# RBSE Class 9 Maths Important Questions Chapter 6 Lines and Angles

Rajasthan Board RBSE Class 9 Maths Important Questions Chapter 6 Lines and Angles Important Questions and Answers.

Rajasthan Board NCERT New Syllabus RBSE Solutions for Class 9 Guide Pdf free download in Hindi Medium and English Medium are part of RBSE Solutions. Here we have given RBSE Class 9th Books Solutions.

## RBSE Class 9 Maths Chapter 6 Important Questions Lines and Angles

I. Multiple Choice Questions:
Choose the correct answer from the given four options.

Question 1.
In the given figure AOB is a straight line. If ∠AOC + ∠BOD = 95°, then ∠COD = ? (a) 95°
(b) 85°
(c) 90°
(d) 55°
(b) 85°

Question 2.
In the adjoining figure, AOB is a straight line: If x : y : z = 4 : 5 : 6, then y = ? (a) 60°
(b) 80°
(c) 48°
(d) 72°
(a) 60°

Question 3.
An exterior angle of a triangle is 105° and its two interior opposite angles are equal. Each of the equal angle is :
(a) 37$$\frac{1^{\circ}}{2}$$
(b) 52$$\frac{1^{\circ}}{2}$$
(c) 72$$\frac{1^{\circ}}{2}$$
(d) 75°
(b) 52$$\frac{1^{\circ}}{2}$$ Question 4.
In the adjoining figure if m ∥ n, then the value of x is: (a) 60°
(b) 55°
(c) 50°
(d) 45°
(b) 55°

Question 5.
In the adjoining figure AB ∥ CD ∥ EF and PQ ∥ RS. If ∠CQP = 60° and ∠RQD = 25° then ∠QRS is equal to: (a) 85°
(b) 135°
(c) 145°
(d) 110°
(c) 145°

Question 6.
In the given figure straight lines AB and CD intersect at O. If ∠AOC = Φ, ∠BOC = θ and θ = 3Φ, then=? (a) 30°
(b) 40°
(c) 45°
(d) 60°
(c) 45°

Question 7.
In the given figure AB ∥ CD, if ∠ABO = 130° and ∠OCD = 1100, then ∠BOC =? (a) 50°
(b) 60°
(c) 70°
(d) 80°
(b) 60°

Question 8.
In the given figure AB ∥ CD, if ∠AOC = 30° and ∠OAB = 100°, then ∠OCD = ? (a) 130°
(b) 150°
(c) 80°
(d) 100°
(a) 130° Question 9.
In the given figure AB ∥ CD. If ∠CAB = 80° and ∠EFC = 25°, then ∠CEF = ? (a) 65°
(b) 55°
(c) 45°
(d) 75°
(b) 55°

Question 10.
In the given figure, AOB is a straight line. If ∠AOC = (∠x - 10), ∠COD = 50° and ∠BOD = (∠C + 20), then ∠AOC = ? (a) 40°
(b) 60°
(c) 80°
(d) 50°
(c) 80°

Question 11.
In ΔABC, side BC is produced to D. If ∠ABC = 40° and ∠ACZ = 120° then ∠A= ? (a) 60°
(b) 40°
(c) 80°
(d) 50°
(c) 80°

Question 12.
In the given figure BO and CO are the bisectors of ∠B and ∠C respectively. If ∠A = 50°, then ∠BOC = ? (a) 130°
(b) 100°
(c) 115°
(d) 120°
(c) 115°

Question 13.
In the given figure if ∠OAB = 75°, ∠OBA = 55° and ∠OCD = 100°, then ∠ODC =? (a) 20°
(b) 25°
(c) 30°
(d)35°
(c) 30°

Question 14.
In the given figure ∠BAC = 30°, ∠ABC = 50° and ∠CDE = 40°. Then ∠AED=? (a) 120°
(b) 100°
(c) 80°
(d) 110°
(a) 120°

Question 15.
In the given figure AB ∥ CD. If ∠APQ = 70° and ∠PRD = 120°,then ∠QPR=? ; (a) 50°
(b) 60°
(c) 40°
(d) 35°
(a) 50° Question 16.
In the given figure, if l ∥ m then x is equal to: (a) 55°
(b) 65°
(c) 75°
(d) 115°
(b) 65°

Question 17.
For what values of x shall we have l ∥ m? (a) x = 35
(b) x = 30
(c) x = 25
(d) x = 20
(c) x = 25

II. Fill in the Blanks :

Question 1.
If the complementary angles are in the ratio 1: 5, then the measure of the angles are _________ and _________ respectively.
15°, 75°

Question 2.
The sum of two adjacent angles is 100° and one of them is 35°, then the other is _________
65°

Question 3.
If two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio 2 : 3, then the greater of the two angles is _________.
108° Question 4.
Lines which are parallel to the same line are _________.
parallel to each other

Question 5.
If two parallel lines are intersected by a transversal, then each pair of alternate angles are _________.
equal.

III. True/False:
State whether the following statements are True or False.

Question 1.
Two congruent angles have same measure.
True

Question 2.
If a transversal intersects two parallel lines, then each pair of interior angles on the same side of transversal is complementary.
False

Question 3.
If the sum of two adjacent angles is 180°, then the non-common arms of the angles form a line.
True

Question 4.
A ray has a finite length.
False

Question 5.
Lines parallel to the same line are parallel to each other.
True Question 6.
If a line is perpendicular to one of the two given parallel lines then it is perpendicular to the other line.
True

Question 7.
If two parallel lines are intersected by a transversal, then each pair of corresponding angles is equal.
True

IV. Match the Columns :

Question 1.
Match the column I with the column II.

 Column I Column II (1) If x° and y° be the measures of two complementary angles, such that 2x = 3y, then x = (i) 45° (2) If an angle is the supplement of itself, then the measure of the angle is (ii) 60° (3) If an angle is the complement of itself, then the measure of the angle is (iii) 54° (4) If x° andy° be the angles forming a linear pair, such that x - y = 60°, then (iv) 90°

 Column I Column II (1) If x° and y° be the measures of two complementary angles, such that 2x = 3y, then x = (iii) 54° (2) If an angle is the supplement of itself, then the measure of the angle is (iv) 90° (3) If an angle is the complement of itself, then the measure of the angle is (i) 45° (4) If x° andy° be the angles forming a linear pair, such that x - y = 60°, then (ii) 60°

V. Very Short Answer Type Questions :

Question 1.
Two lines l and m are perpendicular to the same line n. Are l and m perpendicular to each other? Give reason for your answer.
No, since l and m are perpendicular to the line n‘. ∠1 = ∠2 = 90° (∵ l ⊥ n and min)
It implies that these are corresponding angles.
Hence, l ∥ m.

Question 2.
If one angle is equal to four times of its complement. Find the angle.
Let the angle be x.
Then, x = 4(90° - x)
x = 360° - 4x
⇒ 5x = 360°
⇒ x = $$\frac{360^{\circ}}{5}$$ = 72°
⇒ 360° = 72°

Question 3.
Find the supplement of $$\frac{3}{5}$$ of a right angle.
Given angle $$\frac{3}{5}$$ of a right angle = $$\frac{3}{5}$$ × 90° = 3 × 18° = 54°
Supplement of 54° = An angle of measure (180° - 54°)
= An angle of measure 126°
Hence, the required angle is 126°.

Question 4.
In the figure below, l1 ∥ l2 and a1 ∥ a2. Find the value of x. ∠1 = 4a + 15 (Corresponding angles)
2a = 180 - ∠1 (Corresponding angles)
⇒ 2a = 180° - (4a + 15)
⇒ 2a = 165 - 4x
⇒ 6a = 165 ⇒ x = $$\frac{165}{6}$$ = 27$$\frac{1^{\circ}}{2}$$ Question 5.
Determine the value of x. Here,
3x + 3x + x + 150° = 360°
⇒ 7x + 150° = 360°
⇒ 7x = 360° - 150° = 210°
⇒ x = $$\frac{210^{\circ}}{7}$$ = 30°

Question 1.
In the given figure, if l ∥ n in, then find the value of x.
Draw a line EF such that EF ∥ l ∥ m. Then, ∠FOD = ∠BDO = 44°
(Alternate interior angles)
and ∠FOA = ∠CAO = 65°
(Alternate interior angles)
Now, ∠x = ∠FOD + ∠FOA
= 44° + 650 = 105°

Question 2.
In ΔABC, ∠A = $$\frac{\angle B}{2}=\frac{\angle C}{6}$$ then find the measure of ∠A.
As in a ΔABC,
Given that ∠A + ∠B + ∠C = 180° ...(i)
⇒ ∠B = 2∠A
∠A = (Given) ....(2)
⇒ ∠C = 6∠A ...(3)

Substituting eq. (2) and (3) in (1), we get: ∠A + 2∠A + 6∠A = 180°
9∠A = 180°
∠A = $$\frac{180^{\circ}}{9}$$ = 20°
∠A = 20

Question 3.
Ray Obisects ∠AOB and OF is the ray opposite to 0E. Show that ∠FOB = ∠FOA. ∠FOB + ∠BOE = 180° ............(1) (Linear pair axiom)
∠FOA + ∠AOE = 180°
From equations (1) and (2),
∠FOB + ∠BOE = ∠FQA + ∠AOE
But, ∠BOE = ∠AOE (∵ Ray OE bisects ∠AOB)
∴ From equation (3),
⇒ ∠FOB = ∠FOA.
Hence proved.

Question 4.
In the given figure, AC ⊥ CE and ∠A: ∠B: ∠C = 5:3: 2. Find the value of ∠ECD. Let ∠A = 5x, ∠B = 3x and ¿C =
In ΔABC, we have ∠A + ∠B + ∠C = 180°
⇒ 5x + 3x + 2x = 180°
⇒ x = $$\frac{180^{\circ}}{10}$$ = 18°

∴ ∠A = 5x = 5 × 18° = 90°
∠B = 3x = 3 × 18° = 54° -
∠C = 2x = 2 × 18° = 36°
∠ACD = ∠BAC + ∠ABC
(∵ Exterior angle = Sinn of interior opposite angles)

⇒ ∠ACE + ∠ECD = 90° + 54°
∴ ∠ECD = 54° (∵ ∠ACE = 90°) Question 5.
In the given figure, if AB ∥ CD, then find the value of x. Draw a line EF ∥ AB ∥ CD. Since, AB ∥ EF and AE is a transversal.
∴ ∠BAE + ∠AEF = 180° (Co-interior angles)
⇒ 132° + ∠AEF = 180°
⇒ ∠AEF = 180° - 132° = 48°
Now, ∠FEC = ∠AEC - ∠AEF
= 148° - 48° = 100°

Since, EF ∥ CD and EC is a transversal.
∠FEC = ∠x - 100°
(By corresponding angles axiom)

Question 6.
In the given figure, what value of x will make RF ∥ CD, if AR ∥ CD? Giyen, AB ∥ CD and BC is a transversal.
⇒ ∠ABC = ∠BCD (Alternate interior angles)
⇒ ∠ABC = ∠BCE + ∠ECD
⇒ 65° = 250 + ∠ECD
∠ECD = 65° - 25°
⇒ ∠ECD = 40°
For EF CD, the sum of ∠FEC and ∠ECD should be equal to 180°.
∴ x + ∠ECD = 180° (Co-interior angles)
⇒ x + 40° = 180°
⇒ x = 180° - 40° = 140°
Hence, for x = 140°, EF will be parallel to CD.

Question 7.
In the given figure, QO and RO are the bisectors of Q and R, respectively. If ∠QPR = 75° and ∠PQR = 56°, then find the measure of ∠ORQ. In ΔPQR, we have ∠P + ∠Q + ∠R = 180°
(Since, sum of all the angles of a triangle is 180°.)
75° + 56° + ∠R = 180°
∠R = 180° - (75° + 56°) = 49°
But, ∠ORQ = $$\frac{1}{2}$$∠R (Since, OR is the bisector of ¿R.)
∠ORQ = $$\frac{1}{2}$$(49°) = 24$$\frac{1}{2}$$° VII. Long Answer Type Questions :

Question 1.
If two parallel lines are intersected by a transversal, prove that the bisectors of two pairs of interior angles enclose a rectangle.
Given AB ∥ CD and a transversal ‘f intersects them at E and F respectively.
EG, FG, EH and FH are bisectors of interior angles ∠AEF, ∠CFE, ∠BEF and ∠EFD respectively. To prove : EGFH is a rectangle. -
Proof : AB ∥ CD and transversal‘t’ intersects them at E and F respectively.
⇒ ∠AEF = ∠EFD (Alternate interior angles)
$$\frac{1}{2}$$(∠AEF) = $$\frac{1}{2}$$(∠EFD)
∠GEF = ∠EFH

But, these are alternate interior angles formed when the transversal EF intersects EG and FH.
⇒ EG ∥ FH.
Similarly, EH ∥ GF. Therefore, EGFH is a ∥gm.
Now ray EF stands on line AB.
∠AEF + ∠BEF = 180° (Linear pair)
$$\frac{1}{2}$$(∠AEF) + $$\frac{1}{2}$$(∠BEF) = 90°
⇒∠GEF + ∠HEF = 90°
⇒ ∠GEH = 90°
Thus, EGFH parallelogram has one right angle.
∴ EGFH is a rectangle.
Hence proved.

Question 2.
In the given figure, if TU ∥ SR and TR ∥ SV, then find ∠a and ∠b. Given, TU ∥ RS
∠UTR = ∠SRQ
(By corresponding angles axiom)
⇒ ∠SRQ = 90° (∵ ∠UTR = 90°)

In ΔRPQ, we have :
∠SRQ = ∠RPQ + ∠RQP
(∵ Exterior angle = Sum of interior opposite angles)
⇒ 90° = 50° + b
⇒ b = 90° - 50°
⇒ b = 40°
Also, given that TR ∥ SV
∠UTR = ∠VAU (Alternate interior angles)
∠VAU = 90° (∵ ∠UTR = 90°)
∠VUT = ∠UVA + ∠VAU
(∵ Exterior angle = Sum of interior opposite angles)
⇒ a = 25° + 90°
⇒ a = 115°
Hence, a = 115° and b = 40°. Question 3.
The side BC of a ΔABC is produced, such that D is on ray BC. The bisector of A meet BC in L as shown in figure. Prove that ∠ABC + ∠ACD = 2 ∠ALC.
In ΔABC, we have : ext. ∠ACD = ∠B + ∠A
ext. ∠ACD = ∠B + 2∠A
(∵ AL is the bisector of ∠A)
∴ ∠A = 2∠1
⇒ ∠ACD = ∠B + 2∠1 ..........(1)

In ΔABL, we have :
ext. ∠ALC = ∠B + ∠BAL
⇒ ext. ∠ALC = ∠B + ∠1 ..........(2)
⇒ 2 .∠ALC = 2 ∠B + 2∠1 (Multiplying eq. (2) by 2)

Subtracting (1) from (2), we get:
2∠ALC - ∠ACD = 2∠B - ∠B + 2∠1 - 2∠1
2∠ALC - ∠ACD = ∠B
⇒ ∠ACD + ∠B = 2∠ALC
⇒ ∠ACD + ∠ABC = 2∠ALC.
Hence proved. Last Updated on April 29, 2022, 4:31 p.m.
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Published April 29, 2022