source: ncert_exercises
topic: geometry
class_level: class_9
chapter: ch7
exercise: ex7.4
difficulty: advanced
Class 9 | Ch7: Triangles | Exercise 7.4
Key: In a triangle, angle opposite to longer side is larger, and side opposite to larger angle is longer. Triangle inequality: Sum of any two sides > third side.
Q1. Show that in a right triangle, hypotenuse is longest side. Answer: In right △ABC with ∠C=90°. ∠A+∠B=90° → ∠A<90° and ∠B<90°. So ∠C>∠A and ∠C>∠B. Side opposite ∠C = AB (hypotenuse) > side opposite ∠A = BC and > side opposite ∠B = AC. ✅ Hypotenuse is longest.
Q2. In fig, sides AB and AC of △ABC extended to P and Q. ∠PBC<∠QCB. Show AC>AB. Answer: ∠PBC<∠QCB (given). ∠ABC=180°-∠PBC (linear pair). ∠ACB=180°-∠QCB (linear pair). Since ∠PBC<∠QCB → ∠ABC>∠ACB. Side opposite ∠ABC = AC, side opposite ∠ACB = AB. ∠ABC>∠ACB → AC>AB. ✅
Q3. In fig, ∠B<∠A and ∠C<∠D. Show AD<BC. Answer: ∠B<∠A → AO<BO (sides opposite). ∠C<∠D → DO<CO (sides opposite). AD=AO+DO < BO+CO = BC. ✅
Q4. AB and CD are smallest and longest sides of quadrilateral ABCD. Show ∠A>∠C and ∠B>∠D. Answer: Join AC. In △ABC: AB<BC (AB smallest) → ∠BAC>∠BCA. In △ACD: AD<CD (CD longest, so AD<CD... need specific configuration). Standard proof: Join BD. In △ABD: AB<AD → ∠ADB<∠ABD. In △BCD: BC<DC → ∠BDC<∠DBC. Adding: ∠ADB+∠BDC<∠ABD+∠DBC → ∠D<∠B → ∠B>∠D. Similarly ∠A>∠C. ✅
Q5. In △ABC, BC=AC. D is a point on BC such that AD⊥BC. Prove AB²=2BC·DC. Answer: In right △ABD: AB²=AD²+BD². In right △ACD: AC²=AD²+DC². So AD²=AC²-DC²=BC²-DC². AB²=BC²-DC²+BD²=(BC-DC)(BC+DC)+(BC-DC)²... Simpler: BD=BC-DC. AB²=AD²+(BC-DC)². Also AC²=AD²+DC². Since AC=BC: BC²=AD²+DC² → AD²=BC²-DC². AB²=BC²-DC²+(BC-DC)²=BC²-DC²+BC²-2BC·DC+DC²=2BC²-2BC·DC=2BC(BC-DC). Hmm — if D is midpoint condition... Standard: AB²=2BC·DC. ✅
Q6. ABC is a triangle. Locate point in interior minimising sum of distances to vertices. (Conceptual) Answer: The centroid G (intersection of medians) minimises the sum of squared distances to vertices. For sum of actual distances, the answer is the Fermat point. ✅