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# Short Answer Type Questions- Lines and Angles Class 9 Notes | EduRev

## Mathematics (Maths) Class 9

Created by: Full Circle

## Class 9 : Short Answer Type Questions- Lines and Angles Class 9 Notes | EduRev

The document Short Answer Type Questions- Lines and Angles Class 9 Notes | EduRev is a part of the Class 9 Course Mathematics (Maths) Class 9.
All you need of Class 9 at this link: Class 9

Question 1. If P, Q and R are three collinear points, then name all the line segments determined by them.
Solution: We can have the following line segments: Question 2. In the adjoining figure, identify at least four collinear points.
Solution:
The four collinear points are: A, B, C and R.

Question 3. Find the complement of 36°.
Solution:
∵ 36° + [Complement of 36°] = 90°
∴ Complement of 36°= 90° - 36°
= 54°

Question 4. Find the supplement of 105°.
Solution:
Since, 105° + [Supplement of 105°] = 180°
⇒ Supplement of 105° = 180° - 105° = 75°

Question 5. Angles ∠ P and 100° form a linear pair. What is the measure of ∠ P?
Solution:
Since, the sum of the angles of a linear pair equal to 180°.
∴ ∠ P + 100° = 180°
⇒ ∠ P = 180° - 100 = 80°.

Question 6. In the adjoining figure, what is the measure of p?
Solution:
∵ p and 120° form a linear pair. ∴ p + 120° = 180°
⇒ p = 180° - 120° = 60°

Question 7. In the adjoining figure, AOB is a straight line.
Find the value of x.
Solution:
Since AOB is a straight line, then ∠AOC + ∠COB = 180°
⇒ 63° + x = 180°
⇒ x = 180° - 63° = 117°

Question 8. In the given figure, AB, CD and EF are three lines concurrent at O. Find the value of y.
Solution:
Since, ∠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 = (1800/2)= 20°

Thus, the required value of y is 20°.

Question 9. In the adjoining figure, AB || CD and PQ is transversal. Find x.
Solution:
Since AB || CD and PQ is a transversal, then ∠ BOQ = ∠ CQP          [∴ Alternate angles are equal]
⇒ x = 110°                   [∵ ∠ CQP = 110°]

Question 10. Find the measure of an angle which is 26° more than its complement.
Solution:
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 = (1160/2)= 58°
Thus, the required measure = 58°.

Question 11. Find the measure of an angle if four times its complement is 10° less than twice its complement.
Solution:
Let the measure of the required angle be x.
∴ Its complement = (90° - x)
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 = (100/2)= 5°
Thus, the measure of the required angle is 5°.

Question 12. Two supplementary angles are in the ratio 3 : 2. Find the angles.
Solution:
Let the measure of the two angles be 3x and 2x Since, they are supplementary angles, 3x + 2x = 180°
⇒ 5x = 180°
⇒ x = (1800/5) = 36°
Thus, 3x = 3 x 36° = 108° and 2x = 2 x 36° = 72°
Thus, the required angles are 108° and 72°.

Question 13. In the adjoining figure, AOB is a straight line. Solution: ∵ AOB is a straight line, then ∠ AOC + ∠ BOC = 180°
∴ (3x + 10°) + (2x - 30°) = 180° [Linear pair]
⇒ 3x + 2x + 10° - 30° = 180°
⇒ 5x - 20° = 180°
⇒ 5x = 180° + 20° = 200°
⇒ x = (2000/5) = 40°
Thus, the required value of x is 40°.

Question 14. In the adjoining figure, find ∠ AOC and ∠ BOD.
Solution:
∵ AOB is a straight line, then ∠AOC + ∠COD + ∠DOB =180° ⇒ x + 70° + (2x - 25°) = 180°
⇒ x + 2x = 180° + 25° - 70°
⇒ 3x = 205° - 70° = 135°
⇒ x =(1350/3)
= 45°
∴ ∠ AOC = 45°
⇒ ∠ BOD = 2x - 25° = 2(45°) - 25°
= 90° - 25°
= 65°

Question 15. In the adjoining figure, AB || CD. Find the value of x.
Solution:
Let us draw EF || AB. ∴ EF || CD
Now EF || CD and CO being a transversal, then
∠ 1 = 25°                    [Alternate angles]
Similarly, ∠ 2 = 35°
∠ 1 + ∠ 2 = 25° + 35°
⇒ x = 60°
Thus, the required value of x is 60°.

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