Plane Figures MCQ Quiz - Objective Question with Answer for Plane Figures - Download Free PDF

Let the side length of the square be s meters.

Area of square = s²

Radius of the inscribed circle = s/2

Area of the circle = π x (s/2)² = (π x s²) ÷ 4

Cost of painting:

Inside circle: ₹6/m²

Outside circle (within square): ₹5/m²

Total cost =

⇒ ₹6 x (π x s² ÷ 4) + ₹5 x (s² – π x s² ÷ 4)

⇒ ₹6 x (πs² ÷ 4) + ₹5 x [s² – (πs² ÷ 4)]

⇒ ₹6 x (πs² ÷ 4) + ₹5 x s² x (1 – π ÷ 4)

Given total cost = ₹370.3

Now substitute π ≈ 3.14:

⇒ 6 x (3.14 x s² ÷ 4) + 5 x s² x (1 – 3.14 ÷ 4) = 370.3

⇒ 6 x (0.785s²) + 5 x s² x (1 – 0.785)

⇒ 4.71s² + 5 x s² x 0.215 = 370.3

⇒ 4.71s² + 1.075s² = 370.3

⇒ 5.785s² = 370.3

⇒ s² = 370.3 ÷ 5.785 ≈ 64

⇒ s = √64 = 8 meters

Thus, the correct answer is 8m.

Given:

Scenario 1: Length increased by 5 m, Breadth decreased by 7 m, Area increases by 8 m².

Scenario 2: Length decreased by 5 m, Breadth increased by 8 m, Area increases by 33 m².

Formula used:

Area of rectangle = Length × Breadth (l × b)

Perimeter of rectangle = 2 × (Length + Breadth)

Calculations:

Let the original length of the rectangle be 'l' meters and the original breadth be 'b' meters.

Original Area = l × b

From Scenario 1:

New length = (l + 5) m

New breadth = (b - 7) m

New Area = (l + 5)(b - 7)

Given that the area increases by 8 m²:

(l + 5)(b - 7) = lb + 8

⇒ lb - 7l + 5b - 35 = lb + 8

⇒ -7l + 5b = 8 + 35

⇒ -7l + 5b = 43 (Equation 1)

From Scenario 2:

New length = (l - 5) m

New breadth = (b + 8) m

New Area = (l - 5)(b + 8)

Given that the area increases by 33 m²:

(l - 5)(b + 8) = lb + 33

⇒ lb + 8l - 5b - 40 = lb + 33

⇒ 8l - 5b = 33 + 40

⇒ 8l - 5b = 73 (Equation 2)

Now, we have a system of two linear equations:

1) -7l + 5b = 43

2) 8l - 5b = 73

Add Equation 1 and Equation 2:

(-7l + 5b) + (8l - 5b) = 43 + 73

⇒ -7l + 8l + 5b - 5b = 116

⇒ l = 116

Substitute the value of l into Equation 1:

-7(116) + 5b = 43

⇒ -812 + 5b = 43

⇒ 5b = 43 + 812

⇒ 5b = 855

⇒ b = 171

So, the original length (l) = 116 m and original breadth (b) = 171 m.

Perimeter of the original rectangle = 2 × (l + b)

⇒ Perimeter = 2 × (116 + 171)

⇒ Perimeter = 2 × 287

⇒ Perimeter = 574 m

∴ The perimeter of the original rectangle is 574 m.

Given:

Length of rectangular field = 169 m

Breadth of rectangular field = 154 m

Area of rectangular field = Area of circular field

\(\pi = \dfrac{22}{7}\)

Formula used:

Area of rectangle = Length × Breadth

Area of circle = \(\pi r^2\)

Circumference of circle = \(2\pi r\)

Calculation:

Area of rectangle = 169 × 154

⇒ Area of rectangle = 26026 m²

Area of circle = \(\pi r^2\)

⇒ 26026 = \(\dfrac{22}{7} \times r^2\)

⇒ \(\dfrac{26026 \times 7}{22} = r^2\)

⇒ \(\dfrac{182182}{22} = r^2\)

⇒ r² = 8281

⇒ r = \(\sqrt{8281}\)

⇒ r ≈ 91 m

Circumference of circle = \(2\pi r\)

⇒ Circumference = 2 × \(\dfrac{22}{7}\) × 91

⇒ Circumference = \(\dfrac{4004}{7}\)

⇒ Circumference ≈ 572 m

∴ The correct answer is option (4).

Given:

Sides of right triangle: (x − 13), (x − 26), and x cm

Formula used:

Pythagoras Theorem: Hypotenuse² = Base² + Height²

Area = (1/2) × Base × Height

Calculation:

Hypotenuse = x, Base = (x - 26), Height = (x - 13)

Assume hypotenuse = x

⇒ x² = (x − 13)² + (x − 26)²

⇒ x² = (x² − 26x + 169) + (x² − 52x + 676)

⇒ x² = 2x² − 78x + 845

⇒ Bring all terms to one side:

⇒ x² − 2x² + 78x − 845 = 0

⇒ −x² + 78x − 845 = 0

⇒ x² − 78x + 845 = 0

Solve using quadratic formula:

x = [78 ± √(78² − 4 × 1 × 845)] ÷ 2

x = [78 ± √(6084 − 3380)] ÷ 2

x = [78 ± √2704] ÷ 2

x = [78 ± 52] ÷ 2

⇒ x = (78 + 52)/2 = 130 ÷ 2 = 65 (valid)

⇒ x = (78 − 52)/2 = 26 ÷ 2 = 13 (invalid as sides become 0)

If x = 13, then one side is (x - 13) = (13 - 13) = 0, which is not possible for a triangle. So, x ≠ 13.

So, x = 65

Now, find the lengths of the sides:

Side 1 = x - 13 = 65 - 13 = 52 cm

Side 2 = x - 26 = 65 - 26 = 39 cm

Hypotenuse = x = 65 cm

Area = (1/2) × 39 × 52 = 1014 cm²

∴ Area = 1014 cm²

Given:

Length of the chord = 12 cm

Perpendicular distance from the center of the circle to the chord = 5 cm

Formula Used:

Radius of the circle can be calculated using the relationship between the chord, perpendicular distance, and radius:

\(r^2 = \left(\frac{\text{Chord Length}}{2}\right)^2 + \text{Perpendicular Distance}^2\)

Calculation:

Chord Length / 2 = 12 / 2 = 6 cm

Perpendicular Distance = 5 cm

Substitute values into the formula:

⇒ r² = 6² + 5²

⇒ r2 = 36 + 25

⇒ r2 = 61

Find r:

⇒ r = √61

⇒ r ≈ 7.81 cm

The radius of the circle is approximately 7.81 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

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

⇒ r = 14 cm

∴ The required result will be 14 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.

Area of rhombus = Product of both diagonals/ 2,

⇒ 840 = P × Q /2,

⇒ P × Q = 1680,

Using Pythagorean Theorem we get,

⇒ (P/2)² + (Q/2)² = 37²

⇒ P2 + Q2 = 1369 × 4

⇒ P2 + Q2 = 5476

Using perfect square formula we get,

⇒ (P + Q)² = P² + 2PQ + Q²

⇒ (P + Q)² = 5476 + 2 × 1680

⇒ P + Q = 94

Hence option 4 is correct.

Given:

Total cost of fencing = Rs. 10080

Cost of fencing per metre = Rs. 20

Concept used:

Perimeter = Total cost / Cost per metre

Area of the pavement = area of outer square - area of inner square.

Calculation:

According to the question,

Total cost of fencing = 10080

Perimeter of square = 10080/20 = 504 m

⇒ Side of square = 504/4 = 126 m

According to the diagram,

Breadth of the pavement = 2 × 3m = 6m

Side of inner square = 126 - 6 = 120m

Area of pavement = (126 × 126) - (120 × 120)

⇒ Area of pavement = 1476

Cost of pavement = 1476 × 50 = Rs. 73800.

∴ The cost of pavement is Rs. 73800.

Concept used:

Circumference of a circle = 2 × π × r

Calculation:

The circumference of of the wheel = 2 × 22/7 × 182 = 1144 cm

Here, the lorry can go 66 km in 60 min, then the lorry can cover in 1 min is = 66/60 = 1.1 km = 110000 cm

Now the wheel of the lorry makes the number of revolutions per min is 110000/1144 = 96.15 ~ 96

∴ The correct answer is 96