Mensuration MCQ Quiz - Objective Question with Answer for Mensuration - Download Free PDF

Given:

Diameter of the sphere = 98 cm

Radius (r) = Diameter ÷ 2 = 98 ÷ 2 = 49 cm

Formula used:

Surface Area of a Sphere = 4 × π × r²

Where, r = radius

Calculations:

Surface Area = 4 × π × r²

⇒ Surface Area = 4 × 22/7 × 49²

⇒ Surface Area = 4 × 22/7 × 2401

⇒ Surface Area = 30,184 cm2

∴ The correct answer is option (4).

Given:

Sum of area of square and rectangle = 4284 sq m

Side of square = Breadth of rectangle = x

Perimeter of rectangle = 2(L + B) = 204

Formula used:

Area of square = x²

Area of rectangle = L × x

Calculations:

Total area = x² + Lx = 4284

L + x = 102 ⇒ L = 102 - x

x² + x(102 - x) = 4284

x² + 102x - x² = 4284 ⇒ 102x = 4284 ⇒ x = 42

⇒ L = 102 - 42 = 60

⇒ Area of rectangle = 60 × 42 = 2520 sq m

Cost = 2520 × 2.5 = ₹6300

∴ Total expenditure to cut the grass is ₹6300.

Given:

Area of the square = 144 m²

Formula used:

Area of a square = Side × Side

Perimeter of a square = 4 × Side

Calculation:

Side = √(Area) = √144 = 12 m

Perimeter = 4 × Side = 4 × 12 = 48 m

∴ The perimeter of the square is: 48 m

Given:

Area of rectangle = 35 cm²

Length = 7 cm

Formula used:

Area = Length × Breadth

Perimeter = 2 × (Length + Breadth)

Calculation:

⇒ 35 = 7 × Breadth

⇒ Breadth = 35 ÷ 7 = 5 cm

⇒ Perimeter = 2 × (7 + 5) = 2 × 12 = 24 cm

∴ The perimeter of the rectangle is: 24 cm

Given:

Perimeter of square = 32 cm

Area of another square = x + 10 cm²

Side of another square = 2 × side of earlier square

Formula used:

Perimeter = 4 × side

Area = side²

Calculations:

⇒ side = 32 ÷ 4 = 8 cm

New side = 2 × 8 = 16 cm

Area = 16² = 256 cm²

x + 10 = 256 ⇒ x = 246

Now, x = 246 ⇒ Perimeter = 246 cm

Side = 246 ÷ 4 = 61.5 cm

∴ Side of square whose perimeter is x cm is 61.5 cm.

Given:

Diameter of semicircle = 14√2 cm

Radius = 14√2/2 = 7√2 cm

Total no. of chords = 6

Concept:

Since the chords are equal in length, they will subtend equal angles at the centre. Calculate the area of one sector and subtract the area of the isosceles triangle formed by a chord and radius, then multiply the result by 6 to get the desired result.

Formula used:

Area of sector = (θ/360°) × πr²

Area of triangle = 1/2 × a × b × Sin θ

Calculation:

The angle subtended by each chord = 180°/no. of chord

⇒ 180°/6

⇒ 30°

Area of sector AOB = (30°/360°) × (22/7) × 7√2 × 7√2

⇒ (1/12) × 22 × 7 × 2

⇒ (77/3) cm²

Area of triangle AOB = 1/2 × a × b × Sin θ

⇒ 1/2 × 7√2 × 7√2 × Sin 30°

⇒ 1/2 × 7√2 × 7√2 × 1/2

⇒ 49/2 cm²

∴ Area of shaded region = 6 × (Area of sector AOB - Area of triangle AOB)

⇒ 6 × [(77/3) – (49/2)]

⇒ 6 × [(154 – 147)/6]

⇒ 7 cm²

∴ Area of shaded region is 7 cm²

Formula used

Area = length × breath

Calculation

The garden EFGH is shown in the figure. Where EF = 220 meters & EH = 70 meters.

The width of the path is 4 meters.

Now the area of the path leaving the four colored corners

= [2 × (220 × 4)] + [2 × (70 × 4)]

= (1760 + 560) square meter

= 2320 square meters

Now, the area of 4 square colored corners:

4 × (4 × 4)

{∵ Side of each square = 4 meter}

= 64 square meter

The total area of the path = the area of the path leaving the four colored corners + square colored corners

⇒ Total area of the path = 2320 + 64 = 2384 square meter

∴ Option 4 is the correct answer.

Given:

The width of the path around a square field = 4.5 m

The area of the path = 105.75 m2

Formula used:

The perimeter of a square = 4 × Side

The area of a square = (Side)²

Calculation:

Let, each side of the field = x

Then, each side with the path = x + 4.5 + 4.5 = x + 9

So, (x + 9)² - x² = 105.75

⇒ x2 + 18x + 81 - x2 = 105.75

⇒ 18x + 81 = 105.75

⇒ 18x = 105.75 - 81 = 24.75

⇒ x = 24.75/18 = 11/8

∴ Each side of the square field = 11/8 m

The perimterer = 4 × (11/8) = 11/2 m

So, the total cost of fencing = (11/2) × 100 = Rs. 550

∴ The cost of fencing of the field is Rs. 550

Shortcut TrickIn such types of questions, 

Area of path outside the Square is, 

⇒ (2a + 2w)2w = 105.75

here, a is a side of a square and w is width of a square

⇒ (2a + 9)9 = 105.75

⇒ 2a + 9 = 11.75

⇒ 2a = 2.75

Perimeter of a square = 4a

⇒ 2 × 2a = 2 × 2.75 = 5.50

costing of fencing = 5.50 × 100 = 550

∴ The cost of fencing of the field is Rs. 550

Given : 

Length of an arc of a circle is 4.5π.

Area of ​​the sector circumscribed by it is 27π cm2.

Formula Used : 

Area of sector = θ/360 × πr²

Length of arc = θ/360 × 2πr

Calculation : 

According to question,

⇒ 4.5π = θ/360 × 2πr 

⇒ 4.5 = θ/360 × 2r   -----------------(1)

⇒ 27π = θ/360 × πr2 

⇒ 27 = θ/360 × r2       ---------------(2)

Doing equation (1) ÷ (2)

⇒ 4.5/27 = 2r/πr²

⇒ 4.5/27 = 2/r

⇒ r = (27 × 2)/4.5

⇒ Diameter = 2r = 24

∴ The correct answer is 24.

Given:

The sides of an equilateral triangle are increased by 34%.

Formula used:

Effective increment % = Inc.% + Inc.% + (Inc.²/100) 

Calculation:

Effective increment = 34 + 34 + {(34 × 34)/100}

⇒ 68 + 11.56 = 79.56%

∴ The correct answer is 79.56%.

Given:

The side of the square = 22 cm

Formula used:

The perimeter of the square = 4 × a    (Where a = Side of the square)

The circumference of the circle = 2 × π × r     (Where r = The radius of the circle)

Calculation:

Let us assume the radius of the circle be r

⇒ The perimeter of the square = 4 × 22 = 88 cm

⇒ The circumference of the circle = 2 × π ×  r

⇒ 88 = 2 × (22/7) × r

⇒ 

⇒ r = 14 cm

∴ The required result will be 14 cm.

Given:

The radius of a solid hemisphere is 21 cm.

The ratio of the cylinder's curved surface area to its Total surface area is 2/5.

Formula used:

The curved surface area of the cylinder = 2πRh

The total surface area of cylinder = 2πR(R + h)

The volume of the cylinder = πR²h

The volume of the solid hemisphere = 2/3πr³ 

(where r is the radius of a solid hemisphere and R is the radius of a cylinder)

Calculations:

According to the question,

CSA/TSA = 2/5

⇒ [2πRh]/[2πR(R + h)] = 2/5

⇒ h/(R + h) = 2/5

⇒ 5h = 2R + 2h

⇒ h = (2/3)R .......(1)

The cylinder's volume and the volume of a solid hemisphere are equal.

⇒ πR²h = (2/3)πr³

⇒ R2 × (2/3)R = (2/3) × (21)³

⇒ R³ = (21)3

⇒ R = 21 cm

∴ The radius (in cm) of its base is 21 cm.

The surface area of three faces of a cuboid sharing a vertex are 20 m², 32 m² and 40 m²,

⇒ L × B = 20 sq. Mt

⇒ B × H = 32 sq. Mt

⇒ L × H = 40 sq. Mt

⇒ L × B × B × H × L × H = 20 × 32 × 40

⇒ L²B²H² = 25600

⇒ LBH = 160

∴ Volume = LBH = 160 m³

Given:

Each side of the cube = 8 cm

The rectangular container has a length of 16 cm, breadth of 8 cm, and height of 15 cm

Formula used:

The volume of cube = (Edge)3

The volume of a cuboid = Length × Breadth × Height

Calculation:

The volume of cube = The volume of the rectangular container with a length of 16 cm, breadth of 8 cm, and height of the water level rise

Let, the height of the water level will rise = x cm

So, 8³ = 16 × 8 × x

⇒ 512 = 128 × x

⇒ x = 512/128 = 4

∴ The rise of water level (in cm) is 4 cm

Given:

Radius of the wheel of car = 14 cm

Speed of car = 132 km/hr

Formula Used:

Circumference of the wheel =  

1 km = 1000 m

1m = 100 cm

1hr = 60 mins.

Calculation:

Distance covered by the wheel in one minute =  = 220000 cm.

Circumference of the wheel =  =  = 88 cm

∴ Distance covered by wheel in one revolution = 88 cm

∴ The number of revolutions in one minute =  = 2500.

∴ Therefore the correct answer is 2500.