## Mensuration Worked out Problem for Bank Exam

# Mensuration Worked out Problem

Mensuration Worked out Problem for Bank Exam, SSC, SBI PO and other government exam.

**The area of a rectangular field is 460 square meters. If the length is 15 percent more than the breadth, what is breadth of the rectangular field ?**

a) 15 b) 26 c) 34.5 d) Cannot be determined e) None of these

**Answer :** e) none of these

**Solution:** Let breadth of a rectangle be x

Then length = x + 15x/100

= x + 3x/20

= 23x/20

Area of a rectangle = 460

x × 23x/20 = 460

x × x = 400

x = 20

**If the length of the diagonal AC of a square ABCD is 5.2 cm, then the area of the square is :**

a) 15.12 sq. cm b) 13..52 sq.cm c) 12.62 sq.cm d) 10.00 sq. cm

**Answer:** b) 13.52 sq.cm

**Solution:** since the square is a rhombus and the diagonals of a square are equal

Area of a rhombus = 1/2 × product of diagonals

= 1/2× 5.2×5.2

= 2.6×5.2

= 13.52

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**3.if the diagonal of two squares are in the ratio of 2:5, their area will be in the ratio of**

a) √2:√5 b) 2:5 c) 4:25 d) 4:5

**Answer:** c) 4:25

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**Solution:** let the diagonals of the square be 2x and 5x respectively

Area of one square = 1/2 × 2x × 2x

= 4x^{2}/2

Area of other square = 1/2 × 5x × 5x

= 25x^{2}/2

Ratio of areas of both the square = 4x^{2}/2 25x^{2}/2

= 4/25

= 4:25

**From four corners of a square sheet of side 4 cm, four pieces, each in the shape of arc of a circle with radius 2 cm, are cut out. The area of the remaining portion is**

a) 8 – π b) 16 – 4 π c) 16 – 8 π d) 4 – 2 π

**Answer: **b) 16 – 4 π

**Solution:** area of a square sheet = 4 × 4 = 16 sq. cm

Since the angle between two sides of a square is 90 degree and the radius of arc is 2 cm

S, area of 4 arc will be = 4 × 90/360 π2^{2}

= 4π

Area of remaining portion = 16 – 4π

## Mensuration Worked out Problem for Bank Exam

**The cost of building a fence around a circular field is rupee 7700 at the rate of rupee 14 per foot . what is the area of the circular field ?**

a) 24062.5 b) 23864.4 c) 24644.5 d) Cannot be determined

e) None of these

**Answer:** a) 24062.5

**Solution:** length of a fence = 7700/14

= 550

Circumference of a circle field = 2πr

2πr = 550

2×22/7×r = 550

r = 175/2

Area of circular field = π r^{2}

= 22/7 × 175/2 × 175/2

= 24062.5

**The sides of a triangle are 6 cm, 8 cm and 10 cm. The area (in cm**^{2}) of the triangle formed by joining the mid_points of this triangle is:

a) 12 b) 24 c) 6 d) 18

**Answer:** c) 6

Solution: a = 10, b = 8, c = 6

s= (6 + 8 + 10)/2

S = 12

Area of a triangle ABC = √12(12-10) (12-6) (12-8)

= √12(2) (6)) (4)

= 24

All the four triangles ADE, BDF,CEF and DEF are congruent, so their area are equal

Hence, area of triangle DEF = 24/4

= 6 sq cm.

**If D and E are the mid_points of the side AB and AC respectively of the ΔABC in the figure given here, the shaded region of the triangle is what percent of the whole triangular region?**

a) 50% b) 25% c) 80% d) 75%

**Answer:** d) 75%

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Solution: Since D and E are the mid_ points of AB and AC respectively

So, DE is parallel to BC and 2DE = BC

ΔADE∼ΔABC

area ΔADE/area** Δ**ABC = (DE)^{2}/(BC)^{2}

= (DE)^{2}/(2DE)^{2}

= 1/4

area of shaded part /area **Δ**ABC = ¾

% = 3/4100

= 75%

**If the length of a rectangle is increased by 20% and breadth is decreased by 20% . then its area**

a) Increases by 4% b) Decreases by 4% c) Decreases by 1% d) Remains unchanged

**Answer: **b) decreased by 4%

**Solution:** let the length of rectangle be 10

And its breadth be 10

Then, area = 100

When the length is increased by 20%, new length = 12

When its breadth is decreased by 20%, new breadth = 8

then , new area = 96

So, area decreased by 4%

**If the surface area of a cube is 294 sq. cm, then its volume(in sq. cm) is**

a) 216 b) 125 c) 343 d) 512

**Answer **c) 343

**Solution** surface area of a cube = 294

6a^{2}= 294

a^{2}= 49

a = 7

Volume of a cube = a^{3}

= 7 × 7 × 7

= 343

**A right triangle with sides 3 cm, 4 cm and 5 cm is rotated about the side 3 cm to form a cone, the volume of the cone so formed is**

a) 16 π b) 12 π c) 15 π d) 20 π

**Answer:** a) 16 π

**Solution:**

when a triangle of side 3cm, 4cm and 5cm is rotated about the side 3 cm, then a cone of radius 4 cm and height 3 cm is formed

Volume of a cone = 1/3 π r^{2 }h

= 1/3 π × 4 × 4 × 3

= 16 π

**A wooden box measures 20 cm by 12 cm by 10 cm. Thickness of the wood is 1 cm. Volume of wood to make the box (in cubic cm) is**

a) 960 b) 519 c) 2400 d) 1120

**Answer** a) 960

**Solution** outer dimension of wooden box = 20 cm , 12 cm and 10 cm

Outer volume of box = 20 × 12 × 10

= 2400 cc ( cubic centimeter)

Thickness of wood = 1 cm

Inner dimension of wooden box = 20 – 2 = 18 cm, 12 – 2 = 10 cm and 10 – 2 = 8 cm

Inner volume of box = 18 × 10 × 8

= 1440 cc

Volume of wood in the box = 2400 – 1440 = 960 cc

**Water flows into a tank which is 200m long and 150m wide, through a pipe of cross-section 0.3m × 0.2m at 20 km/hour. Then the time (in hours) for the water level in the tank to reach 8m is**

a) 50 b) 120 c) 150 d) 200

**Answer** d) 200

**This Question asked by Best Bank Question**

**Solution: **speed of water = 20 km/hour

= 20000 m/hour

Area of base = 0.3m × 0.2m

= 0.06 sq. meter

Volume of water flows in 1 hour = area of base× speed of water in 1 hour

= 0.06 × 20000

= 1200 cubic meter

Volume of tank up to 8 m height = 200 × 150 × 8

= 240000 cubic meter

Time to reach the water 8 m high = 240000/1200

= 200 hours

**The base of a right prism is a quadrilateral ABCD. given that AB = 9 cm, BC = 14 cm, CD = 13 cm, DA = 12 cm and angle DAB = 90 degree.**If the volume of the prism be 2070 cm^{3}, then the area of the lateral surface is

a) 720 b) 810 c) 1260 d) 2070

**Answer** a) 720

**Solution** Area of quadrilateral ABCD = area of ΔABD + area of ΔBCD

Since ΔABD is a right angled triangle

BD = √AB^{2} + AD^{2}

= √9^{2} + 12^{2}

= √225

= 15 cm

Area of Δ ABD = 1/2 × 9 × 12

= 54 sq.cm

For ΔBCD, s = (13 + 15 + 14)/2

= 21

area of ΔBCD = √21(21 – 13) (21 – 14) (21 – 15)

= √21 × 8 × 7 × 6

= 84 sq. cm

Area of quadrilateral ABCD = 54 + 84

= 138 sq. cm

Volume of prism = area of base×height

2070 = 138 × h

Height = 2070/138

= 15 cm

Perimeter of quadrilateral ABCD = 9 + 14 + 13 + 12

= 48 cm

Area of lateral surface = perimeter of base×height

= 48 × 15

= 720 sq. cm