HBSE Solutions

Parallel and Intersecting Lines Class 7 Solutions Ganita Prakash Maths Chapter 5✦ Step-by-step

HBSE 7th Class Maths Solutions Chapter 5 Parallel and Intersecting Lines (NCERT Ganita Prakash) — step-by-step solutions for every Figure It Out exercise. Tap Solve step by step on any question.

Class 7 Maths Chapter 5 Parallel and Intersecting Lines Solutions

Key ideas used in this chapter

Linear pairtwo adjacent angles on a line add to 180°
Vertically oppositeequal angles across an intersection
On parallel linescorresponding = equal, alternate = equal
Co-interiorinterior angles on the same side add to 180°

5.1 Across the Line — Figure It Out (Page 108)

Q1Angle pairs
List the linear pairs and vertically opposite angles at two intersecting lines forming ∠a, ∠b, ∠c, ∠d in order around the point.
Linear pairs
(∠a, ∠b), (∠b, ∠c), (∠c, ∠d), (∠d, ∠a)
Vertically opposite
(∠a, ∠c) and (∠b, ∠d)

5.2–5.7 Perpendicular, Parallel & Transversals

ConstructDrawing (Pages 113–119)
Draw perpendicular / parallel lines on dot paper, and a line parallel to l through a point A.
Perpendicular
two lines are perpendicular when they meet at 90° (mark with a small square).
Parallel
parallel lines never meet even when extended (mark with matching arrows).
Parallel through A
Slide a set-square along a ruler placed on l until its edge reaches A, then draw — the new line is parallel to l.

5.8 Alternate Angles — Figure It Out (Pages 123–125)

Q1Find the marked angle
Two parallel lines are cut by a transversal. Find (i) a, given 48° (ii) b, 52° (iii) c, 81° (iv) d, 99° (v) e, 69° (vi) f, with 132° (co-interior) (vii) g, 122° (viii) h, 75° (ix) i, 54° (x) j, 97°.
Alternate = equal
a=48°, b=52°, c=81°, d=99°, e=69°
(vi) co-interior
f + 132° = 180° ⇒ f = 48°
Rest
g=122° (corresponding), h=75°, i=54°, j=97° (alternate)
Q2Find a
(i) with 42° (ii) with 62° (iii) with 110° and 35° (iv) with 67° and 90°.
(i)
180° − 42° = 138°, then alternate ⇒ a = 138°
(ii)
62° carried across, then linear pair ⇒ a = 180° − 62° = 118°
(iii)
∠CBD = 70°, in ∆BCD ∠BCD = 75°, so a (corresponding) = 180°−75° = 105°
(iv)
∠ABE = 180°−67°−90° = 23°, alternate ⇒ a = 23°
Q3Find x and y
(i) with 90° and 65° (ii) with 53° and 78°.
(i)
x = 180° − 90° − 65° = 25°;  y = 180° − 25° = 155°
(ii)
∠GAE = 78° = x + 53° ⇒ x = 25°
Q4Three angles
Given ∠ABC = 45° and ∠IKJ = 78°, find ∠GEH, ∠HEF and ∠FED.
Triangle BEK
∠KEB = 180° − 45° − 78° = 57°
∠HEF
vertically opposite ∠KEB ⇒ 57°
Answers
∠GEH = 45°, ∠HEF = 57°, ∠FED = 78°
Q5Two transversals
AB ∥ CD ∥ EF, EA ⊥ AB, and ∠BEF = 55°. Find x and y.
Step 1
∠CDE = 55° (alternate); then y = 180° − 55° = 125°
Step 2
x = y (corresponding) ⇒ x = 125°. So x = 125°, y = 125°.
Q6Zig-zag angle
Find ∠NOP given ∠LMN = 40°, ∠MNO = 96°, ∠OPQ = 52° (draw lines through N and O parallel to LM and PQ).
Step 1
split ∠MNO: 40° (alternate to LMN) + ∠ANO = 96° ⇒ ∠ANO = 56°
Step 2
∠NOC = 56° (alternate) and ∠COP = 52° (alternate to OPQ)
Answer
∠NOP = 56° + 52° = 108°