Plane Figures MCQ Quiz - Objective Question with Answer for Plane Figures - Download Free PDF
Last updated on Jun 11, 2025
Latest Plane Figures MCQ Objective Questions
Plane Figures Question 1:
The perimeter of the rectangle is 120 cm and the ratio of the length to the breadth of the rectangle is 7:8. Area of square is 4 cm2 more than the area of rectangle. Find the side of square?
Answer (Detailed Solution Below)
Plane Figures Question 1 Detailed Solution
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.
Plane Figures Question 2:
What is the minimum number of square tiles required to completely cover a rectangular floor measuring 5.25 m by 5.10 m?
Answer (Detailed Solution Below)
Plane Figures Question 2 Detailed Solution
Given:
The dimensions of the rectangular floor are:
Length (l) = 5.25 m
Width (w) = 5.10 m
Formula used:
To find the minimum number of square tiles, we calculate the side of the largest square tile that can exactly divide both the length and width of the floor using the HCF (Highest Common Factor).
Number of tiles = Area of floor / Area of one tile
Area of floor = l × w
Area of one tile = (HCF)2
Calculation:
l = 5.25 m = 525 cm
w = 5.10 m = 510 cm
HCF of 525 and 510:
Prime factorization of 525 = 3 × 52 × 7
Prime factorization of 510 = 2 × 3 × 5 × 17
Common factors = 3 × 5 = 15 ⇒ HCF = 15 cm
Area of floor = 525 × 510 = 267750 cm2
Area of one tile = 15 × 15 = 225 cm2
Number of tiles = 267750 / 225
⇒ Number of tiles = 1190
∴ The correct answer is option (1).
Plane Figures Question 3:
The number of triangles in the figure is
Answer (Detailed Solution Below)
Plane Figures Question 3 Detailed Solution
Total number of triangle is:
Hence, the correct answer is "Option 2'.
Plane Figures Question 4:
A field is 1 km long and width
Answer (Detailed Solution Below)
Plane Figures Question 4 Detailed Solution
Given:
The length of the field is 1 km and the width is
Solution:
To calculate the area of a rectangle, we use the formula:
Area = Length × Width
Substituting the given values:
Area = 1 ×
Area =
The area of the field is
Plane Figures Question 5:
The cost of painting a rectangular wall at the rate of Rs.75/square metre is Rs.6825. The length of rectangular wall is equal to the length of a square wall whose area is 196 square metre. Find the breadth of the rectangular wall.
Answer (Detailed Solution Below)
Plane Figures Question 5 Detailed Solution
Given:
Painting rate = Rs.75/m²
Total painting cost = Rs.6825
Area of square wall = 196 m²
Length of rectangular wall = side of square wall
Formula used:
Area = Total Cost / Rate
Area of rectangle = Length × Breadth
Side of square = √Area
Calculations:
Area of rectangular wall = 6825 / 75 = 91 m²
Side of square wall = √196 = 14 m ⇒ Length of rectangular wall = 14 m
Now, 14 × Breadth = 91
⇒ Breadth = 91 / 14 = 6.5 m
∴ The breadth of the rectangular wall is 6.5 metres.
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.
In a circle with centre O, chords PR and QS meet at the point T, when produced, and PQ is a diameter. If
Answer (Detailed Solution Below)
Plane Figures Question 14 Detailed Solution
Download Solution PDFGiven:
∠ROS = 42º
Concept used:
The sum of the angles of a triangle = 180°
Exterior angle = Sum of opposite interior angles
Angle made by an arc at the centre = 2 × Angle made by the same arc at any point on the circumference of the circle
Calculation:
Join RQ and RS
According to the concept,
∠RQS = ∠ROS/2
⇒ ∠RQS = 42°/2 = 21° .....(1)
Here, PQ is a diameter.
So, ∠PRQ = 90° [∵ Angle in the semicircle = 90°]
In ΔRQT, ∠PRQ is an exterior angle
So, ∠PRQ = ∠RTQ + ∠TQR
⇒ 90° = ∠RTQ + 21° [∵ ∠TQR = ∠RQS = 21°]
⇒ ∠RTQ = 90° - 21° = 69°
⇒ ∠PTQ = 69°
∴ The measure of ∠PTQ is 69°
AB is a diameter of a circle with center O. A tangent is drawn at point A. C is a point on the circle such that BC produced meets the tangent at P. If ∠APC = 62º, then find the measure of the minor arc AC(i.e.∠ ABC).
Answer (Detailed Solution Below)
Plane Figures Question 15 Detailed Solution
Download Solution PDFGiven:
AB is a diameter of a circle with a center O
∠APC = 62º
Concept used:
The radius/diameter of a circle is always perpendicular to the tangent line.
Sum of all three angles of a triangle = 180°
Calculation:
Minor arc AC will create angle CBA
∠APC = 62º = ∠APB
∠BAP = 90° (diameter perpendicular to tangent)
In Δ APB,
∠APB + ∠BAP + ∠PBA = 180°
⇒ ∠PBA = 180° - (90° + 62°)
⇒ ∠PBA = 28°
∴ The measure of minor arc AC is 28°
Mistake PointsMeasure of the minor arc AC is asked,
∠ABC marks arc AC,
∴ ∠ABC is the correct angle to show a measure of arc AC
This is a previous year's question, and according to the commission, this is the correct answer.