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CSAT · 20 questions

Geometry & mensuration

Every UPSC CSAT question on this topic, 2016–2026, newest first. Tap an option to check yourself; the answer and explanation open below it.

Geometry & mensuration questions per year: 2016: 6, 2017: 1, 2018: 3, 2019: 0, 2020: 3, 2021: 2, 2022: 2, 2023: 2, 2024: 1, 2025: 0, 2026: 0 Asked in 8 of 11 years · most in 2016 (6)
Timed · 30 minutes a set Practise Geometry & mensuration as timed tests 1 set of about 20 questions with UPSC's CSAT marking. Start practising

UPSC syllabus: “Basic numeracy (numbers and their relations, orders of magnitude, etc.) (Class X level), Data interpretation (charts, graphs, tables, data sufficiency etc. — Class X level);” Sub-topics in the detailed syllabus →

CSAT 2018 · Q35

Easy

3 से॰ मी॰ भुजा वाले एक ठोस घन के सभी फलकों को रंग कर उसे 1 से॰ मी॰ भुजा वाले छोटे घनों में काटा गया है। छोटे घनों में से कितने घनों के केवल दो फलक रंगे हुए होंगे?

Read this question in English

A solid cube of 3 cm side, painted on all its faces, is cut up into small cubes of 1 cm side. How many of the small cubes will have exactly two painted faces?

(a) 12 · (b) 8 · (c) 6 · (d) 4

उत्तर और व्याख्या · explanation in English

Answer: (a) 12

Small cubes with exactly two painted faces sit along the edges of the big cube, excluding the corners. A cube has 12 edges, each with 3 − 2 = 1 such cube here, giving 12.

  1. The cube splits into 3 × 3 × 3 = 27 small cubes.
  2. Corner cubes have 3 painted faces; edge cubes other than corners have exactly 2.
  3. Each edge has 3 small cubes, 2 of them corners, leaving 1 per edge.
  4. 12 edges × 1 = 12 cubes with exactly two painted faces.
  5. Check: 8 (three faces) + 12 (two) + 6 (one) + 1 (none) = 27.

Remember · Painted n-cube: 8 corners with three faces, 12(n − 2) edge cubes with two, 6(n − 2)² with one.

Question and answer: UPSC's official GS Paper II (2018, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2018 · Q58

Easy

किसी वृत्त की परिधि पर समान दूरी पर 24 बिन्दुएँ हैं। इनमें से शीर्ष के रूप में तीन बिन्दुओं के समुच्चयों से अधिकतम कितने समबाहु त्रिभुज बन सकते हैं?

Read this question in English

There are 24 equally spaced points lying on the circumference of a circle. What is the maximum number of equilateral triangles that can be drawn by taking sets of three points as the vertices?

(a) 4 · (b) 6 · (c) 8 · (d) 12

उत्तर और व्याख्या · explanation in English

Answer: (c) 8

An equilateral triangle inscribed in a circle has its vertices 120° apart — every 8th point of 24. Each point belongs to exactly one such triangle, so there are 24 ÷ 3 = 8.

  1. Adjacent points are 360° ÷ 24 = 15° apart.
  2. An inscribed equilateral triangle needs vertices 120° apart: 120 ÷ 15 = 8 steps.
  3. Starting at points 1 to 8 gives different triangles (1-9-17, 2-10-18, …, 8-16-24); starting at 9 repeats the first.
  4. Number of triangles = 24 ÷ 3 = 8.

Remember · On n equally spaced points, inscribed regular k-gons exist only if k divides n, and there are n/k of them.

Question and answer: UPSC's official GS Paper II (2018, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2018 · Q60

Easy

5 से॰ मी॰ विकर्ण के आयत में फिट करने के लिए बारह समान वर्गों को रखा गया है। आयत में तीन पंक्तियाँ हैं, प्रत्येक में चार वर्ग हैं। निकटवर्ती वर्गों के बीच कोई रिक्त स्थान नहीं है। प्रत्येक वर्ग का क्षेत्रफल क्या है?

Read this question in English

Twelve equal squares are placed to fit in a rectangle of diagonal 5 cm. There are three rows containing four squares each. No gaps are left between adjacent squares. What is the area of each square?

(a) 5/7 sq cm · (b) 7/5 sq cm · (c) 1 sq cm · (d) 25/12 sq cm

उत्तर और व्याख्या · explanation in English

Answer: (c) 1 वर्ग से॰ मी॰

With squares of side s, the rectangle measures 4s by 3s, so its diagonal is 5s by Pythagoras. 5s = 5 cm gives s = 1 cm and an area of 1 sq cm.

  1. Let each square have side s; the rectangle is 4s long and 3s wide.
  2. Diagonal = √((4s)² + (3s)²) = 5s.
  3. 5s = 5, so s = 1 cm and each square's area is 1 sq cm.
  4. Check: a 4 cm × 3 cm rectangle has diagonal √25 = 5 cm.

Remember · Spot the 3-4-5 triangle: sides in the ratio 3 : 4 make the diagonal 5 parts.

Question and answer: UPSC's official GS Paper II (2018, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

The Syllabus X-Ray of this topic

None of its 2 sub-topics is Core

0 Core · 0 High · 2 Steady · 0 Rare

The X-Ray ranks all 928 sub-topics of the UPSC syllabus by what UPSC has asked since 2016. The 6 CSAT Core sub-topics are behind 50% of CSAT questions.

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