CBSE Class 6 Mathematics Lines and Angles Practice Paper Worksheet Set 02 with Answers

This is Class 6 Maths Ganita Prakash Chapter 2 Lines and Angles Practice Paper 04, a 40-mark test of 20 questions in Sections A to E, given here with step-by-step answers. It tests the vertex and arms of an angle, straight, right, acute, obtuse and reflex angles, drawing angles of 75°, 82° and 140° with a protractor, angles between clock hands at 2:00 and 4:00, and the angle sum of a triangle.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. In this figure, ∠TER = 80°. What is the measure of ∠BET? What is the measure of ∠SET?

Answer:
B, E and R lie on a straight line, so ∠BER = 180° (a straight angle).
The figure shows ∠SER = 90°, so ∠BES = 180° − 90° = 90°.
∠SET = ∠SER − ∠TER = 90° − 80° = 10°.
∠BET = ∠BES + ∠SET = 90° + 10° = 100°.
Check: ∠BET + ∠TER = 100° + 80° = 180°, the straight angle.
In simple words: ∠SET = 90° − 80° = 10° and ∠BET = 180° − 80° = 100°.

Teacher's Note:
1. Look for the straight line BER first.
2. Always check that the angles on a line add up to 180°.

Question 12. Draw two intersecting lines and label all the angles formed. Name and classify them.
Answer:
Draw two lines AB and CD crossing each other at the point O.

Four angles are formed at O: ∠AOD, ∠DOB, ∠BOC and ∠COA.
In this drawing ∠AOD = ∠BOC = 130°, so they are obtuse angles; ∠COA = ∠DOB = 50°, so they are acute angles.
The angles opposite each other (∠AOD and ∠BOC, ∠COA and ∠DOB) are always equal, and any two neighbouring angles add up to 180°.
If the lines crossed at right angles, all four angles would be 90° (right angles).
In simple words: Two crossing lines make four angles: two equal opposite pairs, usually two acute and two obtuse.

Teacher's Note:
1. Measure one angle with a protractor; the rest follow.
2. Neighbouring angles on a line add up to 180°.

Question 13. Write the steps to draw a 75° angle using a protractor. Name the vertex and arms.
Answer:
Step 1: Draw a ray OA with a ruler.
Step 2: Place the centre of the protractor exactly on O, with its base line along OA.
Step 3: Using the scale that reads 0 at A, find 75° and mark a point B there.
Step 4: Remove the protractor and join O to B. Then ∠AOB = 75°.

Vertex: O. Arms: rays OA and OB.
In simple words: Put the protractor's centre on O, mark 75° on the scale starting at 0, and join it to O.

Teacher's Note:
1. Use the scale whose 0 lies on the first arm.
2. 75° is acute, so the arm should lean less than upright.

Question 14. A triangle has one right angle and another angle of 30°. Find the third angle. What type of triangle is it?
Answer:
The three angles of a triangle add up to 180°.
Third angle = 180° − (90° + 30°) = 180° − 120° = 60°.
The triangle has one right angle (the other two, 30° and 60°, are acute), so it is a right-angled triangle.
All three angles are different, so all three sides are also different: it is a scalene right-angled triangle.
In simple words: The third angle is 60°, and the triangle is right-angled.

Teacher's Note:
1. A triangle is named by its largest angle.
2. A triangle can have at most one right angle.

Section C: Short Answer Questions (3 Marks Each)

Question 15. You are given a straight angle ∠AOB.
(a) draw it and divide into two right angles
(b) then divide each right angle into two equal parts
(c) how many angles of 45° are formed?
Answer:
(a) Draw a straight line AB and mark a point O on it, so ∠AOB = 180°. At O draw a ray OC perpendicular to AB (use the 90° mark of the protractor). This gives two right angles, ∠AOC = ∠COB = 90°.

(b) Half of 90° is 45°. Draw ray OD at 45° inside ∠AOC and ray OE at 45° inside ∠COB.

(c) The small angles are ∠AOD, ∠DOC, ∠COE and ∠EOB, each 45°. So 4 angles of 45° are formed (4 × 45° = 180°).
In simple words: A straight angle splits into 2 right angles, and these split into 4 angles of 45°.

Teacher's Note:
1. Halving 180° twice gives 45°.
2. Count only the smallest angles between neighbouring rays.

Question 16. Draw angles with the following degree measures: (a) 140°, (b) 82°
Answer:
(a) Draw a ray from B. Place the protractor's centre on B with the base along the ray, mark 140° on the scale that starts at 0 on the ray, and join it to B. 140° is obtuse, so the second arm leans back past the upright.
(b) Repeat with 82°. 82° is acute, a little less than a right angle, so the second arm is nearly upright.

In simple words: Use the protractor scale starting at 0 on the first arm and mark 140° and 82°.

Teacher's Note:
1. Check the type first: 140° obtuse, 82° acute.
2. Read the scale that starts at 0 on your base ray.

Question 17. Draw the letter ‘M’ such that the angles on the sides are 40° each and the angle in the middle is 60°.
Answer:
Draw the two slanting middle strokes first so that they meet at the bottom with an angle of 60° (each stroke makes 30° with the upright direction).
At the top end of each slanting stroke, use the protractor to draw an outer stroke making 40° with it.
Since 30° + 10° = 40°, each outer stroke leans 10° outwards from the upright, so the letter spreads slightly at the bottom.

Check with a protractor: 40° at each top corner and 60° at the middle.
In simple words: Make the middle V 60° wide and add outer strokes at 40° to each slanting stroke.

Teacher's Note:
1. If the outer strokes were upright, the middle angle would have to be 80°.
2. Measure every angle again after drawing.

Section D: Long Answer Question (5 Marks)

Question 18. Riya drew a triangle with angles 90°, 45°, and 45°.
(a) identify the type of triangle
(b) draw it with proper markings
(c) draw perpendicular lines from each vertex to the opposite side
(d) how many right angles are formed in the entire construction?
(e) explain how this triangle follows the angle sum property
Answer:
(a) It has a 90° angle, so it is a right-angled triangle. Its other two angles are equal (45° each), so it is also isosceles: a right-angled isosceles triangle.
(b) and (c) Draw BC, make 45° angles at B and C, and let the arms meet at A; then ∠A = 90°. From A draw AD perpendicular to BC.

From B, the perpendicular to the opposite side AC is BA itself, because ∠A = 90°. In the same way, the perpendicular from C to AB is CA itself. So only one new line, AD, is needed.
(d) Right angles: ∠BAC = 90°, and AD makes ∠ADB = 90° and ∠ADC = 90°. So 3 right angles are formed in all. (AD splits ∠A into two angles of 45°.)
(e) ∠A + ∠B + ∠C = 90° + 45° + 45° = 180°, which is exactly the angle sum of any triangle. Each smaller triangle (ABD and ACD) also has 90° + 45° + 45° = 180°.
In simple words: It is a right-angled isosceles triangle with 3 right angles in the construction, and its angles add up to 180°.

Teacher's Note:
1. In a right triangle the two legs are already perpendicular to each other.
2. Count only the angles that really measure 90°.

Section E: Case Study Based Questions (4 Marks Each)

Question 19. Kabir organized a fun classroom challenge during the mathematics exhibition where students had to guess the angles formed between the hour and minute hands of a clock at different times. He asked everyone to calculate the exact angle at specific hours without using a protractor. For example, at 2:00 PM, the angle formed is x°, and at 4:00 PM, the angle formed is y°.
(a) find the values of x and y
(b) which angle is greater?
(c) are both angles acute, right, or obtuse?
(d) draw and label both clock faces showing the angles
Answer:
A full turn of 360° is split into 12 equal hour gaps, so each gap is 360° ÷ 12 = 30°.
(a) At 2:00 the hands are 2 gaps apart: x = 2 × 30° = 60°. At 4:00 they are 4 gaps apart: y = 4 × 30° = 120°.
(b) 120° > 60°, so y, the angle at 4:00 PM, is greater.
(c) 60° is less than 90°, so it is acute; 120° is between 90° and 180°, so it is obtuse.
(d) At 2:00 the minute hand is at 12 and the hour hand at 2; at 4:00 the minute hand is at 12 and the hour hand at 4.

In simple words: Each hour gap is 30°, so 2:00 gives 60° (acute) and 4:00 gives 120° (obtuse).

Teacher's Note:
1. Count the hour gaps between the hands.
2. At exact hours the minute hand is always at 12.

Question 20. In a school science lab, students were conducting an experiment related to angles and light. To observe the effect of sunlight, a window shade was initially pulled open, forming an angle of 95° with the wall. As the intensity of light needed to be increased, the shade was pulled further, creating a larger angle of 185°. This setup helped students understand how angles can influence light entry.
(a) what types of angles are these?
(b) how many degrees was the increase?
(c) draw both positions of the shade
(d) what type of angle will it be if the shade turns 270°?
Answer:
(a) 95° is more than 90° and less than 180°, so it is an obtuse angle. 185° is more than 180° and less than 360°, so it is a reflex angle.
(b) Increase = 185° − 95° = 90° (one right angle).
(c) Both positions share the same vertex (the hinge) and the same arm (the wall):

(d) 270° is more than 180° and less than 360°, so it is a reflex angle (three right angles).
In simple words: 95° is obtuse, 185° and 270° are reflex, and the shade turned through 90° more.

Teacher's Note:
1. Compare each angle with 90°, 180° and 360°.
2. An angle bigger than a straight angle is reflex, not obtuse.