Mathematics Solutions Solutions for Class 8 Maths Chapter 2 Parallel Lines And Transversal are provided here with simple step-by-step explanations. These solutions for Parallel Lines And Transversal are extremely popular among class 8 students for Maths Parallel Lines And Transversal Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Solutions Book of class 8 Maths Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Solutions Solutions. All Mathematics Solutions Solutions for class 8 Maths are prepared by experts and are 100% accurate.
Page No 8:
Question 1:
In the adjoining figure, each angle is shown by a letter. Fill in the boxes with the help of the figure.
Corresponding angles.
(1) ∠p and â
(2) ∠q and â
(3) ∠r and â
(4) ∠s and â
Interior alternate angles.
(5) ∠s and â
(6) ∠w and â
Answer:
Corresponding angles : If the arms on the transversal of a pair of angles are in the same direction and the other arms are on the same side of transversal, then it is called a pair of corresponding angles.
Corresponding angles
(1) ∠p and
(2) ∠q and
(3) ∠r and
(4) ∠s and
Alternate interior angles : A pair of angles which are on the opposite side of the transversal and inside the given lines that are intersected by the transversal.
Interior alternate angles
(5) ∠s and
(6) ∠w and
Page No 8:
Question 2:
Observe the angles shown in the figure and write the following pair of angles.
(1) Interior alternate angles
(2) Corresponding angles
(3) Interior angles
Answer:
(1) Alternate interior angles : A pair of angles which are on the opposite side of the transversal and inside the given lines that are intersected by the transversal.
Interior alternate angles
(a) ∠c and ∠e
(b) ∠b and ∠h
(2) Corresponding angles : If the arms on the transversal of a pair of angles are in the same direction and the other arms are on the same side of transversal, then it is called a pair of corresponding angles.
Corresponding angles
(a) ∠d and ∠h
(b) ∠c and ∠g
(c) ∠a and ∠e
(d) ∠b and ∠f
(3) Interior angles : A pair of angles which are on the same side of transversal and inside the given lines that are intersected by the transversal.
Interior angles
(a) ∠c and ∠h
(b) ∠b and ∠e
Page No 11:
Question 1:
1. Choose the correct alternative.
(1) In the adjoining figure, if line m ⥠line n and line p is a transversal then find x.
(A) 135
(B) 90
(C) 45
(D) 40
(2) In the adjoining figure, if line a ⥠line b and line l is a transversal then find x.
(A) 90
(B) 60
(C) 45
(D) 30
Answer:
(1)
Let us mark the points P and Q on m; R and S on n; A and B on p.
Suppose PQ and AB intersect at M; RS and AB intersect at N.
Since, m||n and p is a transversal, then
m∠QMN + m∠SNM = 180° (Interior angles on the same side of transversal are supplementary)
Substituing the values in the above equation, we get
3x + x = 180°
⇒ 4x = 180°
⇒ x =
∴ x = 45°
So, the correct answer is option (C).
(2)
Let us mark the points P and Q on a; R and S on b; A and B on l.
Suppose PQ and AB intersect at M; RS and AB intersect at N.
Since a||b and l is a transversal, then
m∠RNM = m∠SNB (Vertically opposite angles)
⇒ ∠RNM = 2x
Now, m∠RNM + m∠PMN = 180° (Interior angles on the same side of transversal are supplementary)
⇒ 2x + 4x = 180°
⇒ 6x = 180°
⇒ x =
⇒ x = 30°
So, the correct answer is option (D).
Page No 11:
Question 2:
In the adjoining figure line p ⥠line q. Line t and line s are transversals. Find measure of ∠x and ∠y using the measures of angles given in the figure.
Answer:
Let us mark the points P and Q on p; R and S on q; A and B on t; C and D on s.
Suppose PQ and AB intersect at K; PQ and CD intersect at X.
Suppose RS and AB intersect at L; RS and CD intersect at Y.
Since, AB is a straight line and ray KQ stands on it,
m∠AKX + m∠XKL = 180° (angles in linear pair)
⇒ 40° + m∠XKL = 180°
⇒ m∠XKL = 180° − 40°
⇒ m∠XKL = 140°
Since, p||q and t is a transversal, then
m∠YLB = m∠XKL (Corresponding angles)
⇒ x = 140°
Since, RS and CD are two straight lines intersecting at Y, then
m∠XYL = m∠SYD (Vertically opposite angles)
⇒ m∠XYL = 70°
Since, p||q and s is a transversal, then
m∠KXY + m∠XYL = 180° (Interior angles on same side of transversal are supplementary)
⇒ y + 70° = 180°
⇒ y = 180° − 70°
⇒ y = 110°
Page No 12:
Question 3:
In the adjoining figure. line p ⥠line q. line l ⥠line m. Find measures of ∠a, ∠b and ∠c, using the measures of given angles. Justify your answers.
Answer:
Let us mark the points A and B on p; X and Y on q; P and Q on l; R and S on m.
Suppose AB and XY intersect PQ at K and L respectively.
Suppose AB and XY intersect RS at N and M respectively.
Since, p||q and l is a transversal, then
m∠AKL + m∠XLK = 180° (Interior angles on same side of transversal are supplementary)
⇒ 80° + m∠XLK = 180°
⇒ m∠XLK = 180° − 80°
⇒ m∠XLK = 100°
Since, PQ and XY are straight lines that intersect at L, then
m∠QLM = m∠XLK (Vertically opposite angles)
⇒ a = 100°
Since, l||m and p is a transversal, then
m∠BNR = m∠AKL (Alternate exterior angles)
⇒ c = 80°
Since, p||q and m is a transversal, then
m∠NMY= m∠RNB (Corresponding angles)
⇒ b = c
⇒ b = 80°
Page No 12:
Question 4:
In the adjoining figure, line a ⥠line b. Line l is a transversal. Find the measures of ∠x, ∠y, ∠z using the given information.
Answer:
Let us mark the points A and B on l; K and M on a; L and N on b.
Suppose KM and LN intersect AB at P and Q respectively.
Since, a||b and l is a transversal, then
m∠PQL = m∠APK (Corresponding angles)
⇒ x = 105°
Since, AB and LN are straight lines that intersect at Q, then
m∠BQN = m∠PQL (Vertically opposite angles)
⇒ y = x
⇒ y = 105°
Since, AB is a straight line and ray QN stands on it, then
m∠BQN + m∠PQN = 180° (Angles in linear pair)
⇒ y + m∠PQN = 180°
⇒ 105° + m∠PQN = 180°
⇒ m∠PQN = 180° − 105°
⇒ m∠PQN = 75°
Now, m∠APM = m∠PQN (Corresponding angles)
⇒ z = 75°
Page No 12:
Question 5:
In the adjoining figure, line p ⥠line l ⥠line q. Find ∠x with the help of the measures given in the figure.
Answer:
Let us mark the points A, L and B on p; C, M and D on l; P, N and Q on q.
Since, AB||CD and LM is a transversal intersecting AB at L and CD at M, then
m∠LMD = âm∠ALM (Alternate interior angles)
⇒ m∠LMD = 40°
Since, CD||PQ and MN is a transversal intersecting CD at M and PQ at N, then
m∠DMN = âm∠PNM (Alternate interior angles)
⇒ m∠DMN = 30°
Now, m∠LMD + m∠DMN = 40° + 30°
⇒ m∠LMN = 70°
⇒ x = 70°
Page No 13:
Question 1:
Draw a line l. Take a point A outside the line. Through point A draw a line parallel to line l.
Answer:
Steps of construction :
(1) Draw a line l. Take a point A outside the line l.
(2) Draw a segment AM ⊥ line l.
(3) Take another point N on line l.
(4) Draw a segment NB ⊥ line l, such that l(NB) = l(MA).
(5) Draw a line m passing through the points A and B.
Hence, the line m is the required line that passes through point A and parallel to line l.
Page No 13:
Question 2:
Draw a line l. Take a point T outside the line. Through point T draw a line parallel to line l.
Answer:
Steps of construction :
(1) Draw a line l. Take a point T outside the line l.
(2) Draw a segment MT ⊥ line l.
(3) Take another point N on line l.
(4) Draw a segment NV ⊥ line l, such that l(NV) = l(MT).
(5) Draw a line m passing through the points T and V.
Hence, the line m is the required line that passes through point T and parallel to line l.
Page No 13:
Question 3:
Draw a line m. Draw a line n which is parallel to line m at a distance of 4 cm from it.
Answer:
Steps of construction :
(1) Draw a line m.
(2) Take two points A and B on the line m.
(3) Draw perpendiculars to the line m at A and B.
(4) On the perpendicular lines, take points P and Q at a distance of 4 cm from A and B respectively.
(5) Draw a line n passing through the points P and Q.
So, line n is the required line parallel to the line m at a distance of 4 cm away from it.
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