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.

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**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 36Â°. Solution:** âˆµ 36Â° + [Complement of 36Â°] = 90Â°

âˆ´ Complement of 36Â°= 90Â° - 36Â°

= 54Â°

**Question 4. Find the supplement of 105Â°. Solution:** Since, 105Â° + [Supplement of 105Â°] = 180Â°

â‡’ Supplement of 105Â° = 180Â° - 105Â° = 75Â°

**Question 5. Angles âˆ P and 100Â° form a linear pair. What is the measure of âˆ P? Solution: **Since, the sum of the angles of a linear pair equal to 180Â°.

âˆ´ âˆ P + 100Â° = 180Â°

â‡’ âˆ P = 180Â° - 100 = 80Â°.

Solution:

âˆ´ p + 120Â° = 180Â°

â‡’ p = 180Â° - 120Â° = 60Â°

**Question 7. In the adjoining figure, AOB is a straight line. Find the value of x. Solution:** Since AOB is a straight line, then

âˆ AOC + âˆ COB = 180Â°

â‡’ 63Â° + x = 180Â°

â‡’ x = 180Â° - 63Â° = 117Â°

**Question 8. In the given figure, AB, CD and EF are three lines concurrent at O. Find the value of y. Solution: **Since, âˆ AOE and âˆ BOF and vertically opposite angles

âˆ´ âˆ AOE = âˆ BOF = 5y ..... (1)

Now, CD is a straight line,

â‡’ âˆ COE + âˆ EOA + âˆ AOD = 180Â°

â‡’ 2y + 5y + 2y = 180Â° [From (1)]

â‡’ 9y = 180Â°

â‡’ y = (180^{0}/2)= 20Â°

Thus, the required value of y is 20Â°.

**Question 9. In the adjoining figure, AB || CD and PQ is transversal. Find x. Solution:** Since AB || CD and PQ is a transversal, then

âˆ BOQ = âˆ CQP [âˆ´ Alternate angles are equal]

â‡’ x = 110Â° [âˆµ âˆ CQP = 110Â°]

**Question 10. Find the measure of an angle which is 26Â° more than its complement. Solution: **Let the measure of the required angle be x.

âˆ´ Measure of the complement of xÂ° = (90Â° - x)

â‡’ xÂ° âˆ (90Â° - x) = 26Â°

â‡’ x - 90Â° + x = 26Â°

â‡’ 2x = 26Â° + 90Â° = 116Â°

â‡’ x

Thus, the required measure = 58Â°.

**Question 11. Find the measure of an angle if four times its complement is 10Â° less than twice its complement. Solution:** Let the measure of the required angle be x.

âˆ´ Its complement = (90Â° - x)

Its supplement = (180Â° - x)

According to the condition, 4(Complement of x ) = 2(Supplement of x) - 10

â‡’ 4(90Â° - x) = 2(180Â° - x) - 10Â°

â‡’ 360Â° - 4x = 360Â° - 2x - 10Â°

â‡’ âˆ 4x + 2x = 360Â° - 360Â° - 10Â°

â‡’ âˆ 2x = -10Â°

â‡’ x =

Thus, the measure of the required angle is 5Â°.

**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 = 180Â°

â‡’ 5x = 180Â°

â‡’ x = (180

Thus, 3x = 3 x 36Â° = 108Â° and 2x = 2 x 36Â° = 72Â°

Thus, the required angles are 108Â° and 72Â°.

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

**Solution:** âˆµ AOB is a straight line, then âˆ AOC + âˆ BOC = 180Â°

âˆ´ (3x + 10Â°) + (2x - 30Â°) = 180Â° [Linear pair]

â‡’ 3x + 2x + 10Â° - 30Â° = 180Â°

â‡’ 5x - 20Â° = 180Â°

â‡’ 5x = 180Â° + 20Â° = 200Â°

â‡’ x = (200^{0}/5) = 40Â°

Thus, the required value of x is 40Â°.

**Question 14. In the adjoining figure, find âˆ AOC and âˆ BOD. Solution: **âˆµ AOB is a straight line, then âˆ AOC + âˆ COD + âˆ DOB =180Â°

â‡’ x + 70Â° + (2x - 25Â°) = 180Â°

â‡’ x + 2x = 180Â° + 25Â° - 70Â°

â‡’ 3x = 205Â° - 70Â° = 135Â°

â‡’ x =(135^{0}/3)

= 45Â°

âˆ´ âˆ AOC = 45Â°

â‡’ âˆ BOD = 2x - 25Â° = 2(45Â°) - 25Â°

= 90Â° - 25Â°

= 65Â°

**Question 15. In the adjoining figure, AB || CD. Find the value of x. Solution:** Let us draw EF || AB.

âˆ´ EF || CD

Now EF || CD and CO being a transversal, then

âˆ 1 = 25Â° [Alternate angles]

Similarly, âˆ 2 = 35Â°

Adding

âˆ 1 + âˆ 2 = 25Â° + 35Â°

â‡’ x = 60Â°

Thus, the required value of x is 60Â°.

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