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

Given:

Area of rectangle = 35 cm²

Length = 7 cm

Formula used:

Area = Length × Breadth

Perimeter = 2 × (Length + Breadth)

Calculation:

⇒ 35 = 7 × Breadth

⇒ Breadth = 35 ÷ 7 = 5 cm

⇒ Perimeter = 2 × (7 + 5) = 2 × 12 = 24 cm

∴ The perimeter of the rectangle is: 24 cm

Given:

Perimeter ratio (Rectangle : Square) = 3 : 2

Length of rectangle = (y + 5) cm

Breadth of rectangle = y cm

Area of rectangle = 50 cm²

Area of square = P

P + Q = 89

Formula used:

Area of rectangle = length × breadth

Perimeter of rectangle = 2 × (length + breadth)

Perimeter of square = 4 × side

Area of square = side² = P

Volume of cube = side³ = Q

Calculations:

Area: (y + 5) × y = 50

⇒ y² + 5y = 50

⇒ y² + 5y - 50 = 0

⇒ y = 5 (solving quadratic)

⇒ Length = 10 cm, Breadth = 5 cm

Perimeter of rectangle = 2 × (10 + 5) = 30 cm

Let side of square = x, then perimeter = 4x

Ratio ⇒ 30 : 4x = 3 : 2

⇒ 30/4x = 3/2

⇒ 60 = 12x

⇒ x = 5

Area of square = x² = 25 = P

P + Q = 89 ⇒ Q = 64

Volume of cube = 64 ⇒ side³ = 64

⇒ side = 4

∴ Side length of the cube is 4 cm.

Short Trick:

Area of (ΔAPB + ΔCPD) = Area of (ΔBPC + ΔAPD)

∴ ΔAPB = 11 cm²

Detailed Solution:

Let the side of the perpendicular be a cm and b cm

Drop a perpendicular from p to all sides,

Let the length of perpendicular on side AB be x cm and on side CD be y cm.

Sum of Area of ΔAPB and ΔCPD = 1/2 × (ax + ay) = 1/2 × ah₁

= 1/2 Area of parallelogram ABCD          ----(1)

Let the length of perpendicular on side CB be w cm and on side AD be z cm.

Similarly, Sum of Area of ΔCPB and ΔAPD = 1/2 × (bw + bz) = 1/2 × bh₂ = 1/2 × Area of parallelogram ABCD          ----(2)

As the Area of perpendicular is equal, Equation 1 = Equation 2

⇒ Area of (ΔAPB + ΔCPD) = Area of (ΔBPC + ΔAPD)

∴ ΔAPB = 11 cm²

Given:

Side of a square increased = 10%

Formula used:

Increase = New number - Original number

% increase = (Increase/Original Number) × 100

Area of square = side2

Calculations:

Let the side of the square be a 

Area of a square with side a = a²

After increasing percentage of side a by 10% = a + 10% of a

⇒ a(1 + 10/100) = a(1 + 1/10)

⇒ a(11/10) = 11a/10

Area of the square after increasing side by a = (11a/10)²

⇒ (11a/10) × (11a/10) = 121a²/100

⇒ 1.21a²

Percentage change in area of square =[(1.21a² - a²)/a²] × 100

⇒ [a²(1.21 - 1)/a²] × 100 = 0.21 × 100 

⇒ 21%

∴ Percentage change in area of square is 21%

Alternate Method

Percentage increase in a regular polygon = x + y + (xy/100)

In square length and breadth are same, x = y

Percentage increase = x + x + (xx/100)

⇒ 2x + x²/100 = 2 × 10 + 10 × 10/100

⇒ 20 + 1 = 21%

∴ The required percentage is 21%

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:

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:

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.

Let 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.

Given:

∠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°

Given:

In an isosceles triangle ABC,

AB = AC = 26 cm and BC = 20 cm.

Calculations:

In this triangle ABC,

∆ADC = 90° ( Angle formed by a line from opposite vertex to unequal side at mid point in isosceles triangle is 90°)

So,

AD² + BD² = AB²   (by pythagoras theorem)

⇒ AD² = 576

⇒ AD = 24

Area of triangle = ½(base × height)

⇒ ½(20 × 24)   (Area of triangle = (1/2) base × height)

⇒  240 cm²

∴ The correct choice is option 2.