Students often refer to Ganita Prakash Class 7 Solutions and Class 7 Maths Chapter 5 Parallel and Intersecting Lines NCERT Solutions Question Answer to verify their answers.
NCERT Class 7 Maths Chapter 5 Parallel and Intersecting Lines Solutions Question Answer
Ganita Prakash Class 7 Chapter 5 Solutions Parallel and Intersecting Lines
NCERT Class 7 Maths Ganita Prakash Chapter 5 Parallel and Intersecting Lines Solutions Question Answer
5.1 Across The Line
Figure It Out (Page 108)
Question 1.
List all the linear pairs and vertically opposite angles you observe from the figure:
Solution:
(a) Linear pairs are:
- ∠a and ∠b
- ∠b and ∠c
- ∠c and ∠d
- ∠d and ∠a
(b) Pairs of vertically opposite angles are:
- ∠b and ∠d
- ∠a and ∠c
5.2 Perpendicular Lines, 5.3 Between Lines, 5.4 Parallel and Perpendicular Lines in Paper Folding
Figure It Out (Pages 113 – 114)
Question 1.
Draw some lines perpendicular to the lines given on the dot paper in the given figure.
Solution:
Let the given lines be l, m, n, and p.
In the figure, we draw lines l’, m’ and n’.
Here, l’ is perpendicular to l, m’ is perpendicular to m, and n’ is perpendicular to n.
Given line p is a diagonal of a 3 × 4 rectangle shown by dotted lines.
We draw p’ as the diagonal of a 4 × 3 rectangle shown by dotted lines.
Line p’ is perpendicular to p.
Question 2.
In the given figure, mark the parallel lines using the notation (single arrow, double arrow, etc). Mark the angle between perpendicular lines with a square symbol.
(a) How did you spot the perpendicular lines?
(b) How did you spot the parallel lines?
Solution:
We know that:
(a) Perpendicular lines are a pair of lines that intersect each other at right angles (90°) and
(b) Parallel lines are a pair of lines lying in the same plane and do not intersect each other.
In the given figure, we mark the parallel lines and also the angle between perpendicular lines as shown below:
Question 3.
In the following dot paper, draw different sets of parallel lines. The line segments can be of different lengths but should have dots as end points.
Solution:
On the given dot paper, we draw four sets of parallel lines as shown below:
Question 4.
Using your sense of how parallel lines look, try to draw lines parallel to the line segments on this dot paper.
(a) Did you find it challenging to draw some of them?
(b) Which ones?
(c) How did you do it?
Solution:
On the given dot paper, we draw lines a’, b’, c’, d’, e’, f’, g’, and h’.
Here, a, a’ are parallel, b, b’ are parallel, c, c’ are parallel, d, d’ are parallel, e, e’ are parallel, f, f’ are parallel, g, g’ are parallel and h, h’ are parallel.
(a) Drawing lines parallel to some of the given lines is challenging.
(b) Lines e, f, g, and h.
(c) The lines e, f, g, and h do not pass through the dots. Parallel lines to e, f, g, and h are drawn by considering only the end points of the lines.
Question 5.
In the given figure, which line is parallel to line a-line b or line c? How do you decide this?
Solution:
We extend the given lines a, b, and c as shown in the figure.
In the above figure, we find that lines a and c do not meet after extension, whereas lines a and b will intersect after extension on the left side.
∴ Line c is parallel to line a.
5.5 Transversals, 5.6 Corresponding Angles, 5.7 Drawing Parallel Lines
Figure It Out (Page 119)
Question 1.
Can you draw a line parallel to l, that goes through point A? How will you do it with the tools from your geometry box? Describe your method.
Solution:
We shall draw a line parallel to the given line l and passing through the point A by using a ruler and a set square.
Step 1: Place a ruler along the line as shown in the Figure.
Step 2: Slide a set square along the line l, so that its perpendicular side touches the point A. Draw a line l’ through A as shown in the Figure.
Step 3: Place a ruler along the line l’ and slide a set square along the line l’ so that its perpendicular side touches the point A. Draw a line l” through A as shown in the Figure.
Step 4: Lines l and l” are parallel lines. Line l” is parallel to l and passes through the given point A.
5.8 Alternate Angles
Figure It Out (Pages 123 – 125)
Question 1.
Find the angles marked below.
Solution:
(i) In the figure, parallel lines are intersected by the transversal l.
Here, ∠ABC = 48° and ∠BCD = a°
∠ABC and ∠BCD are alternate angles.
∴ ∠ABC = ∠BCD
∴ 48° = a°
∴ a = 48°
(ii) In the figure, parallel lines are intersected by the transversal l.
Here, ∠ABC = 52° and ∠BCD = b°.
∠ABC and ∠BCD are alternate angles.
∴ ∠ABC = ∠BCD
∴ 52° = b°
∴ b = 52°
(iii) In the figure, parallel lines are intersected by the transversal l.
Here, ∠ABC = 81° and ∠BCD = c°.
∠ABC and ∠BCD are alternate angles.
∴ ∠ABC = ∠BCD
∴ 81° = c°
∴ c = 81°
(iv) In the figure, parallel lines are intersected by the transversal l.
Here, ∠ABC = d° and ∠BCD = 99°.
∠ABC and ∠BCD are alternate angles.
∴ ∠ABC = ∠BCD
∴ d° = 99°
∴ d = 99°
(v) In the figure, parallel lines are intersected by the transversal l.
Here, ∠ABC = e° and ∠BCD = 69°.
∠ABC and ∠BCD are alternate angles.
∴ ∠ABC = ∠BCD
∴ e° = 69°
∴ e = 69°
(vi) In the figure, parallel lines are intersected by the transversal l.
Here, ∠ABC = f° and ∠BCD = 132°.
∠ABC and ∠BCD are interior angles.
∠ABC + ∠BCD = 180°
∴ f°+ 132° = 180°
∴ f + 132 = 180
∴ f = 180 – 132 = 48
∴ f = 48°
(vii) In the figure, parallel lines are intersected by the transversal l.
Here, ∠ABC = g° and ∠BDE = 122°.
∠ABC and ∠BDE are corresponding angles.
∴ ∠ABC = ∠BDE
∴ g° = 122°
∴ g = 122°
(viii) In the figure, parallel lines are intersected by the transversal l.
Here, ∠ABC = 75° and ∠BCD = h°.
∠ABC and ∠BCD are alternate angles.
∴ ∠ABC = ∠BCD
∴ 75° = h°
∴ h = 75°
(ix) In the figure, parallel lines are intersected by the transversal l.
Here, ∠ABC = i° and ∠BCD = 54°.
∠ABC and ∠BCD are alternate angles.
∴ ∠ABC = ∠BCD
∴ i° = 54°
∴ i = 54°
(x) In the figure, parallel lines are intersected by the transversal l.
Here, ∠ABC = 97° and ∠BCD = j°.
∠ABC and ∠BCD are alternate angles.
∴ ∠ABC = ∠BCD
∴ 97° = j°
∴ j = 97°
Question 2.
Find the angle represented by a.
Solution:
(i) In the figure, parallel lines are intersected by the transversal l.
Here, ∠ACB = 42° and ∠CEF = a.
∠ACB and ∠BCE are linear angles.
∴ ∠ACB + ∠BCE = 180°
⇒ 42° + ∠BCE = 180°
⇒ ∠BCE – 180° – 42° = 138°
And ∠BCE and ∠CEF are alternate angles.
∴ ∠BCE = ∠CEF
⇒ 138° = a
⇒ a = 138°
(ii) In the figure, parallel lines l, l’ are intersected by parallel transversals m and m’.
Here, ∠ABC = 62° and ∠DEE = a.
∠ABC and ∠BCE are alternate angles.
∴ ∠ABC = ∠BCE
⇒ 62° = ∠BCE
∠BCE and ∠GED are corresponding angles.
∴ ∠BCE = ∠GED
⇒ 62° = ∠GED
∠GED and ∠DEF are linear angles.
∴ ∠GED + ∠DEF = 180°
62° + a = 180°
⇒ a = 180° – 62° = 118°.
(iii) In the figure, parallel lines l, m, and n are intersected by the transversal p.
Here, ∠ABC = 110°, ∠BDC = 35° and ∠DEF = a.
∠ABC and ∠CBD are linear angles.
∴ ∠ABC + ∠CBD = 180°
⇒ 110° + ∠CBD = 180°
⇒ ∠CBD = 180° – 110° = 70°
In ∆BCD, we have
∠CBD + ∠BCD + ∠CDB = 180°.
∴ 70° + ∠BCD + 35° = 180°
⇒ ∠BCD = 180° – 105° = 75°
∠GCB and ∠BCD are linear angles.
∴ ∠GCB + ∠BCD = 180°
⇒ ∠GCB + 75° = 180°
⇒ ∠GCB = 180° – 75° = 105°
∠GCB and ∠DEF are corresponding angles.
∴ ∠GCB = ∠DEF
⇒ 105° = a
⇒ a = 105°
(iv) In the figure, parallel lines l and m are intersected by the transversals n and p.
We have ∠ABC = 67°, ∠DBE = 90° and ∠BED = a.
∠ABC, ∠ABE, and ∠DBE are linear angles.
∴ ∠ABC + ∠ABE + ∠DBE = 180°
⇒ 67° + ∠ABE + 90° = 180°
⇒ ∠ABE = 180° – 157° = 23°
∠ABE and ∠BED are alternate angles.
∴ ∠ABE = ∠BED
⇒ 23° = a
⇒ a = 23°.
Question 3.
In the figures below, what angles do x and y stand for?
Solution:
(i) In the figure, parallel lines l and m are intersected by transversals n and p.
We have ∠ABG = 90°, ∠BDG = 65°, ∠EDF = x and ∠DGH = y.
∠ABG and ∠DBG are linear angles.
∴ ∠ABG + ∠DBG = 180°
90° + ZDBG =180°
⇒ ∠DBG = 180° – 90° = 90°
And ∠DBG and ∠BDE are alternate angles.
∴ ∠DBG = ∠BDE
⇒ 90° = ∠BDE
∠EDF, ∠BDE, and ∠BDG are linear angles.
∴ ∠EDF + ∠BDE + ∠BDG = 180°
⇒ x + 90° + 65° = 180°
⇒ x = 180° – 155° = 25°
∠EDF and ∠BGD are corresponding angles.
∴ ∠EDF = ∠BGD
⇒ x = ∠BGD
⇒ ∠BGD = 25°
∠BGD and ∠DGH are linear angles.
∴ ∠BGD + ∠DGH = 180°
⇒ 25° + y = 180°
⇒ y = 180° – 25° = 155°
∴ x = 25° and y = 155°.
(ii) In the figure, parallel lines l and m are intersected by transversals n and p.
We have ∠GAD = x, ∠ABC = 53°, and ∠ACF = 78°.
∠DAE and ∠ABC are corresponding angles.
∴ ∠DAE = ∠ABC
⇒ ∠DAE = 53°
∠GAE and ∠ACF are corresponding angles.
∴ ∠GAE = ∠ACF
∠GAE = 78°
Also, ∠GAE = ∠GAD + ∠DAE
∴ 78° = x + 53°
⇒ x = 78° – 53° = 25°
Question 4.
In the figure below, ∠ABC = 45° and ∠IKJ = 78°. Find angles ∠GEH, ∠HEF, and ∠FED.
Solution:
In the figure, parallel lines l and m are intersected by transversals n and p.
Lines n and p intersect at E.
We are to find ∠GEH, ∠HEF, and ∠FED.
∠BKE and ∠IKJ are vertically opposite angles.
∴ ∠BKE = ∠IKJ
⇒ ∠BKE = 78°
∠EBK and ∠ABC are vertically opposite angles.
∴ ∠EBK = ∠ABC
⇒ ∠EBK = 45°
In ∆BEK, we have
∠EBK + ∠BKE + ∠KEB = 180°.
45° + 78° + ∠KEB = 180°
⇒ ∠KEB = 180° – 123° = 57°
∠HEF and ∠KEB are vertically opposite angles.
∴ ∠HEF = ∠KEB
⇒ ∠HEF = 51°
∠IKJ and ∠GEJ are corresponding angles.
∴ ∠IKJ = ∠GEJ
⇒ 78° = ∠GEJ
∠GEJ and ∠FED are vertically opposite angles.
∴ ∠GEJ = ∠FED
⇒ 78° = ∠FED
∠ABC and ∠BED are corresponding angles.
∴ ∠ABC = ∠BED
⇒ 45° = ∠BED
∠BED and ∠GEH are vertically opposite angles.
∴ ∠BED = ∠GEH
⇒ 45° = ∠GEH
We have ∠GEH = 45°, ∠HEF = 57° and ∠FED = 78°.
Question 5.
In the figure below, AB is parallel to CD, and CD is parallel to EF. Also, EA is perpendicular to AB. If ∠BEF = 55°, find the values of x and y.
Solution:
In the figure, parallel lines l, m, and n are intersected by transversals p and q. Also, line p is perpendicular to line l.
We have ∠DEF = 55°, ∠GBE = x, ∠HDE = y, and ∠IAB = 90°.
∠CDE and ∠DEF are alternate angles.
∴ ∠CDE = ∠DEF
⇒ ∠CDE = 55°
∠CDE and ∠HDE are linear angles.
∴ ∠CDE + ∠HDE = 180°
⇒ 55° + y = 180°
⇒ y = 180° – 55° = 125°
∠HDE and ∠GBE are corresponding angles.
∴ ∠HDE = ∠GBE
⇒ y = x
⇒ x = 125°
∴ We have x = 125° and y = 125°.
Question 6.
What is the measure of angle ∠NOP in the figure below?
[Hint: Draw lines parallel to LM and PQ through points N and O]
Solution:
Using a ruler and a set square, draw a line l parallel to LM and passing through point N.
Also, draw a line m parallel to PQ and passing through point O.
We have ∠LMN = 40°, ∠MNO = 96°, ∠NOP = a° and ∠OPQ = 52°.
∠LMN and ∠MNA are alternate angles.
∴ ∠LMN = ∠MNA
⇒ 40° = ∠MNA
Also, ∠MNO = ∠MNA + ∠ANO
∴ 96° = 40° + ∠ANO
⇒ ∠ANO = 96° – 40° = 56°
∠ANO and ∠NOC are alternate angles.
∠ANO = ∠NOC
⇒ 56° = ∠NOC
∠COP and ∠OPQ are alternate angles.
∴ ∠COP = ∠OPQ
⇒ ∠COP = 52°
Also, ∠NOP = ∠NOC + ∠COP
∴ a° = 56° + 52° = 108°.