# IBPS RRB Mensuration Quiz 2

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# IBPS RRB Mensuration Quiz 2

### Introduction

Mensuration  is a topic in Geometry which is a branch of mathematics. Mensuration deals with length, area, and volume 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 like IBPS RRB, RRB ALP/Technical Exams/Junior Engineer Recruitment Exams, SSC CGL, SSC CHSL, IBPS, SBI PO, SBI Clerks, CAT and etc.

### Quiz

1. The area of a sector of a circle of radius 36 cm is 72π$${cm}^{2}$$. The length of the corresponding arc of the sector is

A. π cm
B. 2π cm
C. 3π cm
D. 4π cm

Explanation:
Given that, radius (r) = 36 cm

And, Area of sector = 72π $${cm}^{2}$$

⇒ πr2θ = $$\frac{72π}{360°}$$

∴ θ = $$\frac{72π × 360°}{π{r}^{2}}$$

= 72 × 360° =$$\frac{20°}{36 × 36}$$

Now, length of arc = $$\frac{πrΘ}{180°}$$

= π × 36 × 20° = 4π 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

A. $${a}^{2}$$
B. 2$${a}^{2}$$
C. 3$${a}^{2}$$
D. 4$${a}^{2}$$

Explanation:

Given that, Diameter = 2a

For inscribed square,

Diameter of circle = Diagonal of inner square

For circumscribed square,

Diameter of circle = Side of outer square

∴ Area of inner square = $$\frac{1}{2}$$ $${(diagonal)}^{2}$$ = $$\frac{1}{2}$$ × $${(2a)}^{2}$$ = 2$${a}^{2}$$

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

Now, Required difference = 4$${a}^{2}$$ – 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 – z equal to ?

A. 48 $${cm}^{2}$$
B. 32 $${cm}^{2}$$
C. 0
D. None of these

Explanation:

Given that, AB = 6 cm and AC = 8 cm

In ΔABC, by Pythagoras theorem,

BC = $${AB}^{2}$$ +$${AC}^{2}$$ = 62 + 82 = 100 = 10 cm

Now, Area of that semi-circle which diameter is AB = $$\frac{{π(3)}^{2}}{2}$$
∴ x = 9π $${cm}^{2}$$
2 Similarly, Area of that semi-circle which diameter is AC = $$\frac{{π(4)}^{2}}{2}$$
∴ y = 16π $${cm}^{2}$$
2 Similarly, Area of that semi-circle which diameter is BC = $$\frac{{π(5)}^{2}}{2}$$

∴ z = $$\frac{25π}{2}$$ $${cm}^{2}$$

Now, x + y – z = ($$\frac{9π}{2}$$ + $$\frac{16π}{2}$$) – $$\frac{25π}{2}$$ = 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

A. 2 units
B. 3 units
C. 6 units
D. infinity

Explanation:

Perimeter of all triangles

= (3 × 1) + (3 × 0.5) + (3 × 0.25) + (3 × 0.125)

= 3 + 1.5 + 0.75 + 0.375 = 5.625 ≈ 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 ΔACD ?

A. $$\frac{π}{2}$$
B. π
C.$$\frac{π}{4}$$
D.

Explanation:

Required ratio

= Area of circle : Area of ΔACD

= π$$\frac{{r}^{2}}{ 1 × 2r × r}$$ = $$\frac{π}{2}$$

1. The area of an isosceles ΔABC with AB = AC and altitude AD = 3 cm is 12 sq cm. What is its perimeter ?

A. 18 cm
B. 16 cm
C. 14 cm
D. 12 cm

Explanation:

Given, Area of triangle ABC = 12 $${cm}^{2}$$

∴ $$\frac{1}{2}$$ × h × b = 12

⇒ $$\frac{1}{2}$$ × 3 × b = 12

⇒ b = 8 cm

Here, BD =

CD = $$\frac{b}{2}$$ = $$\frac{8}{2}$$ = 4 cm

In right-angled ΔABD, 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 × 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 ?

A. 17.5 $${cm}^{2}$$
B. 18.5 $${cm}^{2}$$
C. 20.5 $${cm}^{2}$$
D. 21.5 $${cm}^{2}$$

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 – 2 Area of semi-circle
= $${a}^{2}$$ – 2 × $$\frac{1}{3}$$ π$${r}^{2}$$

= (10)2 – 2 × $$\frac{1}{2}$$ × $$\frac{22}{7}$$ × $${(5)}^{2}$$

= 100 – 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

A. 52 $${cm}^{2}$$
B. 56 $${cm}^{2}$$
C. 60 $${cm}^{2}$$
D. 64 $${cm}^{2}$$

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

∴ Area of rectangle = l × b

⇒ 144 = 4x × 9x = 36$${x}^{2}$$

⇒ $${x}^{2}$$ = 4

⇒ x = 2

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

∴ 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 ?

A. 12.56 $${cm}^{2}$$
B. 14.56 $${cm}^{2}$$
C. 16.76 $${cm}^{2}$$
D. 22.56 $${cm}^{2}$$

Explanation:
Area of parallelogram = Base × Height = 8.06 × 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 ΔCOD is

A. 40 $${cm}^{2}$$
B. 45 $${cm}^{2}$$
C. 50 $${cm}^{2}$$
D. 80 $${cm}^{2}$$

Explanation:

Given that, CD = 20 cm

And, area of rectangle ABCD = 100 $${cm}^{2}$$

⇒ AD × CD = 100

⇒ AD × 20 = 100

∵ AD = OP = 5 cm

∴ Area of ΔCOD = $$\frac{1}{2}$$ × CD × OP = $$\frac{1}{2}$$ × 20 × 5 = 50 $${cm}^{2}$$

1. A piece of wire 78 cm long is bent in the form of an isosceles triangle. If the 1. ratio of one of the equal sides to the base is 5 : 3, then what is the length of the base?

A. 16 cm
B. 18 cm
C. 20 cm
D. 30 cm

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

By given condition,

Perimeter of isosceles triangle = Length of wire

⇒ 5x + 5x + 3x = 78

⇒ 13x = 78

⇒ x = 6 cm

∴ Length of base = 3x = 3 × 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 π = 3.14)

A. 88.78
B. 84.78
C. 67.74
D. 57.78

Explanation:
Given, r = 9 cm

The angle made by the minute hand in 20 min, θ = 120°

∴ The area swept by the minute hand in 20 min

= θ × $$\frac{π{r}^{2}}{360°}$$

= $$\frac{120° × 3.14 × 9 × 9}{360°}$$

= 84.78 $${cm}^{2}$$

3. If the area of a ΔABC 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 ?

A. 18 cm
B. 14 cm
C. 12 cm
D. 8 cm

Explanation:
Given, AB = 9 cm

Let the length of altitude to AB = l cm

By given condition,

Area of ΔABC = Area of square

∴ $$\frac{1}{2}$$ × Base × Altitude = $${(Side length)}^{2}$$

⇒ $$\frac{1}{2}$$ × 9 × l = 36

⇒ l = $$\frac{36 × 2}{9}$$ = 8 cm

4. What is the area of an equilateral triangle having altitude equal to 2√3 cm?

A. √3 sq cm
B. 2√3 sq cm
C. 3√3 sq cm
D. 4√3 sq cm

Explanation:
We know that,

altitude of an equilateral triangle

= $$\frac{√3}{2}$$ a

⇒ 2√3 = $$\frac{√3}{2}$$ a

⇒ a = 4 cm

∴ Area of equilateral triangle

= $$\frac{√3}{4}$$ $${a}^{2}$$ = $$\frac{√3}{4}$$ × $${(4)}^{2}$$ = 4√3 $${cm}^{2}$$

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

A. 48π sq cm
B. 50π sq cm
C. 100π sq cm
D. 200π sq cm

Explanation:

In right-angled ΔABC,

AC = $$\sqrt{{AB}^{2} + {BC}^{2}}$$ = $$\sqrt{{16}^{2} + {12}^{2}}$$ = $$\sqrt{400}$$ = 20 cm
Diameter of circumcircle = 20 cm

∴ Radius of circumcircle (r) = 10 cm

Area of circumcircle = π$${r}^{2}$$ = π × $${(10)}^{2}$$ = 100π $${cm}^{2}$$