Class 9 Mathematics Part 2 (Geometry) – Daily Revision MCQ (English Medium)
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Points P, Q and R are on a line. d(P, Q) = 7, d(Q, R) = 4 and d(P, R) = 3. Which point is between the other two?
Chapter 1: Basic Concepts in Geometry
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Answer: C. Point R
d(P, R) + d(R, Q) = 3 + 4 = 7 = d(P, Q), so R is between P and Q (P - R - Q).Two lines are cut by a transversal and a pair of alternate angles formed are congruent. What can we conclude about the two lines?
Chapter 2: Parallel Lines
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Answer: A. They are parallel
Alternate angles test: if a pair of alternate angles is congruent, the two lines are parallel.In ΔABC, side AB ≅ side AC and m∠B = 50°. What is m∠A?
Chapter 3: Triangles
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Answer: D. 80°
Isosceles triangle theorem: ∠C ≅ ∠B = 50°, so ∠A = 180° − (50° + 50°) = 80°.To construct ΔABC with BC, ∠B and AB + AC given, point D is marked on the ray from B so that BD = AB + AC. On which line does vertex A then lie?
Chapter 4: Constructions of Triangles
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Answer: B. The perpendicular bisector of seg CD
BA + AD = BA + AC gives AD = AC, so A is equidistant from C and D and lies on the perpendicular bisector of seg CD.In ΔPQR, M and N are the midpoints of sides PQ and PR. If QR = 14 cm, what is MN?
Chapter 5: Quadrilaterals
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Answer: C. 7 cm
Midpoint theorem: the segment joining the midpoints of two sides is half the third side, so MN = 14 ÷ 2 = 7 cm.In parallelogram ABCD, m∠A = 70°. What is m∠B?
Chapter 5: Quadrilaterals
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Answer: A. 110°
Adjacent angles of a parallelogram are supplementary, so ∠B = 180° − 70° = 110°.The centre of the incircle of a triangle is the point of concurrence of the triangle's —
Chapter 6: Circle
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Answer: B. angle bisectors
The incircle touches all three sides; its centre I is where the bisectors of the angles of the triangle meet.What is the equation of the line parallel to the Y-axis that lies 3 units to the right of it?
Chapter 7: Co-ordinate Geometry
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Answer: D. x = 3
Every point on that line has x co-ordinate 3; a line parallel to the Y-axis has an equation of the form x = a, here x = 3.What is the value of 2 sin 30° + tan 45°?
Chapter 8: Trigonometry
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Answer: C. 2
sin 30° = ½ and tan 45° = 1, so 2 × ½ + 1 = 2.What is the surface area of a sphere of radius 3.5 cm? (π = 22/7)
Chapter 9: Surface Area and Volume
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Answer: D. 154 cm²
Surface area = 4πr² = 4 × 22/7 × 3.5 × 3.5 = 154 cm².
Class 9: MCQ in other subjects
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All questions are written by PadhaiNotes in our own words from the chapters of the official Maharashtra State Board Balbharati textbook (ebalbharati.in). Each question shows its chapter.