Mathematics Class 7 — Ganita…
🏠 Home Log in
TOPIC 27 OF 31

Triangles Chapter Exercises

🎓 Class 7 Mathematics CBSE Theory Ch 7 — A Tale of Three Intersecting Lines ⏱ ~35 min
🌐 Language: [gtranslate]

This MCQ module is based on: Triangles Chapter Exercises

This mathematics assessment will be based on: Triangles Chapter Exercises
Targeting Class 7 level in Geometry, with Basic difficulty.

Upload images, PDFs, or Word documents to include their content in assessment generation.

7.5 Types of Triangles

Triangles can be classified in two useful ways — by side lengths and by angle measures.

Classification by Sides

Equilateral (3 equal) Isosceles (2 equal) Scalene (all different)
Fig 7.10 — Equilateral, isosceles, scalene.

Classification by Angles

A triangle may also be classified by its largest angle:

  • Acute-angled triangle — all three angles are less than \(90^\circ\).
  • Right-angled triangle — exactly one angle is \(90^\circ\).
  • Obtuse-angled triangle — one angle is greater than \(90^\circ\).
Acute (all < 90°) Right (one = 90°) Obtuse (one > 90°)
Fig 7.11 — Acute, right, obtuse triangles.
Fact
A right-angled triangle can never be obtuse, and vice-versa. But a right/obtuse triangle can also be isosceles (e.g., 90°, 45°, 45°).

End-of-Chapter Exercises — Figure it Out

Q1. Construct \(\triangle ABC\) with BC = 5 cm, AB = 6 cm, CA = 5 cm. Construct an altitude from A to BC.
Draw BC = 5 cm. Arcs of 6 cm at B and 5 cm at C intersect at A. Drop a perpendicular from A to BC using a set-square to locate its foot D on BC; AD is the altitude.
Q2. Construct \(\triangle TRY\) with RY = 4 cm, TR = 7 cm and \(\angle R = 140^\circ\). Construct the altitude from T to RY.
Because \(\angle R\) is obtuse, T lies outside the strip above RY. After locating T, the foot of the altitude from T falls on the extension of RY beyond R (altitude lies outside the triangle).
Q3. Construct a right-angled triangle \(\triangle ABC\) with \(\angle B = 90^\circ\), AC = 5 cm, AB = 3 cm. How many different triangles can exist with these measures?
BC is determined by Pythagoras: \(BC = \sqrt{5^2-3^2}=4\) cm. Steps: draw AB = 3 cm, erect a perpendicular at B, mark C so that AC = 5 cm. Only one such triangle (up to reflection).
Q4. By considering construction, explore: can a triangle be both right-angled and obtuse? Can it be right-angled and isosceles? Can it be right-angled and equilateral?
Right + obtuse: impossible (sum would exceed 180°). Right + isosceles: possible (45°, 45°, 90°). Right + equilateral: impossible (equilateral ⇒ each angle 60°).
Q5. Check whether a triangle exists with sides 3 cm, 4 cm and 7 cm.
\(3+4=7\), not > 7. No — the figure is degenerate.
Q6. Find the altitude length from A in an equilateral triangle of side 6 cm.
Altitude \(= \frac{\sqrt3}{2}\times 6 = 3\sqrt3 \approx 5.196\) cm.

Summary

  • A triangle has three vertices, three sides and three angles.
  • Compass construction simplifies triangle drawing when side lengths are known (SSS).
  • A set of three lengths forms a triangle only if the sum of the two smaller ones exceeds the longest — the triangle inequality.
  • An equilateral triangle has all three sides equal (and each angle 60°); an isosceles triangle has at least two sides equal.
  • The sum of the three interior angles of any triangle is \(180^\circ\) — the angle sum property.
  • An exterior angle equals the sum of the two opposite interior angles.
  • An altitude is the perpendicular segment from a vertex to the opposite side. Every triangle has three altitudes; they meet at the orthocentre.
  • Triangles are classified by sides (equilateral/isosceles/scalene) and by angles (acute/right/obtuse).
Puzzle: Shortest Path in a Box
Materials: Small cardboard box, string, pencil.
  1. Mark a spider at one corner of the box and a fly at the opposite (farthest) corner.
  2. The spider walks only on the outer surfaces. Trace at least two different paths with string.
  3. Unfold the box (flatten the faces) and draw a straight line between the two corners on the flattened net — measure its length.
  4. Compare this straight-line distance with your earlier paths.
The shortest path on the surface is the straight line seen after unfolding the box — an application of the triangle inequality in two dimensions.

Competency-Based Questions

Scenario: A surveyor marks three landmarks L, M, N. She measures LM = 12 m, MN = 9 m, and \(\angle M = 90^\circ\). She claims the triangle is uniquely determined.
Q1. Find LN.
L3 Apply
Right-angled at M ⇒ \(LN = \sqrt{12^2+9^2}=\sqrt{225}=15\) m.
Q2. What type of triangle (by angles) is LMN?
L4 Analyse
Right-angled (angle M = 90°). It is also scalene (all three sides 9, 12, 15 differ).
Q3. The surveyor's assistant claims that two sides and an angle always uniquely fix a triangle. Evaluate this.
L5 Evaluate
Only partially true. If the angle is included between the two sides (SAS), the triangle is uniquely fixed. If the angle is not included (SSA), two different triangles may exist — the ambiguous case.
Q4. Propose an original construction specification for an isosceles right triangle with legs of 4 cm, and describe how you would verify the third side equals \(4\sqrt 2\).
L6 Create
Draw AB = 4 cm. Erect a perpendicular at B of length 4 cm to mark C. Join AC. Measure AC with a ruler — approximately 5.66 cm, matching \(4\sqrt2 \approx 5.657\).

Assertion–Reason Questions

A: A triangle with sides 5, 5, 5 is both equilateral and isosceles.
R: Isosceles means "at least two sides equal"; equilateral means "all three equal".
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
(a) — three equal sides satisfy both definitions.
A: The altitudes of an obtuse triangle cannot all lie inside the triangle.
R: For the vertex opposite the obtuse angle, the perpendicular to the opposite side meets that side inside the triangle.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
(b) — A is true (two altitudes fall outside), and R by itself is also true but only describes one altitude; it does not explain why the other two are outside.
A: If two angles of a triangle are 60° each, the triangle is equilateral.
R: Angle sum 180° forces the third angle to be 60°, and equal angles imply equal opposite sides.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
(a) — correct and R precisely explains A.

Frequently Asked Questions

How do you classify a triangle with sides 5, 5, 5?
All three sides are equal, so it is an equilateral triangle. Its interior angles are each 60 degrees.
Can a triangle have two obtuse angles?
No. Since interior angles sum to 180 degrees and each obtuse angle is greater than 90 degrees, two obtuse angles would exceed 180 degrees, which is impossible.
What is the summary of Chapter 7 Triangles?
A triangle has three sides and three angles whose sum is 180 degrees. The triangle inequality must hold for any three side lengths. SSS construction builds triangles, and altitudes, medians and angle bisectors lead to important triangle centres.
How do exterior angle problems connect to the angle sum?
Because an exterior angle plus its adjacent interior angle is 180 degrees, it equals the sum of the other two interior angles. Many exercises use this shortcut.
What is a common exercise on triangle inequality?
You are asked to decide whether three given lengths can form a triangle. Verify by checking that each pair of sides sums to more than the third side.
Keyword

AI Tutor
Mathematics Class 7 — Ganita Prakash
Ready
Hi! 👋 I'm Gaura, your AI Tutor for Triangles Chapter Exercises. Take your time studying the lesson — whenever you have a doubt, just ask me! I'm here to help.