Master Chapter 7: Triangles & Congruence (Class 9 Revision for Competitive Exams)
Welcome, Aspirants! Chapter 7 (Triangles) is one of the most vital chapters in Class 9 geometry, serving as a foundational pillar for competitive exams like SSC CGL, CHSL, GD Constable, RRB NTPC, and Group D. Questions on congruence criteria (SAS, ASA, SSS, RHS), properties of isosceles triangles, and triangle inequalities are frequently asked in these exams.
This handpicked practice quiz features high-yield Previous Year Questions (PYQs) designed to sharpen your conceptual understanding and boost your exam-day problem-solving speed using smart shortcut tricks.
Q1. In ΔABC and ΔPQR, it is given that AB = AC, ∠C = ∠P, and ∠B = ∠Q. The two triangles are:
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Answer: A (Isosceles but not congruent)
Step-by-Step Explanation:
1. In ΔABC, given AB = AC. This means ∠B = ∠C (angles opposite to equal sides are equal). Thus, ΔABC is an isosceles triangle.
2. We are given ∠C = ∠P and ∠B = ∠Q. Since ∠B = ∠C, we get ∠P = ∠Q. Therefore, ΔPQR is also an isosceles triangle (sides opposite to equal angles are equal: PR = QR).
3. However, no information is given regarding the actual side lengths of ΔPQR compared to ΔABC. Only angles are equal, which guarantees similarity (AAA) but not congruence because side lengths could differ.
Shortcut Trick: For congruence, at least one corresponding side length condition must be explicitly matching (e.g., ASA, SAS, SSS). Since only angles are corresponding, they cannot be proven congruent!
Q2. [SSC GD & CHSL PYQ] In ΔABC, if AB = AC and ∠B = 65°, what is the value of ∠A?
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Answer: B (50°)
Step-by-Step Explanation:
1. Since AB = AC, the angles opposite to these sides must be equal: ∠C = ∠B.
2. Given ∠B = 65°, therefore ∠C = 65°.
3. By the angle sum property of a triangle: ∠A + ∠B + ∠C = 180°.
4. ∠A + 65° + 65° = 180° ⇒ ∠A + 130° = 180° ⇒ ∠A = 50°.
Shortcut: Vertex angle of an isosceles triangle = 180° - 2 × (Base Angle) = 180° - 130° = 50°.
Q3. [RRB NTPC PYQ] Which of the following set of side lengths can form a triangle?
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Answer: C (5 cm, 6 cm, 10 cm)
Step-by-Step Explanation:
1. According to the Triangle Inequality Theorem (Class 9 Chapter 7 property), the sum of the lengths of any two sides of a triangle must be strictly greater than the length of the third side.
2. Let's test the options by summing the two smaller sides:
- Option A: 2 + 3 = 5 (5 < 6) → Cannot form a triangle.
- Option B: 3 + 4 = 7 (7 < 8) → Cannot form a triangle.
- Option C: 5 + 6 = 11 (11 > 10) → Can form a triangle!
- Option D: 1.5 + 2.5 = 4.0 (4 < 4.5) → Cannot form a triangle.
Shortcut: Just check if (Smallest Side + Middle Side) > Largest Side. If yes, it's a valid triangle!
Q4. [SSC CGL PYQ] In ΔABC, the bisectors of interior angles ∠B and ∠C intersect at point O. If ∠A = 70°, then what is the measure of ∠BOC?
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Answer: B (125°)
Step-by-Step Explanation:
1. This is a very popular competitive exam property derived from angle bisectors in a triangle.
2. Formula: ∠BOC = 90° + ½(∠A).
3. Substitute the value of ∠A:
∠BOC = 90° + ½(70°)
∠BOC = 90° + 35° = 125°.
Shortcut: For interior angle bisectors, always use: 90 + A/2. (For exterior bisectors, it is 90 - A/2).
Q5. [SSC CHSL PYQ] If AD is a median of ΔABC and G is the centroid, and the length of AD is 12 cm, then find the length of AG.
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Answer: C (8 cm)
Step-by-Step Explanation:
1. The centroid (G) of a triangle divides each median (AD) in the ratio 2:1 starting from the vertex (i.e., AG : GD = 2 : 1).
2. Total parts = 2 + 1 = 3 parts.
3. Given total median length AD = 12 cm. Therefore, 3 parts = 12 cm ⇒ 1 part = 4 cm.
4. Length of AG = 2 parts = 2 × 4 = 8 cm.
Shortcut: Length of vertex segment (AG) = ⅔ × Total Median Length = ⅔ × 12 = 8 cm.
Excellent effort! Keep practicing these concepts. Triangles and Congruence are high-scoring topics if you master the basic properties and shortcuts. Make sure to bookmark this page for rapid revision before your SSC, RRB, or GD exams!
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