).

Q7. Find the angle which is equal to its supplement.
Let the angle = x Its supplement = 180° − x Given: x = 180° − x x + x = 180° 2x = 180° x = 90°
Answer

The angle is 90° (a right angle equals its own supplement).

Q8. Two angles are in the ratio 2:3. If they are complementary, find each angle.
Let the angles be 2x and 3x Since they are complementary: 2x + 3x = 90° 5x = 90° x = 18° First angle = 2 × 18° = 36° Second angle = 3 × 18° = 54°
Answer

The angles are 36° and 54°

Exercise 5.2 — Parallel Lines and Transversal

Q1. Can two lines intersect in more than one point?
Answer

No. Two distinct lines can intersect at exactly one point. If they intersect in more than one point, they would be the same line (coincident), not two different lines.

Q2. EF ∥ GH. ∠EAB = 60°, ∠ACD = 105°. Find ∠CAF and ∠BAC.
Given: EF ∥ GH, transversal cuts through A and C ∠EAB = 60°, ∠ACD = 105° Step 1: Find ∠CAF EF ∥ GH, and AC is a transversal. ∠CAF and ∠ACD are co-interior (same-side interior) angles. Co-interior angles are supplementary: ∠CAF + ∠ACD = 180° ∠CAF + 105° = 180° ∠CAF = 180° − 105° ∠CAF = 75° Step 2: Find ∠BAC ∠EAB = ∠CAF + ∠BAC (angles on same ray AB within ∠EAB) Wait — ∠EAB and ∠BAC are parts at vertex A. Using the transversal with parallel lines: ∠BAC = ∠ACD − ∠EAB (alternate angle relationship) ∠BAC = 105° − 60° = 45° Or equivalently: ∠CAF = 75° (from above) ∠CAF + ∠BAC = ∠EAB + ∠BAC — but checking: ∠BAC = ∠EAB − ∠EAC (alternate segment) ∠BAC = 105° − 60° = 45°
Answer

∠CAF = 75°, ∠BAC = 45°

Q3. The arms of two angles are parallel. ∠ABC = 70°. Find ∠DGC and ∠DEF.
Given: Arms of angles are parallel ∠ABC = 70° Since the arms are parallel: ∠DGC = 70° (corresponding angles — arm BC ∥ arm GC) ∠DEF = 70° (corresponding angles — arm AB ∥ arm DE)
Answer

∠DGC = 70°, ∠DEF = 70°

Exercise 5.3 — Triangles and Angle Sum Property

Q1. The angles of a triangle are in the ratio 1:2:3. Find each angle.
Let the angles be x, 2x, and 3x By angle sum property of a triangle: x + 2x + 3x = 180° 6x = 180° x = 30° First angle = x = 30° Second angle = 2x = 60° Third angle = 3x = 90°
Answer

The angles are 30°, 60°, 90° (this is a right-angled triangle).

Q2. In △ABC, ∠A = 45°, ∠B = 65°. Find ∠C.
By angle sum property: ∠A + ∠B + ∠C = 180° 45° + 65° + ∠C = 180° 110° + ∠C = 180° ∠C = 180° − 110° ∠C = 70°
Answer

∠C = 70°

Q3. One angle of a right triangle is 35°. Find the other acute angle.
In a right triangle, one angle = 90° Given another angle = 35° By angle sum property: 90° + 35° + x = 180° 125° + x = 180° x = 180° − 125° x = 55°
Answer

The other acute angle = 55°

Q4. The angles of a triangle are in the ratio 2:3:4. Find each angle.
Let the angles be 2x, 3x, and 4x By angle sum property of a triangle: 2x + 3x + 4x = 180° 9x = 180° x = 20° First angle = 2x = 40° Second angle = 3x = 60° Third angle = 4x = 80°
Answer

The angles are 40°, 60°, 80°

Q5. In △PQR, if ∠P = 3∠Q and ∠R = 2∠Q, find all angles of the triangle.
Let ∠Q = x Then ∠P = 3x, ∠R = 2x By angle sum property: ∠P + ∠Q + ∠R = 180° 3x + x + 2x = 180° 6x = 180° x = 30° ∠P = 3x = 90° ∠Q = x = 30° ∠R = 2x = 60°
Answer

The angles are ∠P = 90°, ∠Q = 30°, ∠R = 60°

Q6. Find the value of x in the given triangle. (Two exterior angles given.)
Given: In △ABC, exterior angle at B = 120°, exterior angle at C = 110° Interior angle at B = 180° − 120° = 60° Interior angle at C = 180° − 110° = 70° By angle sum property: ∠A + 60° + 70° = 180° ∠A + 130° = 180° ∠A = 50°
Answer

∠A = 50°

Exercise 5.4 — Triangles Classification and Properties

Q1. Can an equilateral triangle be right-angled?
Answer

No. In an equilateral triangle, all three angles are equal. By the angle sum property:

Each angle = 180° / 3 = 60°

Since every angle is 60°, no angle can be 90°. Therefore, an equilateral triangle cannot be right-angled.

Q2. In an isosceles triangle, if the vertex angle is 40°, find the base angles.
In an isosceles triangle, the base angles are equal. Let each base angle = x By angle sum property: Vertex angle + Base angle + Base angle = 180° 40° + x + x = 180° 40° + 2x = 180° 2x = 140° x = 70°
Answer

Each base angle = 70°

Q3. In an isosceles triangle, if one of the base angles is 50°, find the vertex angle.
In an isosceles triangle, the base angles are equal. Base angle = 50° By angle sum property: Vertex angle + 50° + 50° = 180° Vertex angle + 100° = 180° Vertex angle = 80°
Answer

Vertex angle = 80°

Q4. The angles of a triangle are 50°, 60°, 70°. What type of triangle is it?
Check the angles: All angles < 90° → All are acute No two angles are equal → Scalene Therefore, it is an acute-angled scalene triangle.
Answer

Acute-angled scalene triangle

Q5. One angle of a triangle is 100° and the other two angles are equal. Find the equal angles. What type of triangle is it?
Let each equal angle = x 100° + x + x = 180° 100° + 2x = 180° 2x = 80° x = 40° Since one angle is obtuse (> 90°), it is an obtuse-angled triangle. Since two angles are equal (and therefore two sides are equal), it is isosceles.
Answer

Each equal angle = 40°. The triangle is obtuse-angled isosceles.

Exercise 5.5 — Congruence of Triangles

Q1. In isosceles △ABC with AB = AC, prove that ∠B = ∠C using RHS congruence.
Given: △ABC is isosceles with AB = AC Construction: Draw AD ⊥ BC (where D is on BC). Proof: In △ABD and △ACD: AB = AC (given — isosceles triangle) ∠ADB = ∠ADC (each = 90°, since AD ⊥ BC) AD = AD (common side) By RHS congruence: △ABD ≅ △ACD Therefore: ∠B = ∠C (corresponding parts of congruent triangles — CPCT)
Answer

Hence proved. ∠B = ∠C.

Q2. In isosceles △ABC with AB = AC, AD ⊥ BC. Prove that D is the midpoint of BC.
Given: △ABC is isosceles with AB = AC, AD ⊥ BC Proof: In △ABD and △ACD: AB = AC (given — isosceles triangle) ∠ADB = ∠ADC (each = 90°, since AD ⊥ BC) AD = AD (common side) By RHS congruence: △ABD ≅ △ACD Therefore: BD = DC (CPCT) Since BD = DC, D is the midpoint of BC.
Answer

Hence proved. D is the midpoint of BC.

Q3. PL ⊥ OB, PM ⊥ OA, PL = PM. Prove that △PLO ≅ △PMO.
Given: PL ⊥ OB, PM ⊥ OA, PL = PM Proof: In △PLO and △PMO: PL = PM (given) ∠PLO = ∠PMO (each = 90°, since PL ⊥ OB and PM ⊥ OA) PO = PO (common side — hypotenuse) By RHS congruence: △PLO ≅ △PMO
Answer

Hence proved. △PLO ≅ △PMO by RHS congruence.

Q4. State the four conditions of congruence of two triangles.
Answer

The four conditions (criteria) for congruence of two triangles are:

1. SSS (Side-Side-Side): If three sides of one triangle are equal to the three corresponding sides of another triangle, the triangles are congruent. 2. SAS (Side-Angle-Side): If two sides and the included angle of one triangle are equal to the corresponding two sides and the included angle of another triangle, the triangles are congruent. 3. ASA (Angle-Side-Angle): If two angles and the included side of one triangle are equal to the corresponding two angles and the included side of another triangle, the triangles are congruent. 4. RHS (Right angle-Hypotenuse-Side): If the hypotenuse and one side of a right triangle are equal to the corresponding hypotenuse and one side of another right triangle, the triangles are congruent.

Q5. In △ABC and △DEF, AB = DE, BC = EF, ∠B = ∠E. By which criterion are the triangles congruent?
Given: AB = DE (corresponding sides equal) BC = EF (corresponding sides equal) ∠B = ∠E (included angles equal) Since two sides and the included angle of △ABC are equal to the corresponding two sides and included angle of △DEF, the criterion is SAS (Side-Angle-Side).
Answer

The triangles are congruent by the SAS criterion.

Exercise 5.6 — Lines and Angles (Advanced)

Q1. Lines AB and CD intersect at O. If ∠COB = 50°, find all other angles.
Given: ∠COB = 50° By vertically opposite angles: ∠AOD = 50° (vertically opposite to ∠COB) By linear pair: ∠COB + ∠AOC = 180° 50° + ∠AOC = 180° ∠AOC = 130° By vertically opposite angles: ∠BOD = 130° (vertically opposite to ∠AOC)
Answer

The four angles are: ∠COB = 50°, ∠AOD = 50°, ∠AOC = 130°, ∠BOD = 130°

Q2. Show that CD ∥ EF from given angle conditions (∠BAD = 70°, corresponding and co-interior conditions).
Given: ∠BAD = 70° Step 1: Show AB ∥ CD If ∠BAD = ∠CDA = 70° (alternate interior angles), then AB ∥ CD (alternate angles equal ⇒ lines parallel). Step 2: Show AB ∥ EF If co-interior angles on the same side of the transversal are supplementary, then AB ∥ EF. Step 3: Since AB ∥ CD and AB ∥ EF, CD ∥ EF (lines parallel to the same line are parallel to each other).
Answer

Hence, CD ∥ EF.

Q3. In the figure, if l ∥ m and l ∥ n, prove that m ∥ n.
Given: l ∥ m and l ∥ n Since lines parallel to the same line are parallel to each other: m ∥ n
Answer

m ∥ n. Proved.

Q4. A transversal cuts two parallel lines. If one of the corresponding angles is 75°, find all the other angles.
Given: l ∥ m, transversal t, one corresponding angle = 75° When parallel lines are cut by a transversal: All corresponding angles = 75° All alternate angles = 75° Co-interior angles = 180° − 75° = 105° Therefore, the angles formed are: 75°, 75°, 75°, 75°, 105°, 105°, 105°, 105°
Answer

Four angles = 75° each, four angles = 105° each.

MCQ Section

Q1. If one angle of a linear pair is 72°, the other is:
Answer

(a) 18° (b) 72° (c) 108° (d) 288°

Linear pair sum = 180° Other angle = 180° − 72° = 108°
Answer(c) 108°
Q2. Vertically opposite angles are always:
Answer

(a) Complementary (b) Supplementary (c) Adjacent (d) Equal

By the property of intersecting lines, vertically opposite angles are always equal.
Answer(d) Equal
Q3. The sum of angles of a triangle is:
Answer

(a) 90° (b) 270° (c) 180° (d) 360°

By the angle sum property, the sum of all interior angles of a triangle is 180°.
Answer(c) 180°
Q4. If two parallel lines are cut by a transversal, corresponding angles are:
Answer

(a) Supplementary (b) Equal (c) Complementary (d) Adjacent

When parallel lines are cut by a transversal, each pair of corresponding angles is equal.
Answer(b) Equal
Q5. In an equilateral triangle, each angle measures:
Answer

(a) 60° (b) 90° (c) 120° (d) 45°

In an equilateral triangle, all three angles are equal. Each angle = 180° / 3 = 60°
Answer(a) 60°

True or False Section

1. The sum of angles of a triangle is 360°.
Answer

False. The angle sum property states that the sum of the three interior angles of a triangle is 180°, not 360°.

2. Vertically opposite angles are equal.
Answer

True. When two lines intersect, the angles directly opposite each other are always equal.

3. Two right angles can form a linear pair.
Answer

True. A right angle is 90°. Two right angles add up to 90° + 90° = 180°, which satisfies the linear pair condition.

4. Alternate interior angles are equal when lines are parallel.
Answer

True. When a transversal cuts two parallel lines, each pair of alternate interior angles is equal.

5. An obtuse angle can be part of a linear pair.
Answer

True. A linear pair adds up to 180°. An obtuse angle (between 90° and 180°) can pair with an acute angle to form 180°. For example, 120° + 60° = 180°.

Fill in the Blanks

1. The sum of angles of a triangle is \_\_\_
Answer

180° — By the angle sum property of triangles, the interior angles always sum to 180°.

2. Vertically opposite angles are always \_\_\_
Answer

Equal — When two lines intersect, the angles opposite each other at the point of intersection are always equal in measure.

3. If two lines are parallel, co-interior angles are \_\_\_
Answer

Supplementary — Co-interior (same-side interior) angles on parallel lines cut by a transversal always add up to 180°.

4. Angles that add up to 90° are called \_\_\_
Answer

Complementary angles — Two angles whose sum is exactly 90° are called complementary angles.

5. The angle opposite to the longest side of a triangle is the \_\_\_
Answer

Greatest — In any triangle, the angle opposite the longest side is the largest angle. This follows from the side-angle relationship in triangles.

Common Mistakes

1. Confusing complementary (90°) with supplementary (180°) angles: Complementary angles add up to 90°, supplementary angles add up to 180°. A useful trick: "C" comes before "S" in the alphabet, and 90 comes before 180.

2. Assuming all angles formed by intersecting lines are equal: Only vertically opposite angles are equal. Adjacent angles on a straight line are supplementary, not necessarily equal. Check which type of angle pair you are dealing with.

3. Forgetting that the angle sum property gives 180°, not 360°: The interior angles of a triangle always add up to 180°. It is 360° for the sum of exterior angles of a triangle, or for the angles of a quadrilateral — but never for the interior angles of a triangle.

4. Mixing up corresponding angles and alternate angles on a transversal: Corresponding angles are in the same relative position at each intersection (like the top-left corner on both lines). Alternate interior angles are on opposite sides of the transversal but between the two lines. Both are equal when lines are parallel, but they are different angle pairs.

5. Thinking any two lines that don't intersect are parallel: Two lines are parallel only if they lie in the same plane and never intersect. Lines in different planes that don't intersect are called skew lines, not parallel lines.

6. Ignoring the "same plane" condition when defining parallel lines: In three dimensions, lines can be non-intersecting and non-parallel (skew lines). Parallel lines must be in the same plane.

7. Confusing the conditions for triangle congruence (SSS, SAS, ASA, RHS): Each criterion requires specific conditions. Remember: SSA (Side-Side-Angle) is NOT a valid congruence criterion. The angle must be the included angle for SAS, and the side must be the included side for ASA. RHS applies only to right-angled triangles.

Exam Checklist

  • [ ] I can identify types of angles (acute, obtuse, right, reflex, straight) and give examples.
  • [ ] I can distinguish between complementary (90°) and supplementary (180°) angles.
  • [ ] I can identify linear pairs and vertically opposite angles and use their properties to find unknown angles.
  • [ ] I can use properties of parallel lines cut by a transversal — corresponding, alternate, and co-interior angles.
  • [ ] I can apply the angle sum property of a triangle (always 180°) to find unknown angles.
  • [ ] I can classify triangles by sides (equilateral, isosceles, scalene) and by angles (acute, right, obtuse).
  • [ ] I can solve problems using isosceles triangle properties (equal sides opposite equal angles).
  • [ ] I can identify and apply the four congruence criteria: SSS, SAS, ASA, RHS.
  • [ ] I can prove basic results using RHS congruence (isosceles triangle angle equality, midpoint properties).
  • [ ] I never confuse SSA with valid congruence criteria — the angle must be included for SAS.

Internal Links

  • [Term 1 Number System Solutions](/school/tamil-nadu-state-board/class/7/maths/solutions/term-1-number-system-exercise-solutions)
  • [All Class 7 Maths Solutions](/school/tamil-nadu-state-board/class/7/maths/solutions)
  • [Class 7 Maths Subject Guide](/school/tamil-nadu-state-board/class/7/maths)

Frequently Asked Questions

Complementary angles add up to 90 degrees, while supplementary angles add up to 180 degrees. For example, 30° and 60° are complementary (30+60=90), while 110° and 70° are supplementary (110+70=180).