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

Find Length and Breadth of the Rectangle

Let length= 9x, breadth= 5x.

Perimeter = 2(9x + 5x) = 280

2(14x) = 280 ⟹ 28x = 280 ⟹ x = 10

Thus,

Length = 9x = 90 cm, Breadth = 5x = 50 cm

Area of rectangle = 90 × 50 = 4500 cm².

The top surface of the cylinder is a circle with area πr²

Given:

πr2 = 4500 ⟹ r2 = 4500π

Given: Radius r = 120% of height h, so:

r = 1.2h ⟹ h = r / 1.2 = 5r / 6

Now, let's find the Curved Surface Area of Cylinder:

Curved surface area = 2πrh 

Substituteh = 5r / 6:

CSA = 2πr(5r/6) = 10πr2 / 6 = 5πr2 / 3

Since πr2 = 4500:

CSA = 5 x 4500 / 3 = 7500 cm2

Thus, the correct answer is 7500 cm2.

Given:

Height of cone = 2 × Radius of base circle.

Radius of base circle = r.

Total Surface Area (TSA) of cone = Area of base + Curved Surface Area.

Formula Used:

Area of base = πr².

Curved Surface Area = πr × l, where l is the slant height.

Slant height (l) = √(r² + h²).

Total Surface Area = πr² + πr × l.

Ratio = Area of base / Total Surface Area.

Calculation:

Height (h) = 2r.

Slant height (l) = √(r² + (2r)²)

⇒ l = √(r² + 4r²)

⇒ l = √(5r²)

⇒ l = r√5.

Curved Surface Area = πr × r√5

⇒ Curved Surface Area = πr²√5.

Total Surface Area = πr² + πr²√5.

⇒ Total Surface Area = πr²(1 + √5).

Area of base = πr².

Ratio = Area of base / Total Surface Area

⇒ Ratio = πr² / [πr²(1 + √5)]

⇒ Ratio = 1 / (1 + √5).

Given:

Radius of the circle = 14 cm

Radius of the semi-circle = 14 cm

Formula Used:

Perimeter of a circle = 2 × π × radius

Perimeter of a semi-circle = π × radius + diameter

Calculation:

Perimeter of the circle = 2 × π × 14

Perimeter of the circle = 28π

Perimeter of the semi-circle = π × 14 + 2 × 14

Perimeter of the semi-circle = 14π + 28

Ratio of the perimeters = (Perimeter of the circle) / (Perimeter of the semi-circle)

⇒ Ratio = (28π) / (14π + 28)

⇒ Ratio = 28π / (14(π + 2))

⇒ Ratio = 2π / (π + 2)

Approximating π = 3.14:

Ratio = (2 × 3.14) / (3.14 + 2)

⇒ Ratio = 6.28 / 5.14

⇒ Ratio ≈ 1.22 : 1

Correct Ratio ≈ 12 : 9

The ratio of their perimeters is 12 : 9.

Given:

Perimeter of the rectangle = 120 cm

Ratio of length to breadth = 7 : 8

Area of the square is 4 cm² more than the area of the rectangle.

Formula used:

Perimeter of rectangle = 2 × (Length + Breadth)

Area of rectangle = Length × Breadth

Area of square = Side²

Calculations:

Let the length = 7x and the breadth = 8x (from the ratio 7:8).

Perimeter = 2 × (Length + Breadth) = 2 × (7x + 8x) = 120

⇒ 2 × 15x = 120

⇒ 30x = 120

⇒ x = 4

Length = 7x = 7 × 4 = 28 cm

Breadth = 8x = 8 × 4 = 32 cm

Area of rectangle = Length × Breadth = 28 × 32 = 896 cm²

Let the side of the square be 's'.

Area of square = s²

We are given that the area of the square is 4 cm² more than the area of the rectangle:

s² = 896 + 4

s² = 900

⇒ s = √900

⇒ s = 30 cm

∴ The side of the square is 30 cm.

Given:

Perimeter of rectangle = 620 mm

Formula Used:

Perimeter of rectangle, P = 2(l + w)

Area of rectangle, A = l × w

For a given perimeter, the maximum area of a rectangle is achieved when it is a square (i.e., length = width).

Calculation:

⇒ 620 = 2(l + w)

⇒ l + w = 620 / 2

⇒ l + w = 310

For maximum area, l = w

⇒ l = 310 / 2

⇒ l = 155 mm

⇒ w = 155 mm

Maximum Area = l × w

⇒ Maximum Area = 155 × 155

⇒ Maximum Area = 24025 mm²

The maximum area of the rectangle with perimeter 620 mm is 24025 mm².

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

⇒ \(r = {{88\ \times\ 7 }\over {22\ \times \ 2}}\)

⇒ 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

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 = \(2\pi r\) 

1 km = 1000 m

1m = 100 cm

1hr = 60 mins.

Calculation:

Distance covered by the wheel in one minute = \(\frac{132 \times 1000 \times 100}{60}\) = 220000 cm.

Circumference of the wheel = \(2\pi r\) = \(2\times \frac{22}{7} \times 14\) = 88 cm

∴ Distance covered by wheel in one revolution = 88 cm

∴ The number of revolutions in one minute = \(\frac{220000}{88}\) = 2500.

∴ Therefore the correct answer is 2500.