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

Calculation

ATQ, 2 ×(2πr) = 176

Or, 2πr = 88

Or, πr = 44

So, r = 14 cm

Volume of the cylinder = [22/7] r2h = 3528π

So, 196h = 3528

So, h = 18

Side of the square = 18 × [4 / 3] = 24 cm

Required area = 24 × 24 = 576 cm

Given:

breadth = 5 m

diameter = 14 m

height = 24 m

Rate = Rs. 25/m

Formula used:

CSA(Cone) = 22/7 x r x l

l² = h² + r² 

r = radius of the cone/tent(here)

h = slant height

CSA = Curved Surface Area

Solution:

r = 14/2 = 7 m

l = 

l = 25 m

CSA = 22/7 x 7 x 25 

CSA = 550 m² 

Cost of cloth required = 550 x 25 = Rs. 13750

Hence, the correct option is 2.

Given:

Edge of solid iron cube = 24 cm

Thickness of rectangular sheet = 2 mm = 0.2 cm

Ratio of length (l) to breadth (b) = 6:5

Formula Used:

Volume of cube = Volume of rectangular sheet

Volume of cube = Edge³

Volume of rectangular sheet = Length × Breadth × Thickness

l / b = 6 / 5

Calculation:

Volume of cube = Edge³

⇒ Volume = 24³

⇒ Volume = 13824 cm³

Volume of rectangular sheet = l × b × 0.2

⇒ 13824 = l × b × 0.2

⇒ l × b = 13824 / 0.2

⇒ l × b = 69120

l / b = 6 / 5

⇒ l = 6k, b = 5k

⇒ l × b = 6k × 5k

⇒ 69120 = 30k²

⇒ k² = 69120 / 30

⇒ k² = 2304

⇒ k = √2304

⇒ k = 48

l = 6k = 6 × 48 = 288 cm

b = 5k = 5 × 48 = 240 cm

l + b = 288 + 240

⇒ l + b = 528 cm

The correct answer is option 2 (528 cm).

Given:

Radius of cylinder = 3 inches.

Height of cylinder = 8 inches.

Formula Used:

Volume of cylinder = πr²h

Volume of hemisphere = (2/3)πr³

Calculation:

Volume of cylinder = π × 3² × 8

⇒ Volume of cylinder = π × 9 × 8

⇒ Volume of cylinder = 72π cubic inches

Volume of hemisphere = (2/3)π × 3³

⇒ Volume of hemisphere = (2/3)π × 27

⇒ Volume of hemisphere = 18π cubic inches

Number of hemispheres = Volume of cylinder / Volume of hemisphere

⇒ Number of hemispheres = 72π / 18π

⇒ Number of hemispheres = 4

The number of hemispheres formed is 4.

Given:

Edge of the wooden cube = 14 cm.

Formula Used:

Volume of the cube = Edge³

Volume of the cone = (1/3) × π × r² × h

Maximum volume of the cone when it is carved from a cube: r = h/2, where h = edge of the cube.

Volume of the removed portion = Volume of the cube - Volume of the cone.

Calculation:

Volume of the cube = Edge³

⇒ Volume of the cube = 14³

⇒ Volume of the cube = 2744 cm³

Radius (r) of the cone = h / 2 = 14 / 2 = 7 cm

Height (h) of the cone = 14 cm

Volume of the cone = (1/3) × π × r² × h

⇒ Volume of the cone = (1/3) × 22/7 × 7² × 14

⇒ Volume of the cone = (1/3) × 22 × 49 × 2

⇒ Volume of the cone = (1/3) × 2156

⇒ Volume of the cone = 718.67 cm³

Volume of the removed portion (V) = Volume of the cube - Volume of the cone

⇒ V = 2744 - 718.67

⇒ V = 2025.33 cm³

3V = 3 × 2025.33

⇒ 3V = 6076 cm³

The value of 3V is 6076 cm³.

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.