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JEE Main Geometry & Mensuration Questions

Practise geometry & mensuration (mathematics) for the Joint Entrance Examination (Main): 10 multiple-choice questions, each with the correct answer and a worked explanation. Click an option to check it, or “New questions” for another set.

Geometry & Mensuration practice

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  1. Q1.Find the distance between (0, -3) and (8, 12).

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    Answer: D — 17

    Distance = √((Δx)² + (Δy)²) = √(8² + 15²) = 17.

  2. Q2.Find the distance between (3, 4) and (8, 16).

    Show answer & explanation

    Answer: D — 13

    Distance = √((Δx)² + (Δy)²) = √(5² + 12²) = 13.

  3. Q3.Find the volume of a cone with radius 7 cm and height 17 cm (take π = 22/7).

    Show answer & explanation

    Answer: C — 872.67 cm³

    Using ⅓πr²h with π = 22/7: 872.67 cm³.

  4. Q4.Find the volume of a cylinder with radius 3.5 cm and height 20 cm (take π = 22/7).

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    Answer: A — 770 cm³

    Using πr²h with π = 22/7: 770 cm³.

  5. Q5.Find the distance between (-2, 5) and (4, 13).

    Show answer & explanation

    Answer: C — 10

    Distance = √((Δx)² + (Δy)²) = √(6² + 8²) = 10.

  6. Q6.Find the area of a circle of radius 7 cm (take π = 22/7).

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    Answer: D — 154 cm²

    Using πr² with π = 22/7: 154 cm².

  7. Q7.Find the volume of a sphere of radius 3.5 cm (take π = 22/7).

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    Answer: B — 179.67 cm³

    Using (4/3)πr³ with π = 22/7: 179.67 cm³.

  8. Q8.Find the distance between (3, -5) and (11, 10).

    Show answer & explanation

    Answer: D — 17

    Distance = √((Δx)² + (Δy)²) = √(8² + 15²) = 17.

  9. Q9.A right triangle has legs 20 cm and 21 cm. Find the hypotenuse.

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    Answer: C — 29 cm

    By Pythagoras, hypotenuse = √(20² + 21²) = √841 = 29.

  10. Q10.Find the distance between (5, 2) and (11, 10).

    Show answer & explanation

    Answer: A — 10

    Distance = √((Δx)² + (Δy)²) = √(6² + 8²) = 10.

Free JEE Main Geometry & Mensuration Questions by pGento

How to approach geometry & mensuration questions

Memorise core formulas (circle πr², triangle ½bh, Pythagoras, distance √((x₂−x₁)² + (y₂−y₁)²)). Sketch the figure — most errors come from misreading which length is given.

About JEE Main

Joint Entrance Examination (Main) — conducted by National Testing Agency (NTA). India’s national engineering entrance for NITs, IIITs and other institutes, and the qualifier for JEE Advanced.

Original practice questions written for this site in the style and difficulty of each exam’s syllabus. They are not past papers and the site is not affiliated with any exam body. Exam patterns, marking schemes and syllabi change — always check the latest official notification.

JEE Main practice topics

TopicSectionQuestion source
AlgebraMathematicsCalculated (new numbers every set)
Geometry & MensurationMathematicsCalculated (new numbers every set)
Probability & CombinatoricsMathematicsCalculated (new numbers every set)
PhysicsScienceCalculated + question bank
ChemistryScienceCalculated + question bank

Frequently asked questions

Find the distance between (0, -3) and (8, 12).
Answer: D — 17. Distance = √((Δx)² + (Δy)²) = √(8² + 15²) = 17.
Find the distance between (3, 4) and (8, 16).
Answer: D — 13. Distance = √((Δx)² + (Δy)²) = √(5² + 12²) = 13.
Find the volume of a cone with radius 7 cm and height 17 cm (take π = 22/7).
Answer: C — 872.67 cm³. Using ⅓πr²h with π = 22/7: 872.67 cm³.
Are these real past exam papers?
No. Every question is an original practice question written in the style, syllabus and difficulty of the exam, with a worked explanation. The site is not affiliated with any exam body.
JEE Main Geometry & Mensuration Questions

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