## Mensuration for SSC CGL Question Set 27 with answers

Solve 10 questions on mensuration for SSC CGL Set 27 in 15 minutes. Verify from answers. Learn to solve the questions easy and quick from paired solutions.

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### 10 questions on Mensuration for SSC CGL Set 27 - Answering time 15 mins

**Problem 1.**

The area of a rectangle is thrice that of a square. The length of the rectangle is 20 cm and the breadth $\frac{3}{2}$ times the length of a side of the square. The side length of the square (in cm) is,

- 10
- 20
- 60
- 30

**Problem 2.**

The perimeter of a rhombus is 40 cm and its height is 5 cm. Its area in (cm$^2$) is,

- 45
- 60
- 55
- 50

**Problem 3.**

The sum of the length, breadth and height of a rectangular parallelepiped is 24 cm and the length of the diagonal is 15 cm. Then its total surface area is,

- 351 cm$^2$
- 256 cm$^2$
- 265 cm$^2$
- 315 cm$^2$

**Problem 4.**

In the figure given below two shaded regions are formed in a circle of radius $a$.

The area of the shaded region is,

- $\frac{a^2}{2}(\frac{\pi}{2} - 1)$
- $a^2(\pi - 1)$
- $\frac{a^2}{2}(\pi - 1)$
- $a^2(\frac{\pi}{2} - 1)$

**Problem 5.**

Three circles of radii $a$, $b$ and $c$ touch each other externally. The area of the triangle formed by the three centres is,

- $ab + bc + ca$
- $(a + b +c)\sqrt{ab + bc + ca}$
- $\sqrt{(a + b + c)abc}$
- none of the above

**Problem 6.**

The circumference of a circle is 11 cm. The area of a sector of the circle subtending an angle of $60^0$ at the periphery (take $\pi = \frac{22}{7})$ is,

- $1\displaystyle\frac{29}{48}$ cm$^2$
- $2\displaystyle\frac{27}{48}$ cm$^2$
- $2\displaystyle\frac{29}{48}$ cm$^2$
- $1\displaystyle\frac{27}{48}$ cm$^2$

**Problem 7.**

If the difference between the areas of the circumcircle and incircle of an equilateral triangle is 44 cm$^2$, then the area of the triangle (in cm$^2$, take $\pi = \frac{22}{7}$), is,

- $28$ cm$^2$
- $21$ cm$^2$
- $7\sqrt{3}$ cm$^2$
- $14\sqrt{3}$ cm$^2$

**Problem 8.**

If area of an equilateral triangle is $A$ and its height is $b$, the value of $\displaystyle\frac{b^2}{A}$ is,

- $\displaystyle\frac{1}{3}$
- $3$
- $\sqrt{3}$
- $\displaystyle\frac{1}{\sqrt{3}}$

**Problem 9.**

ABCD is a parallelogram. P and Q are the mid-points of sides BC and CD respectively. If the area of the $\triangle ABC$ is 12 cm$^2$, the area of $\triangle APQ$ is,

- 9 cm$^2$
- 12 cm$^2$
- 10 cm$^2$
- 8 cm$^2$

** Problem 10.**

A wire when bent in the form of a square encloses an area of 484 sq cm. What will be the enclosed area when the same wire is bent into the form of a circle?

- 616 sq cm
- 693 sq cm
- 462 sq cm
- 539 sq cm

Know how to solve the questions quickly at,

**Mensuration for SSC CGL Solution set 27.**

### Answers to the 10 questions on mensuration for SSC CGL Set 27

**Problem 1.** Answer. Option a: 10.

** Problem 2.** Answer. Option d : 50.

** Problem 3.** Answer. Option a: 351 cm$^2$.

** Problem 4.** Answer. Option d: $a^2(\frac{\pi}{2} - 1)$.

** Problem 5.** Answer. Option c: $\sqrt{(a + b + c)abc}$.

** Problem 6.** Answer. Option a : $1\displaystyle\frac{29}{48}$ cm$^2$.

** Problem 7.** Answer. Option d: $14\sqrt{3}$ cm$^2$.

** Problem 8.** Answer. Option c: $\sqrt{3}$.

** Problem 9.** Answer. Option a: 9 cm$^2$.

** Problem 10.** Answer. Option a: 616 sq cm.

### 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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