Plane Figures MCQ Quiz - Objective Question with Answer for Plane Figures - Download Free PDF
Last updated on Jul 19, 2025
Latest Plane Figures MCQ Objective Questions
Plane Figures Question 1:
The side of square is 11/19 of the perimeter of a rectangle, Area of square is 484 m2. Ratio of length and breadth of rectangle is 10:9. Area of rectangle is [x + 60] m2. Find the value of x?
Answer (Detailed Solution Below)
Plane Figures Question 1 Detailed Solution
Calculation
Area of square = 484
⇒ side = √484 = 22 m
Side = 11/19 of perimeter of rectangle
Let perimeter = P
⇒ 11/19P = 22
⇒ P = 38
Ratio L:B = 10:9
⇒ L = 10k, B = 9k
Perimeter = 2(L + B) = 2×19k = 38
⇒ k = 1
⇒ L = 10, B = 9
⇒ Area = 90
Area = x + 60 ⇒ x = 90 - 60 = 30
Plane Figures Question 2:
A semi-circle has a circumference of 144 cm, Area of full circle is equal to perimeter of a square. Side of square is equal to area of rectangle. Breadth of rectangle is 14cm. Find the perimeter of rectangle?
Answer (Detailed Solution Below)
Plane Figures Question 2 Detailed Solution
Given:
Circumference of semicircle = 144 cm
Breadth of rectangle = 14 cm
Formula used:
Semi-circle circumference = πr + 2r = r(π + 2)
Area of circle = πr2 = Perimeter of square
Perimeter of square = 4 × side
Area of rectangle = length × breadth = side of square
Perimeter of rectangle = 2 × (length + breadth)
Calculations:
144 = r(π + 2) ⇒ r × (22/7 + 2) = 144
⇒ r × (36/7) = 144 ⇒ r = 144 × 7 / 36 = 28
Area of circle = πr2 = 22/7 × 28 × 28 = 2464 cm2
Perimeter of square = 2464 ⇒ side = 2464 ÷ 4 = 616
Area of rectangle = side = 616, breadth = 14
⇒ length = 616 ÷ 14 = 44
Perimeter of rectangle = 2 × (44 + 14) = 2 × 58 = 116 cm
∴ Perimeter of rectangle = 116 cm
Plane Figures Question 3:
Three times the length of a rectangle is equal to seven times its breadth. The area of the rectangle is 2352 sq. cm. What is the area of a square whose perimeter is equal to that of the rectangle?
Answer (Detailed Solution Below)
Plane Figures Question 3 Detailed Solution
3 × Length = 7 × Breadth ⇒ Length : Breadth = 7 : 3 ⇒ Let Breadth = 3k, Length = 7k
Area of Rectangle = Length × Breadth = (7k) × (3k) = 21k² = 2352
⇒ k² = 2352 ÷ 21 = 112 ⇒ k = √112 = 4√7
⇒ Breadth = 3k = 3 × 4√7 = 12√7
⇒ Length = 7k = 7 × 4√7 = 28√7
Perimeter of Rectangle = 2(L + B) = 2(28√7 + 12√7) = 2 × 40√7 = 80√7
Square with same perimeter:
Let side of square = s ⇒ 4s = 80√7 ⇒ s = 20√7
Area of square = s² = (20√7)² = 400 × 7 = 2800 sq cm
Thus, the correct answer is 2800 sq cm.
Plane Figures Question 4:
Length of the rectangle is (x + 5) m and the breadth is (x - 3) m. side of the square is (3x + y). If the perimeter of the rectangle is 64 m and value of y is 1/3rd of x then finds the area of the square?
Answer (Detailed Solution Below)
Plane Figures Question 4 Detailed Solution
Given:
Length = x + 5, Breadth = x − 3
Perimeter of rectangle = 64
Side of square = 3x + y, y = x/3
Formula used:
Perimeter = 2(L + B) = 64
Area of square = side²
Calculations:
⇒ 2[(x + 5) + (x − 3)] = 64
⇒ 2(2x + 2) = 64
⇒ 2x + 2 = 32
⇒ 2x = 30 ⇒ x = 15
⇒ y = x/3 = 15/3 = 5
⇒ Side of square = 3x + y = 45 + 5 = 50
⇒ Area = 50² = 2500 m²
∴ Area of the square = 2500 m²
Plane Figures Question 5:
The sides of a rectangular field are 11 m and 14 m long. Its area is equal to the area of a circular field. What is the circumference (in m) of the circular field?
Answer (Detailed Solution Below)
Plane Figures Question 5 Detailed Solution
Given:
Length of rectangle (L) = 11 m
Breadth of rectangle (B) = 14 m
Area of rectangle = Area of circular field
Formula used:
Area of rectangle = L × B
Area of circle = π × r2
Circumference of circle = 2 × π × r
Calculation:
Area of rectangle = 11 × 14
⇒ Area = 154 m2
Area of circle = π × r2
⇒ 154 = (22/7) × r2
⇒ r2 = (154 × 7) / 22
⇒ r2 = 49
⇒ r = √49
⇒ r = 7 m
Circumference of circle = 2 × π × r
⇒ Circumference = 2 × (22/7) × 7
⇒ Circumference = 44 m
∴ The correct answer is option (4).
Top Plane Figures MCQ Objective Questions
Six chords of equal lengths are drawn inside a semicircle of diameter 14√2 cm. Find the area of the shaded region?
Answer (Detailed Solution Below)
Plane Figures Question 6 Detailed Solution
Download Solution PDFGiven:
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
There is a rectangular garden of 220 metres × 70 metres. A path of width 4 metres is built around the garden. What is the area of the path?
Answer (Detailed Solution Below)
Plane Figures Question 7 Detailed Solution
Download Solution PDFFormula 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.
The width of the path around a square field is 4.5 m and its area is 105.75 m2. Find the cost of fencing the field at the rate of Rs. 100 per meter.
Answer (Detailed Solution Below)
Plane Figures Question 8 Detailed Solution
Download Solution PDFGiven:
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
The length of an arc of a circle is 4.5π cm and the area of the sector circumscribed by it is 27π cm2. What will be the diameter (in cm) of the circle?
Answer (Detailed Solution Below)
Plane Figures Question 9 Detailed Solution
Download Solution PDFGiven :
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.
If the side of an equilateral triangle is increased by 34%, then by what percentage will its area increase?
Answer (Detailed Solution Below)
Plane Figures Question 10 Detailed Solution
Download Solution PDFGiven:
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%.
A wire is bent to form a square of side 22 cm. If the wire is rebent to form a circle, then its radius will be:
Answer (Detailed Solution Below)
Plane Figures Question 11 Detailed Solution
Download Solution PDFGiven:
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.
How many revolutions per minute a wheel of car will make to maintain the speed of 132 km per hour? If the radius of the wheel of car is 14 cm.
Answer (Detailed Solution Below)
Plane Figures Question 12 Detailed Solution
Download Solution PDFGiven:
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 =
Circumference of the wheel =
∴ Distance covered by wheel in one revolution = 88 cm
∴ The number of revolutions in one minute =
∴ Therefore the correct answer is 2500.
One side of a rhombus is 37 cm and its area is 840 cm2. Find the sum of the lengths of its diagonals.
Answer (Detailed Solution Below)
Plane Figures Question 13 Detailed Solution
Download Solution PDFLet P and Q be the lengths of diagonals of the rhombus,
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.
Answer (Detailed Solution Below)
Plane Figures Question 14 Detailed Solution
Download Solution PDFGiven:
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.
The wheel of a lorry has radius 182 cm. The number of revolutions (approximately) per minute the lorry wheel will make is _______ (if the speed of the lorry is 66 km/h).
Answer (Detailed Solution Below)
Plane Figures Question 15 Detailed Solution
Download Solution PDFConcept 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