Test: Parallel Lines And Transversals


15 Questions MCQ Test Mathematics (Maths) Class 9 | Test: Parallel Lines And Transversals


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This mock test of Test: Parallel Lines And Transversals for Class 9 helps you for every Class 9 entrance exam. This contains 15 Multiple Choice Questions for Class 9 Test: Parallel Lines And Transversals (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Parallel Lines And Transversals quiz give you a good mix of easy questions and tough questions. Class 9 students definitely take this Test: Parallel Lines And Transversals exercise for a better result in the exam. You can find other Test: Parallel Lines And Transversals extra questions, long questions & short questions for Class 9 on EduRev as well by searching above.
QUESTION: 1

Line A makes an angle of 30 degrees with the line B, also line C makes an angle of 30 degrees at the other end with line B on the opposite side. Then, ______.

Solution:
QUESTION: 2

In the figure below, the angle a = 150°. The other angle in the figure which is also 150° is:

Solution:
QUESTION: 3

If a transversal intersects two parallel lines, then each pair of corresponding angles is ………

Solution: By using corresponding angle axiom to find out the relation between the alternate interior angle when a transversal two parallel lines .In fig transversal PS intersect parallel lines AB and CD at points Q and R . angle BQR = Angle QRC andAQR = angle QRD we know that angle PQA = angle QRC ……1corresponding angle axiom angle PQA = angle BQR now , from 1 and 2 you may conclude angle BQR= angle QRCsimilarly,. angle AQR= angle QRD
QUESTION: 4

In the figure, PQ || MN, the value of x will be

Solution: First Mr parallel EF and RO is transversal line angle MRO=angle ROF=(alt.int .angle)toh angle MRO=55 degree and angle ROF =angle 1 toh 55=angle 1 is angle 1 =55 then second EF parallel PQ and OS is transversal line toh angle 2 = angle OSP is = (apt.int angle)toh angle 2 = 38 degree and Angle ROS = x hoga woo toh jnte hoge aap log toh x= angle 1 +angle 2 is toh angle 1hoga 55 or angle 2 hoga 38 toh x = 93
QUESTION: 5

The figure below shows a parallelogram. What is the measure of ∠b?

Solution:
QUESTION: 6

In the figure if l║m, n║p and ∠1 = 85°, then ∠2 is equal to

Solution:
QUESTION: 7

In the given figure, PQ||RS and EF||QS. If ∠PQS = 60°, then the measure of ∠RFE is:

Solution:

60°+ QSR=180°[co interior angles]

QSR=120°

RFE=QSR

So,RFE=120°

QUESTION: 8

In the above figure AB || CD. Also, ∠EAB = 84° and ∠ECD = 126°. Then, the value of “x” is

Solution:
QUESTION: 9

Observe the given figure and choose the correct statement

Solution: IN THIS CORRECT ANSWER IS D . BECAUSE IF WE CONSIDERED AB AND CD AS 2 LINE AND BC AND AD AS TWO TRANSVERSAL . THEN ANGLE 'B' = 65 AND ANGLE C = 115 THEN ANGLE B + C = 180 DEGREE BUT THEY ARE ANGLES ON THE SAME SIDE AND THEY ARE SUPPLYMENTRY BUT THIS ONLY HAPPENS IN A PARRALLEL LINES THERE AB || CD .
QUESTION: 10

In the figure, AB║CD. If ∠2 = (2x + 30)°, ∠4 = (x + 2y)° and ∠6 = (3y+10)° the measure of ∠5 is

Solution:
QUESTION: 11

In fig., if m||n and ∠a : ∠b = 2 : 3, the measure of ∠h is

Solution:
QUESTION: 12

In the figure, AB || CD and ∠F = 30° the value of ∠ECD is

Solution: In triangle AFE A+F+E=180 (angle sum property of triangle) 90+30+e=180 e=60 fea=bec=60 bec+dec=180. (co interior) dec=120degree.
QUESTION: 13

If two parallel lines are intersected by a transversal then, pair of alternate interior angles are:

Solution:
QUESTION: 14

In the adjoining figure, ABCD is a parallelogram in which ∠A = 72°. Calculate ∠B, ∠C and ∠D.

Solution:

ABCD is a parallelogram and ∠ A = 72°

We know that opposite angles of a parallelogram are equal

∴ ∠A = ∠C and ∠B = ∠D

∴ ∠C = 72°

∠A and ∠B are adjacent angles

i.e., ∠A + ∠B = 180°

⇒ ∠B = 180° – ∠A

⇒∠B = 180° – 72° = 108°

∠B = ∠D =108°

Hence, ∠B = ∠D = 108° and ∠C = 72°

QUESTION: 15

Two parallel lines l and m are such that A, C are the points on l and B, D are the points on m. AB and CD are perpendicular to both l and m. If AB = 4 cm, then CD =

Solution:

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