Class 9 Exam  >  Class 9 Notes  >  Mathematics (Maths) Class 9  >  Short Answer Type Questions: Lines & Angles

Class 9 Maths Chapter 6 Question Answers - Lines and Angles

Q.1. If P, Q, and R are three collinear points, then name all the line segments determined by them.

Ans.Class 9 Maths Chapter 6 Question Answers - Lines and Angles

We can have the following line segments:

Class 9 Maths Chapter 6 Question Answers - Lines and Angles


Q.2. In the adjoining figure, identify at least four collinear points.

Class 9 Maths Chapter 6 Question Answers - Lines and Angles

Ans. The four collinear points are A, B, C, and R.
 

Q.3. Find the complement of 36°.
Ans. ∵ 36° + [Complement of 36°] = 90°
⇒ Complement of 36°= 90° - 36° = 54°

Q.4. Find the supplement of 105°.
Ans.  105° + [Supplement of 105°] = 180°
⇒ Supplement of 105° = 180° - 105° = 75°


Q.5. Angles ∠ P and 100° form a linear pair. What is the measure of ∠ P?
Ans. 
∵ The sum of the angles of a linear pair equal to 180°.
∴ ∠ P + 100° = 180° ⇒ ∠ P = 180° - 100 = 80°.

Q.6. In the adjoining figure, what is the measure of p?Class 9 Maths Chapter 6 Question Answers - Lines and AnglesAns. ∵ p and 120° form a linear pair.
∴ p + 120° = 180° ⇒ p = 180° - 120° = 60°

Q.7. In the adjoining figure, AOB is a straight line. Find the value of x.Class 9 Maths Chapter 6 Question Answers - Lines and AnglesAns. ∵ AOB is a straight line.
∴ ∠AOC + ∠COB = 180°
⇒ 63° + x = 180° ⇒ x = 180° - 63° = 117°


Q.8. In the given figure, AB, CD, and EF are three lines concurrent at O. Find the value of y.Class 9 Maths Chapter 6 Question Answers - Lines and AnglesAns. ∵ ∠AOE and ∠BOF and vertically opposite angles.
∴ ∠AOE = ∠BOF = 5y ..... (1)
Now, CD is a straight line,
⇒ ∠COE + ∠EOA + ∠AOD = 180°
⇒ 2y + 5y + 2y = 180° [From (1)]
⇒ 9y = 180°⇒ y = (180°/2)= 20°
Thus, the required value of y is 20°.
 

Q.9. In the adjoining figure, AB || CD and PQ is transversal. Find x.Class 9 Maths Chapter 6 Question Answers - Lines and AnglesAns. ∵ AB || CD and PQ is a transversal.
∴ ∠ BOQ = ∠ CQP [∵ Alternate angles are equal]
⇒ x = 110° [∵ ∠ CQP = 110°]
 

Q.10. Find the measure of an angle that is 26° more than its complement.
Ans. 
Let the measure of the required angle be x.
∴ Measure of the complement of x° = (90° - x)
⇒ x° - (90° - x) = 26°
⇒ x - 90° + x = 26°
⇒ 2x = 26° + 90° = 116°
⇒ x = (116°/2) = 58°
Thus, the required measure = 58°.


Q.11. Find the measure of an angle if four times its complement is 10° less than twice its complement.
Ans. Let the measure of the required angle be x.
∴ Its complement = (90° - x) and Its supplement = (180° - x)
According to the condition:

4(Complement of x) = 2(Supplement of x) - 10

⇒ 4(90° - x) = 2(180° - x) - 10°
⇒ 360° - 4x = 360° - 2x - 10°
⇒ 4x - 2x = 360° - 360° + 10°
⇒ 2x = 10° ⇒ x = (10°/2)= 5°
Thus, the measure of the required angle is 5°.


Q.12. Two supplementary angles are in the ratio 3:2. Find the angles.
Ans. Let the measure of the two angles be 3x and 2x.
∵ They are supplementary angles.
∴ 3x + 2x = 180°
⇒ 5x = 180° ⇒ x = (180°/5) = 36°
∴ 3x = 3 x 36° = 108° and 2x = 2 x 36° = 72°
Thus, the required angles are 108° and 72°.


Q.13. In the adjoining figure, AOB is a straight line.

Class 9 Maths Chapter 6 Question Answers - Lines and Angles

Ans. ∵ AOB is a straight line.
∴ ∠ AOC + ∠ BOC = 180°
⇒ (3x + 10°) + (2x - 30°) = 180° [Linear pair]
⇒ 3x + 2x + 10° - 30° = 180°
⇒ 5x - 20° = 180°
⇒ 5x = 180° + 20° = 200° ⇒ x = (200°/5) = 40°
Thus, the required value of x is 40°.


Q.14. In the adjoining figure, find ∠ AOC and ∠ BOD.Class 9 Maths Chapter 6 Question Answers - Lines and AnglesAns. ∵ AOB is a straight line.
∴ ∠AOC + ∠COD + ∠DOB =180°
⇒ x + 70° + (2x - 25°) = 180°
⇒ x + 2x = 180° + 25° - 70°
⇒ 3x = 205° - 70° = 135° ⇒ x = (135°/3) = 45°
∴ ∠ AOC = 45°
⇒ ∠ BOD = 2x - 25° = 2 (45°) - 25° = 90° - 25° = 65°


Q.15. In the adjoining figure, AB || CD. Find the value of x.Class 9 Maths Chapter 6 Question Answers - Lines and AnglesAns. Let us draw EF || AB and pass through point O.
∴ EF || CD and CO is a transversal.
⇒ ∠ 1 = 25° [Alternate angles]
Similarly, ∠ 2 = 35°
Adding, ∠ 1 + ∠ 2 = 25° + 35° ⇒ x = 60°
Thus, the required value of x is 60°.

The document Class 9 Maths Chapter 6 Question Answers - Lines and Angles is a part of the Class 9 Course Mathematics (Maths) Class 9.
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FAQs on Class 9 Maths Chapter 6 Question Answers - Lines and Angles

1. What are the basic types of angles in geometry?
Ans. The basic types of angles in geometry are acute angles (less than 90 degrees), right angles (exactly 90 degrees), obtuse angles (greater than 90 degrees but less than 180 degrees), straight angles (exactly 180 degrees), and reflex angles (greater than 180 degrees).
2. How do you calculate the sum of angles in a triangle?
Ans. The sum of the interior angles in a triangle is always 180 degrees. This means that if you have a triangle with angles A, B, and C, the equation A + B + C = 180° holds true.
3. What is the difference between complementary and supplementary angles?
Ans. Complementary angles are two angles whose sum is 90 degrees, while supplementary angles are two angles whose sum is 180 degrees. For example, if one angle is 30 degrees, its complement would be 60 degrees, and its supplement would be 150 degrees.
4. What are parallel lines and how do they relate to angles?
Ans. Parallel lines are lines in a plane that never meet and are always the same distance apart. When a transversal crosses parallel lines, it creates corresponding angles, alternate interior angles, and alternate exterior angles which have specific relationships, such as being equal or supplementary.
5. How can I identify vertical angles?
Ans. Vertical angles are formed when two lines intersect. They are opposite each other and are always equal in measure. For instance, if two lines intersect and create angles of 40 degrees and 140 degrees, the angles opposite to these (the vertical angles) will also measure 40 degrees and 140 degrees, respectively.
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