NCERT Solutions for Class 7 Maths Chapter 7: A Tale of Three Intersecting Lines – Ganita Prakash

Complete NCERT Solutions for Class 7 Maths Chapter 7 “A Tale of Three Intersecting Lines” from the Ganita Prakash textbook, covering the triangle inequality, triangle construction, angle sum property, and the exterior angle theorem. These Class 7 Mathematics Chapter 7 solutions are also useful as quick revision notes before exams.

7.1 What is a Triangle?

What happens when three vertices lie on a straight line?
Answer: When three points lie on a straight line, they are called collinear. They no longer form a triangle, since collinear points do not enclose any area — they simply lie along the same straight path.

7.2 Constructing a Triangle When Its Sides Are Given

General method (SSS construction): Draw one side as the base. From each endpoint of the base, draw an arc with radius equal to one of the remaining two given side lengths. The point where the two arcs intersect is the third vertex. Join it to both ends of the base to complete the triangle.

This method works for constructing triangles with sides such as 4,4,6 / 3,4,5 / 1,5,5 / 4,6,8 / 3.5,3.5,3.5 (all cm) — each of these sets satisfies the triangle inequality, so all are constructible.

Isosceles/equilateral triangles from circles: Connecting two points on a circle to the centre always forms an isosceles triangle, since any two radii of the same circle are equal in length. When two equal circles intersect, connecting their centres and an intersection point forms an equilateral triangle (since the radius equals the distance between centres). Note that every equilateral triangle is also a special case of an isosceles triangle.

7.3 The Triangle Inequality

Can a triangle with sides 3 cm, 4 cm, and 8 cm be constructed?
Answer: No — since 3 + 4 = 7, which is less than 8, the arcs would never meet, so the triangle cannot exist.

What about sides 3 cm, 3 cm, and 7 cm?
Answer: No. The direct path (7 cm) is longer than the roundabout path via the third vertex (3+3 = 6 cm), so this triangle is not possible.

The Triangle Inequality Rule: A triangle can exist only if the sum of any two sides is strictly greater than the third side. Equivalently, it’s enough to check that the sum of the two smaller sides exceeds the largest side.

Figure It Out: Checking Side Sets

Testing whether the following can be triangle side-lengths:

• (a) 2, 2, 5 — Cannot form a triangle (5 > 2+2 = 4)
• (b) 3, 4, 6 — Can form a triangle (all sums exceed the third side)
• (c) 2, 4, 8 — Cannot form a triangle (8 > 2+4 = 6)
• (d) 5, 5, 8 — Can form a triangle (8 < 5+5 = 10)
• (e) 10, 20, 25 — Can form a triangle (25 < 10+20 = 30)
• (f) 10, 20, 35 — Cannot form a triangle (35 > 10+20 = 30)
• (g) 24, 26, 28 — Can form a triangle (28 < 24+26 = 50)

More side-set checks: (a) 1, 100, 100 — exists (100 < 1+100=101) (b) 3, 6, 9 — does not exist (9 = 3+6 exactly, a degenerate case, not strictly less) (c) 1, 1, 5 — does not exist (5 > 1+1=2) (d) 5, 10, 12 — exists (all sums exceed the third side).

Does an equilateral triangle exist for any side length? Answer: Yes — for any positive side length x, x < x+x = 2x always holds, so an equilateral triangle always exists regardless of size.

Finding the valid range for a third side: if two sides are a and b (a < b), the third side c must satisfy (b − a) < c < (a + b). For example: with sides 1 and 100, the third side must be strictly between 99 and 101. With sides 5 and 5, the third side must be strictly between 0 and 10. With sides 3 and 7, the third side must be strictly between 4 and 10.

7.4 Angle Sum Property and Triangle Construction

SAS Construction (Side-Angle-Side)

Construct a triangle when two sides and the included angle are given — e.g., 3 cm, 75°, 7 cm: draw the 7 cm base, construct a 75° angle at one end, then mark the second side (3 cm) along the new arm and join to complete the triangle. The same method applies for 6 cm/25°/3 cm and 3 cm/120°/8 cm.

ASA Construction (Angle-Side-Angle)

Construct a triangle when two angles and the included side are given — e.g., 75°, 5 cm, 75°: draw the 5 cm base and construct 75° angles at both ends; the point where the two new arms meet is the third vertex. Similarly for 25°/3 cm/60° and 120°/6 cm/30°. All three angle-pairs given (150°, 85°, and 150° total for the two given angles) sum to less than 180°, so each construction is valid and a unique triangle results.

Angle Sum Property

The three interior angles of any triangle always sum to 180°. Given two angles, the third can always be found by subtracting their sum from 180°: for example, with angles 60° and 70°, the third angle = 180° − 60° − 70° = 50°. This third angle does not depend on the length of the base — only on the other two angles.

Figure It Out: Finding the Third Angle

• 36°, 72° → third angle = 180 − 36 − 72 = 72°
• 150°, 15° → third angle = 180 − 150 − 15 = 15°
• 90°, 30° → third angle = 180 − 90 − 30 = 60°
• 75°, 45° → third angle = 180 − 75 − 45 = 60°

Can a triangle have all angles equal to 70°? No — since 70+70+70 = 210 ≠ 180. If two angles are 70°, the third must be 180 − 140 = 40°. For a triangle with all three angles equal, each angle must be 180÷3 = 60° — this is the equilateral triangle.

If angle B = angle C, and angle A = 50°, find B and C: B + C = 180 − 50 = 130°, so B = C = 130÷2 = 65° each.

Checking Valid Angle Pairs

Which second angle allows a triangle to exist, given a fixed first angle? Rule: the second angle must be strictly less than (180° − first angle).

• 30° → possible if second angle < 150° (e.g., 60°, 90°); not possible if ≥ 150° (e.g., 160°, 170°)
• 70° → possible if second angle < 110°; not possible if ≥ 110°
• 54° → possible if second angle < 126°; not possible if ≥ 126°
• 144° → possible if second angle < 36°; not possible if ≥ 36°

Testing specific pairs: (a) 35°,150° → sum=185° — not possible (b) 70°,30° → sum=100°, third=80° — possible (c) 90°,85° → sum=175°, third=5° — possible (d) 50°,150° → sum=200° — not possible.

Exterior Angles

Exterior Angle Theorem: the exterior angle of a triangle equals the sum of the two remote (non-adjacent) interior angles. Example: if angle A = 50° and angle B = 60°, then angle ACB (interior) = 180 − 50 − 60 = 70°, so the exterior angle ACD = 180 − 70 = 110° — which indeed equals angle A + angle B = 50+60 = 110°, confirming the theorem.

7.5 Constructions Related to Altitudes

Why is the paper-folded crease (altitude) always perpendicular to the base? Because the shortest distance from a point (the opposite vertex) to a line (the base) is always the perpendicular segment; folding the paper physically constructs this shortest path, guaranteeing perpendicularity.

Constructing an altitude in an obtuse triangle: when one angle (e.g., angle R) is obtuse (like 140°), the foot of the altitude from the opposite vertex falls outside the triangle’s base — the base line must be extended before the perpendicular can be dropped using a set-square.

How many right triangles exist with a fixed hypotenuse (e.g., AC = 5 cm) and the right angle at B? Answer: Infinitely many — since angle A and angle C can vary (as long as they sum to 90°), vertex B can lie anywhere on the semicircle with AC as diameter, producing infinitely many differently-shaped right triangles.

7.6 Types of Triangles

Why can’t an acute-angled triangle be defined as “a triangle with one acute angle”? Because every triangle — even right or obtuse triangles — always has at least two acute angles (since only one angle can be 90° or more for the angle sum to stay at 180°). An acute triangle must have all three angles acute.

Can an equilateral triangle be right-angled or obtuse-angled? No — every angle in an equilateral triangle is fixed at exactly 60°, so it can never contain a 90° or obtuse angle.

Isosceles right triangle: one 90° angle and two equal 45° angles (90+45+45=180). Isosceles obtuse triangle: one obtuse angle (e.g., 120°) and two equal smaller angles (e.g., 30° each; 120+30+30=180).

Extra Questions: Class 7 Maths Chapter 7
Revision Notes: Class 7 Maths Chapter 7