Mensuration Â is a topic inÂ **Geometry**Â which is a branch of mathematics. Mensuration deals withÂ **length, area, and volume**Â of different kinds of shape- both 2D and 3D. The article IBPS RRB Mensuration Quiz 2 provides information about Mensuration, an important topic of Mathematics Consists of different types of Mensuration questions with solutions useful for candidates preparing for different competitive examinations likeÂ **IBPS RRB, RRB ALP/Technical Exams/Junior Engineer Recruitment Exams, SSC CGL, SSC CHSL, IBPS, SBI PO, SBI Clerks, CAT and etc**.

**Answer**: Option: D

**Explanation**:

Given that, radius (r) = 36 cm

And, Area of sector = 72Ï€ \({cm}^{2}\)

â‡’ Ï€r2Î¸ = \(\frac{72Ï€}{360Â°}\)

âˆ´ Î¸ = \(\frac{72Ï€ Ã— 360Â°}{Ï€{r}^{2}}\)

= 72 Ã— 360Â° =\(\frac{20Â°}{36 Ã— 36}\)

Now, length of arc = \(\frac{Ï€rÎ˜}{180Â°}\)

= Ï€ Ã— 36 Ã— 20Â° = 4Ï€ cm

**2. A square is inscribed in a circle of diameter 2a and another square is circumscribing circle. The difference between the areas of outer and inner squares is**

**Answer**: Option: B

**Explanation**:

For inscribed square,

Diameter of circle = Diagonal of inner square

For circumscribed square,

Diameter of circle = Side of outer square

âˆ´ Area of inner square = \(\frac{1}{2}\) \({(diagonal)}^{2}\) = \(\frac{1}{2}\) Ã— \({(2a)}^{2}\) = 2\({a}^{2}\)

And, Area of outer square = \({(side)}^{2}\) = \({(2a)}^{2}\) = 4\({a}^{2}\)

Now, Required difference = 4\({a}^{2}\) â€“ 2\({a}^{2}\) = 2\({a}^{2}\)

**3. ABC is a triangle right angled at A. AB = 6 cm and AC = 8 cm. Semi-circles drawn (outside the triangle) on AB, AC and BC as diameters which enclose areas x, y and z square units, respectively. What is x + y â€“ z equal to ?**

**Answer**: Option: C

**Explanation**:

In Î”ABC, by Pythagoras theorem,

BC = \({AB}^{2}\) +\({AC}^{2}\) = 62 + 82 = 100 = 10 cm

Now, Area of that semi-circle which diameter is AB = \(\frac{{Ï€(3)}^{2}}{2}\)

âˆ´ x = 9Ï€ \({cm}^{2}\)

2 Similarly, Area of that semi-circle which diameter is AC = \(\frac{{Ï€(4)}^{2}}{2}\)

âˆ´ y = 16Ï€ \({cm}^{2}\)

2 Similarly, Area of that semi-circle which diameter is BC = \(\frac{{Ï€(5)}^{2}}{2}\)

âˆ´ z = \(\frac{25Ï€}{2}\) \({cm}^{2}\)

Now, x + y â€“ z = (\(\frac{9Ï€}{2}\) + \(\frac{16Ï€}{2}\)) â€“ \(\frac{25Ï€}{2}\) = 0

**4. Consider an equiateral triangle of a side of unit length. A new equilateral triangle is formed by joining the mid-points of one, then a third equilateral triangle is formed by joining the mid-points of second. The process is continued. The perimeter of all triangles, thus formed is**

**Answer**: Option: C

**Explanation**:

= (3 Ã— 1) + (3 Ã— 0.5) + (3 Ã— 0.25) + (3 Ã— 0.125)

= 3 + 1.5 + 0.75 + 0.375 = 5.625 â‰ˆ 6 units

**5. If AB and CD are two diameters of a circle of radius r and they are mutually perpendicular, then what is the ratio of the area of the circle to the area of the Î”ACD ?**

**Answer**: Option: B

**Explanation**:

= Area of circle : Area of Î”ACD

= Ï€\(\frac{{r}^{2}}{ 1 Ã— 2r Ã— r}\) = \(\frac{Ï€}{2}\)

**Answer**: Option: A

**Explanation**:

âˆ´ \(\frac{1}{2}\) Ã— h Ã— b = 12

â‡’ \(\frac{1}{2}\) Ã— 3 Ã— b = 12

â‡’ b = 8 cm

Here, BD =

CD = \(\frac{b}{2}\) = \(\frac{8}{2}\) = 4 cm

In right-angled Î”ABD, by pythagoras theorem,

AB = \({AD}^{2}\) + \({BD}^{2}\)

a = \({3}^{2}\) + \({4}^{2}\) = 25 = 5 cm

Now, perimeter of an isosceles triangle = 2a + b = 2 Ã— 5 + 8 = 18 cm

**2. What is the area between a square of side 10 cm and two inverted semi-circular, cross-sections each of radius 5 cm inscribed in the square ?**

**Answer**: Option: D

**Explanation**:

Given, each side of a square (a) = 10 cm

and, radius of each semi-circular (r) = 5 cm

Area between square and semi-circules = Area of square â€“ 2 Area of semi-circle

= \({a}^{2}\) â€“ 2 Ã— \(\frac{1}{3}\) Ï€\({r}^{2}\)

= (10)2 â€“ 2 Ã— \(\frac{1}{2}\) Ã— \(\frac{22}{7}\) Ã— \({(5)}^{2}\)

= 100 â€“ 78.5 = 21.5 \({cm}^{2}\)

**3. The perimeter of a rectangle having area equal to 144 cm2 and sides in the ratio 4 : 9 is**

**Answer**: Option: A

**Explanation**:

Let, length of rectangle (l) = 4x and breadth of rectangle (b) = 9x

âˆ´ Area of rectangle = l Ã— b

â‡’ 144 = 4x Ã— 9x = 36\({x}^{2}\)

â‡’ \({x}^{2}\) = 4

â‡’ x = 2

Now, l = 4x = 4 Ã— 2 = 8 cm and b = 9x = 9 Ã— 2 = 18 cm

âˆ´ Perimeter of rectangle = 2(l + b) = 2 (8 + 18) = 52 \({cm}^{2}\)

**4. One side of a parallelogram is 8.06 cm and its perpendicular distance from opposite side is 2.08 cm. What is the approximate area of the parallelogram ?**

**Answer**: Option: C

**Explanation**:

Area of parallelogram = Base Ã— Height = 8.06 Ã— 2.08 = 16.76 \({cm}^{2}\)

**5. In the figure given below, the area of rectangle ABCD is 100 sq cm, O is any point on AB and CD = 20 cm. Then, the area of Î”COD is**

**Answer**: Option: C

**Explanation**:

And, area of rectangle ABCD = 100 \({cm}^{2}\)

â‡’ AD Ã— CD = 100

â‡’ AD Ã— 20 = 100

â‡’ AD = 5 cm

âˆµ AD = OP = 5 cm

âˆ´ Area of Î”COD = \(\frac{1}{2}\) Ã— CD Ã— OP = \(\frac{1}{2}\) Ã— 20 Ã— 5 = 50 \({cm}^{2}\)

**Answer**: Option: B

**Explanation**:

Let the sides of isosceles triangle is 5x, 5x and 3x cm respectively.

By given condition,

Perimeter of isosceles triangle = Length of wire

â‡’ 5x + 5x + 3x = 78

â‡’ 13x = 78

â‡’ x = 6 cm

âˆ´ Length of base = 3x = 3 Ã— 6 = 18 cm

**2. The length of a minute hand of a wall clock is 9 cm. What is the area swept (in cm2) by the minute hand in 20 min ? (take Ï€ = 3.14)**

**Answer**: Option: B

**Explanation**:

Given, r = 9 cm

The angle made by the minute hand in 20 min, Î¸ = 120Â°

âˆ´ The area swept by the minute hand in 20 min

= Î¸ Ã— \(\frac{Ï€{r}^{2}}{360Â°}\)

= \(\frac{120Â° Ã— 3.14 Ã— 9 Ã— 9}{360Â°}\)

= 84.78 \({cm}^{2}\)

**3. If the area of a Î”ABC is equal to area of square of sde length 6 cm, then what is the length of the altitude to AB, where AB = 9 cm ?**

**Answer**: Option: D

**Explanation**:

Given, AB = 9 cm

Let the length of altitude to AB = l cm

By given condition,

Area of Î”ABC = Area of square

âˆ´ \(\frac{1}{2}\) Ã— Base Ã— Altitude = \({(Side length)}^{2}\)

â‡’ \(\frac{1}{2}\) Ã— 9 Ã— l = 36

â‡’ l = \(\frac{36 Ã— 2}{9}\) = 8 cm

**4. What is the area of an equilateral triangle having altitude equal to 2âˆš3 cm?**

**Answer**: Option: D

**Explanation**:

We know that,

altitude of an equilateral triangle

= \(\frac{âˆš3}{2}\) a

â‡’ 2âˆš3 = \(\frac{âˆš3}{2}\) a

â‡’ a = 4 cm

âˆ´ Area of equilateral triangle

= \(\frac{âˆš3}{4}\) \({a}^{2}\) = \(\frac{âˆš3}{4}\) Ã— \({(4)}^{2}\) = 4âˆš3 \({cm}^{2}\)

**5. If a circle circumscribes a rectangle with side 16 cm and 12 cm, then what is the area of the circle ?**

**Answer**: Option: C

**Explanation**:

In right-angled Î”ABC,

AC = \(\sqrt{{AB}^{2} + {BC}^{2}}\) = \(\sqrt{{16}^{2} + {12}^{2}}\) = \(\sqrt{400}\) = 20 cm

Diameter of circumcircle = 20 cm

âˆ´ Radius of circumcircle (r) = 10 cm

Area of circumcircle = Ï€\({r}^{2}\) = Ï€ Ã— \({(10)}^{2}\) = 100Ï€ \({cm}^{2}\)