practice question on Mensuration (Aptitude) for NDA , Bank , GMAT , SAT , GATE

Mensuration

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Question no. 1

The side and the height of a rhombus are 13 and 20 cms respectively. Find the area.

looks_one 260 cm²

looks_two 275 cm²

looks_3 290 cm²

looks_4 None of these

Option looks_one 260 cm²

Solution :
side = 13 , height = 20
area = height * side = 20*13 = 260 cm²

Question no. 2

If the ratio of areas of two squares is 9 : 1, the ratio of their perimeter is:

looks_one 9:1

looks_two 3:4

looks_3 3:1

looks_4 1:3

Option looks_3 3:1

Solution :

Question no. 3

A circle and a rectangle have the same perimeter. The sides of the rectangle are 18 cm and 26 cm. What is the area of the circle ?

looks_one 88 cm²

looks_two 154 cm²

looks_3 1250 cm²

looks_4 616 cm²

option looks_4 616 cm²

Solution :
r is the radius of circle and a ,b are length and breadth of rectangle
2πr = 2(a+b)
πr = a+ b = 18 + 26 = 44
(22/7)r = 44 ⇒ r = 14 cm
Area = πr² = (22/7)*14*14 = 616 cm²

Question no. 4

If the perimeter and diagonal of a rectangle are 14 and 5 cms respectively, find its area.

looks_one 12 cm²

looks_two 16 cm²

looks_3 20 cm²

looks_4 24 cm²

Option looks_one 12 cm²

Solution :
let a , b are dimension of rectangle
diagonal = (a² + b²)^1/2 = 5 ⇒ a² + b² =25
2(a+b)= 14 ⇒ a+b = 7
Squaring both sides
(a+b)² = 49
a² + b² + 2ab = 49
25 + 2ab = 49
ab = 12 cm²

Question no. 5

The area of a triangle is 615 m². If one of its sides is 123 metre, find the length of the perpendicular dropped on that side from opposite vertex.

looks_one 15

looks_two 12

looks_3 10

looks_4 None of these

Option looks_3 10

Solution :
(base) b = 123
height (prependicular from the base) h = ?
area = 0.5 * b * h
615 = 0.5 *123 *h
h = 10 m

Question no. 6

A horse is tethered to one corner of a rectangular grassy field 40 m by 24 m with a rope 14 m long. Over how much area of the field can it graze?

looks_one 154 m²

looks_two 308 m²

looks_3 150 m²

looks_4 None of these

Option looks_one 154 m²

Solution :

Question no. 7

When the circumference and area of a circle are numerically equal, then the diameter is numerically equal to

looks_one area

looks_two circumference

looks_3 4

looks_4 2π

Option looks_3 4

Solution :
r is the radius of circle
2πr = πr²
2r = r²
r = 2
D = 2r = 4

Question no. 8

If the area of a circle decreases by 36%, then the radius of a circle decreases by

looks_one 20%

looks_two 18%

looks_3 36%

looks_4 64%

Option looks_two 18%

Solution :
A = πr² Change in area w.r.t radius
ΔA = 2Δr
0.36 = 2Δr
Δr = 0.18

Question no. 9

The area of a square field is 576 km². How long will it take for a horse to run around at the speed of 12 km/h ?

looks_one 12 h

looks_two 10 h

looks_3 8 h

looks_4 6 h

Option looks_3 8 h

Solution :
let a is the side of square
a² = 576
a = 24 km
perimeter(a horse to run around) = 4a = 4*24 = 96 km
time taken by the horse to cover 96 km = distance / speed = 96/12 = 8h

Question no. 10

A cylindrical bucket of height 36 cm and radius 21 cm is filled with sand. The bucket is emptied on the ground and a conical heap of sand is formed, the height of the heap being 12 cm. The radius of the heap at the base is :

looks_one 63 cm

looks_two 53 cm

looks_3 56 cm

looks_4 66 cm

Option looks_one 63 cm

Solution :
Volume of bucket = volume of heap

Question no. 11

The altitude drawn to the base of an isosceles triangle is 8 cm and the perimeter is 32 cm. The area of the triangle is

looks_one 72 cm²

looks_two 60 cm²

looks_3 66 cm²

looks_4 None of these

Option looks_4 None of these

Solution :

Let the sides of isosceles triangle = a,a,c
perimeter = 2a + c = 32
c = 32 -2a
Altitude divides the base into two equal parts
Apply the pythagoras theorem in ABD ,
a² = c²/4 + 64 ⇒ 4a² = c² + 256
put the value of 2a = 32-c in above equation
(32 - c)² = c² + 256
1024 + c² - 64c = c² + 256
64c = 768 ⇒ c = 12
Area = 0.5 * B * h = 0.5 * 8*12 = 48 cm²

Question no. 12

Four equal circles are described about the four corners of a square so that each touches two of the others. If a side of the square is 14 cm, then the area enclosed between the circumferences of the circles is

looks_one 24 cm²

looks_two 42 cm²

looks_3 154 cm²

looks_4 196 cm²

Option looks_two 42 cm²

Solution :

Let's circle radius is r ,
2r = a( side of square )
r = 7
Area enclosed between the circumferences of the circles = a² - 4(1/4)πr²
= (14)² - (22/7)*7*7 = 196 - 154 = 42 cm²

Question no. 13

A rectangular tank measuring 5m × 4.5 m × 2.1 m is dug in the centre of the field measuring 13.5 m × 2.5. The earth dug out is spread evenly over the remaining portion of a field. How much is the level of the field raised ?

looks_one 4.0

looks_two 4.1

looks_3 4.2

looks_4 None of these

Option looks_3 4.2

Solution :
Volume of tank = 5 × 4.5 × 2.1 m³ = 47.25 m³
Area of field where mud is spread evenly = area of field - area of center field = 13.5 * 2.5 - 5 * 4.5 = 11.25 m²
Height of field * Area = volume of mud
Height of field = volume of mud / area = 47.25/11.25 = 4.2 m

Question no. 14

A cow is tethered in the middle of a field with a 14 feet long rope. If the cow grazes 100 sq. ft. per day, then approximately what time will be taken by the cow to graze the whole field?

looks_one 2 days

looks_two 6 days

looks_3 18 days

looks_4 24 days

Option looks_two 6 days

Solution :
Area grazed By cow = πr² = (22/7)*14 * 14 = 616 sq ft
Approximate time to graze (if 100 sq ft in a day ) = 616/100 = 6.16 = 6 hr (approximate)

Question no. 15

If the volume of a sphere is divided by its surface area, the result is 27 cms. The radius of the sphere is

looks_one 9 cm

looks_two 27 cm

looks_3 81 cm

looks_4 243 cm

Option looks_3 81 cm

Solution :