## Mensuration questions with answers for SSC CGL

Solve 10 mensuration questions, some of which are tough, in 15 minutes in SSC CGL Question Set 26. Verify from answers. Clear doubts from linked solutions.

Answers and the link to the quick easy solutions are at the end.

### 10 Mensuration questions SSC CGL Set 26 - Answering time 15 mins

**Problem 1.**

Three circles of radii 5.5 cm, 4.5 cm and 3.5 cm touch each other externally. The perimeter of the triangle formed by joining the centres of the circles (in cm) is,

- $27\pi$
- $27$
- $13.5$
- $\pi(3.5^2 + 4.5^2 + 5.5^2)$

**Problem 2.**

Three circles of diameter 10 cm each are bound together by a rubber band as shown below,

The length of the rubber band (in cm) in stretched condition as shown, will be,

- $30 + \pi$
- $30$
- $10\pi$
- $60 + 20\pi$

**Problem 3.**

A circular wire of radius 42 cm is bent in the form of a rectangle whose sides are in a ratio of 6 : 5. The smaller side of the rectangle is, (take $\pi =\frac{22}{7}$)

- 30 cm
- 60 cm
- 36 cm
- 25 cm

**Problem 4.**

Two equal maximum sized circular plates are cut-off from a circular paper-sheet of circumference 352 cm. The circumference of each circular plate is,

- 176 cm
- 180 cm
- 165 cm
- 150 cm

**Problem 5.**

If length and perimeter of a rectangle are in the ratio 5 : 16, then its length and breadth will be in the ratio,

- 5 : 11
- 5 : 4
- 5 : 3
- 5 : 8

**Problem 6.**

From a point in the interior of an equilateral triangle, the perpendicular distances of the sides are, $\sqrt{3}$ cm, $2\sqrt{3}$ cm and $5\sqrt{3}$ cm. The perimeter (in cm) of the triangle is,

- 48
- 64
- 24
- 32

**Problem 7.**

Through each vertex of a triangle, a line parallel to the opposite side is drawn. The ratio of the perimeter of the new triangle thus formed with the original triangle is,

- 3 : 2
- 4 : 1
- 5 : 3
- 2 : 1

**Problem 8.**

A right triangle with side lengths 3 cm, 4 cm and 5 cm is rotated about the side 3 cm as rotation axis, to form a cone. The volume of the cone so formed is,

- $16\pi$ cm$^3$
- $20\pi$ cm$^3$
- $15\pi$ cm$^3$
- $12\pi$ cm$^3$

**Problem 9.**

An equilateral triangle of side 6 cm has its corners cut-off to form a regular hexagon. Area (in cm$^2$) of this hexagon will be,

- $3\sqrt{3}$
- $6\sqrt{3}$
- $5\displaystyle\frac{\sqrt{3}}{2}$
- $3\sqrt{6}$

** Problem 10.**

The ratio of sides of a triangle is 3 : 4 : 5 and area of the triangle is 72 square units. Then the area of an equilateral triangle with same perimeter as the first triangle (in square units) is,

- $48\sqrt{3}$
- $96$
- $60\sqrt{3}$
- $32\sqrt{3}$

Learn how to solve these questions quick and easy go through the companion solution set,

**SSC CGL level Solution Set 26 on Mensuration****.**

### Answer to the Mensuration questions SSC CGL Set 26

**Problem 1. Answer:** Option b: 27.

**Problem 2. Answer:** Option a : $30 + 10\pi$.

**Problem 3. Answer:** Optopn b: 60cm.

**Problem 4. Answer:** Option a: 176 cm.

**Problem 5. Answer:** Option c: 5 : 3.

**Problem 6. Answer:** Option a : 48.

**Problem 7. Answer:** Option d: 2 : 1.

**Problem 8. Answer:** Option a: $16\pi$ cm$^3$.

**Problem 9. Answer:** Option b: $6\sqrt{3}$.

**Problem 10. Answer:** Option a: $48\sqrt{3}$.

### Other related guideline, question set and solution set on SSC CGL mensuration

* 7 steps for sure success in SSC CGL tier 1 and tier 2 competitive tests* or

*to access all the valuable student resources that we have created specifically for SSC CGL, but*

**section on SSC CGL****generally for any hard MCQ test.**

**Tutorials that you should refer to**

*Basic concepts on Geometry 1 lines points and triangles*

*Basic concepts on Geometry 2 quadrilaterals polygons squares*

*Basic and rich concepts on Geometry 3 Circles*

*Basic and rich Geometry concepts part 4 Arc angle subtending concept proof*

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**SSC CGL level Question Set 42 on Mensuration 4**

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**SSC CGL level Question Set 27 on Mensuration 2**

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