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

Total number of triangle is:

Hence, the correct answer is "Option 2'.

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.

Calculation:

Let the sides of the rectangular field be 5x and 4x. The area of the rectangle is:

Area = 5x × 4x = 500

⇒ 20x² = 500

⇒ x² = 25

⇒ x = 5.

Thus, the length = 5x = 25 m, and the breadth = 4x = 20 m.

The perimeter of the rectangular field = 2 × (25 + 20) = 90 m.

The sides of the triangular field are in the ratio 3 : 2 : 4. Let the sides be 3y, 2y, and 4y.

The perimeter of the triangle is: 3y + 2y + 4y = 9y.

Since the perimeter is 90 m, we have: 9y = 90 ⇒ y = 10.

The longest side of the triangle is: 4y = 4 × 10 = 40 m.

∴ The longer side of the triangular field is 40 meters.

Given:

Parallel sides: a = 48 cm, b = 20 cm

Non-parallel sides: c = 26 cm, d = 30 cm

Formula used:

Area of trapezium = ½ × (a + b) × height (h)

Height (h) found using: h = √(c² - m²), where m = ((a - b)² + c² - d²) / (2 × (a - b))

Calculations:

a - b = 48 - 20 = 28

m = [28² + 26² - 30²] / (2 × 28)

m = (784 + 676 - 900) / 56 = (560) / 56 = 10

h = √(26² - 10²) = √(676 - 100) = √576 = 24 cm

Area = ½ × (48 + 20) × 24 = ½ × 68 × 24 = 34 × 24 = 816 cm²

∴ Area of the trapezium = 816 cm².

Given:

Triangle sides: a = 28 cm, b = 45 cm, c = 53 cm

π = 3.14

Formula used:

Area of triangle = √(s(s - a)(s - b)(s - c)), where s = semi-perimeter = (a + b + c)/2

Area of circle = π × r², where r = inradius = Area of triangle / s

Area excluding circle = Area of triangle - Area of circle

Calculations:

s = (28 + 45 + 53)/2

⇒ s = 63 cm

Area of triangle = √(s(s - a)(s - b)(s - c))

⇒ Area of triangle = √(63 × (63 - 28) × (63 - 45) × (63 - 53))

⇒ Area of triangle = √(63 × 35 × 18 × 10)

⇒ Area of triangle = √396900

⇒ Area of triangle = 630 cm²

Inradius (r) = Area of triangle / s

⇒ r = 630 / 63

⇒ r = 10 cm

Area of circle = π × r²

⇒ Area of circle = 3.14 × 10²

⇒ Area of circle = 314 cm²

Area excluding circle = Area of triangle - Area of circle

⇒ Area excluding circle = 630 - 314

⇒ Area excluding circle = 316 cm²

∴ The correct answer is option (3).

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:

∠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:

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.