6 + Interview Questions in MIX in MENSURATION Page 1 InterviewSolution

1.	The perimeter of one face of a cube is 32 cm. Its volume must be1). 800 cm32). 300 cm33). 125 cm34). 400 cm3
Answer» Let the side of the cube be ‘a’. We know that, PERIMETER of one face of a cube = 4a ⇒ 4a = 32 ⇒ a = 32/4 = 8 cm We know that, Volume of the cube = a³ = 8³ = 512 cm³

1.

The perimeter of one face of a cube is 32 cm. Its volume must be1). 800 cm32). 300 cm33). 125 cm34). 400 cm3

Answer»

Let the side of the cube be ‘a’.

We know that,

PERIMETER of one face of a cube = 4a

⇒ 4a = 32

⇒ a = 32/4 = 8 cm

We know that,

Volume of the cube = a³ = 8³ = 512 cm³

Discussion

2.	1). 2826.42). 1256.63). 2433.54). 3925.5
Answer» spherical shell has an inner RADIUS of 10 cm. The outer radius is 50% more in length than inner radius. ⇒ Outer radius = 10 + (10 × (50/100)) = 15 cm After shell is cut into two equal halves, each of the halves will be a hemispherical shell. A hemispherical shell has three SURFACES: Outer curved SURFACE, inner curved surface and planar circular ring SEPARATING the two curved surfaces. Surface area of outer curved surface = 2 × π × square of outer radius Surface area of inner curved surface = 2 × π × square of inner radius Surface area of planar circular ring = π × (square of outer radius – square of inner radius) ∴ Total surface area of each of the halves = (2 × 3.14 × 15 × 15) + (2 × 3.14 × 10 × 10) + (3.14 × (15 × 15 – 10 × 10)) = 1413 + 628 + 392.5 = 2433.5 square cm

2.

1). 2826.42). 1256.63). 2433.54). 3925.5

Answer»

spherical shell has an inner RADIUS of 10 cm. The outer radius is 50% more in length than inner radius.

⇒ Outer radius = 10 + (10 × (50/100)) = 15 cm

After shell is cut into two equal halves, each of the halves will be a hemispherical shell.

A hemispherical shell has three SURFACES: Outer curved SURFACE, inner curved surface and planar circular ring SEPARATING the two curved surfaces.

Surface area of outer curved surface = 2 × π × square of outer radius

Surface area of inner curved surface = 2 × π × square of inner radius

Surface area of planar circular ring = π × (square of outer radius – square of inner radius)

∴ Total surface area of each of the halves = (2 × 3.14 × 15 × 15) + (2 × 3.14 × 10 × 10) + (3.14 × (15 × 15 – 10 × 10))

= 1413 + 628 + 392.5 = 2433.5 square cm

Discussion

3.	1). 62 : 552). 55 : 623). 31 : 194). 59 : 55
Answer» Solution : Given :ABCD is a TRAPEZIUM with AE and BF are perpendicular to CD side of trapezium and, AD= BC = 25cm Inside the trapezium there is a circle of radius 3.5cm and if we clear notice the diagram diameter of the circle will give us the LENGTH of EF,AB,AE,BF. Diameter of circle (2r) = length of EF So, EF = 2× 3.5 EF = AB = AE = BF = 7cm Length of DE =√(AD² - AE²) Length of DE =√(25² - 7²) Length of DE =√(625 - 49) Length of DE =√576 = 24cm Since BC = AD and BF = AE DE = FC = 24cm Area of trapezium ABCD = (AE/2)(AB + DC) AE =7cm , AB = 7cm , DC = DE + EF + FC = 24 + 7 + 24 = 55cm Area of trapezium ABCD = (7/2)(7 + 55) Area of trapezium ABCD = (3.5)(62) Area of trapezium ABCD =217cm² Area of trapezium ABCE = (BF/2)(AB + EC) BF = 7cm , AB = 7cm , EC = EF + FC = 7 + 24 = 31cm Area of trapezium ABCE = (7/2)(7 + 31) Area of trapezium ABCE = (3.5)(38) Area of trapezium ABCE = 133cm² Required ratio: Area of trapezium ABCD : Area of trapezium ABCE = 217 : 133 = 31 : 19 So, the correct option is 3).31 : 19

3.

1). 62 : 552). 55 : 623). 31 : 194). 59 : 55

Answer»

Solution :

Given :ABCD is a TRAPEZIUM with AE and BF are perpendicular to CD side of trapezium and,

AD= BC = 25cm

Inside the trapezium there is a circle of radius 3.5cm and if we clear notice the diagram diameter of the circle will give us the LENGTH of EF,AB,AE,BF.

Diameter of circle (2r) = length of EF

So, EF = 2× 3.5

EF = AB = AE = BF = 7cm

Length of DE =√(AD² - AE²)

Length of DE =√(25² - 7²)

Length of DE =√(625 - 49)

Length of DE =√576 = 24cm

Since BC = AD and BF = AE

DE = FC = 24cm

Area of trapezium ABCD = (AE/2)(AB + DC)

AE =7cm , AB = 7cm , DC = DE + EF + FC = 24 + 7 + 24 = 55cm

Area of trapezium ABCD = (7/2)(7 + 55)

Area of trapezium ABCD = (3.5)(62)

Area of trapezium ABCD =217cm²

Area of trapezium ABCE = (BF/2)(AB + EC)

BF = 7cm , AB = 7cm , EC = EF + FC = 7 + 24 = 31cm

Area of trapezium ABCE = (7/2)(7 + 31)

Area of trapezium ABCE = (3.5)(38)

Area of trapezium ABCE = 133cm²

Required ratio:

Area of trapezium ABCD : Area of trapezium ABCE

= 217 : 133

= 31 : 19

So, the correct option is 3).31 : 19

Discussion

4.	1). 14 cm2). 16 cm3). 18 cm4). 20 cm
Answer» <P>For a RHOMBUS with side ‘a’ and diagonal ‘p’ and ‘q’, we can write, ⇒ 4a² = p² + q² Given, side = a = 10 cm, diagonal = p = 12 cm ⇒ 4(10)² = (12)² + q² ⇒ q² = 400 – 144 = 256 ⇒ q = √256 = 16 cm ∴ LENGTH of other diagonal = 16 cm

4.

1). 14 cm2). 16 cm3). 18 cm4). 20 cm

Answer»

<P>For a RHOMBUS with side ‘a’ and diagonal ‘p’ and ‘q’, we can write,

⇒ 4a² = p² + q²

Given, side = a = 10 cm, diagonal = p = 12 cm

⇒ 4(10)² = (12)² + q²

⇒ q² = 400 – 144 = 256

⇒ q = √256 = 16 cm

∴ LENGTH of other diagonal = 16 cm

Discussion

5.	The perimeter of a rectangle and equilateral triangle are same. Also, one of the sides of the rectangle is equal to the side of the triangle. The ratio of the area of the rectangle and the triangle is:1). √2 : 12).3). √3 : 14). 1 : √3
Answer» LET length of rectangle = x units and breadth = y units ∴ side of triangle = y units Since their PERIMETERS are equal, ⇒ 2X + 2y = 3y ⇒ 2x = y …………(i) ∴ Area of rectangle = xy Area of triangle = y²√3/4 Required RATIO $( = \frac{{xy}}{{\frac{{{y^2}\sqrt 3 }}{4}}}\; = \frac{{4X}}{{y\sqrt 3 }} = \frac{{4x}}{{2x\sqrt 3 }} = \frac{2}{{\sqrt 3 }})$

5.

The perimeter of a rectangle and equilateral triangle are same. Also, one of the sides of the rectangle is equal to the side of the triangle. The ratio of the area of the rectangle and the triangle is:1). √2 : 12).3). √3 : 14). 1 : √3

Answer»

LET length of rectangle = x units and breadth = y units

∴ side of triangle = y units

Since their PERIMETERS are equal,

⇒ 2X + 2y = 3y

⇒ 2x = y …………(i)

∴ Area of rectangle = xy

Area of triangle = y²√3/4

Required RATIO $( = \frac{{xy}}{{\frac{{{y^2}\sqrt 3 }}{4}}}\; = \frac{{4X}}{{y\sqrt 3 }} = \frac{{4x}}{{2x\sqrt 3 }} = \frac{2}{{\sqrt 3 }})$

Discussion

6.	1). 4 metres2). 1200 metres3). 12 metres4). 400 metres
Answer» Let the maximum length of wire that can be DRAWN is L cm. Radius of wire = 0.6 mm/2 = 0.3 mm = 0.03 cm Volume of Copper will be same in both cases. Shape of wire is cylindrical. ⇒ Volume of sphere = Volume of cylinder ⇒ (4/3) × 3.14 × 3 × 3 × 3 = 3.14 × 0.03 × 0.03 × L ⇒ L = 40000 ∴ Maximum length of wire that can be drawn = 40000 cm = 400 metres

6.

1). 4 metres2). 1200 metres3). 12 metres4). 400 metres

Answer»

Let the maximum length of wire that can be DRAWN is L cm.

Radius of wire = 0.6 mm/2 = 0.3 mm = 0.03 cm

Volume of Copper will be same in both cases. Shape of wire is cylindrical.

⇒ Volume of sphere = Volume of cylinder

⇒ (4/3) × 3.14 × 3 × 3 × 3 = 3.14 × 0.03 × 0.03 × L

⇒ L = 40000

∴ Maximum length of wire that can be drawn = 40000 cm = 400 metres

Discussion

Explore topic-wise InterviewSolutions in .

The perimeter of one face of a cube is 32 cm. Its volume must be1). 800 cm32). 300 cm33). 125 cm34). 400 cm3

1). 2826.42). 1256.63). 2433.54). 3925.5

1). 62 : 552). 55 : 623). 31 : 194). 59 : 55

1). 14 cm2). 16 cm3). 18 cm4). 20 cm

The perimeter of a rectangle and equilateral triangle are same. Also, one of the sides of the rectangle is equal to the side of the triangle. The ratio of the area of the rectangle and the triangle is:1). √2 : 12).3). √3 : 14). 1 : √3

1). 4 metres2). 1200 metres3). 12 metres4). 400 metres