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

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.

Given:

Perimeter of rectangle = 620 mm

Formula Used:

Perimeter of rectangle, P = 2(l + w)

Area of rectangle, A = l × w

For a given perimeter, the maximum area of a rectangle is achieved when it is a square (i.e., length = width).

Calculation:

⇒ 620 = 2(l + w)

⇒ l + w = 620 / 2

⇒ l + w = 310

For maximum area, l = w

⇒ l = 310 / 2

⇒ l = 155 mm

⇒ w = 155 mm

Maximum Area = l × w

⇒ Maximum Area = 155 × 155

⇒ Maximum Area = 24025 mm²

The maximum area of the rectangle with perimeter 620 mm is 24025 mm².

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).

Given:

Original order: One 12-inch circular pizza

Formula used:

Area of a circle = \(\pi r^2\) or \(\pi (\frac{d}{2})^2\) = \(\frac{\pi d^2}{4}\), where r = radius, d = diameter

Calculations:

Original pizza area (12-inch diameter):

⇒ Area = \(\frac{\pi \times 12^2}{4}\) = \(\frac{144\pi}{4}\) = \(36\pi\) square inches

Option 1: Six 4-inch pizzas

⇒ Area of one 4-inch pizza = \(\frac{\pi \times 4^2}{4}\) = \(\frac{16\pi}{4}\) = \(4\pi\) square inches

⇒ Total area = 6 × \(4\pi\) = \(24\pi\) square inches

Option 2: Four 6-inch pizzas

⇒ Area of one 6-inch pizza = \(\frac{\pi \times 6^2}{4}\) = \(\frac{36\pi}{4}\) = \(9\pi\) square inches

⇒ Total area = 4 × \(9\pi\) = \(36\pi\) square inches

Option 3: Seven 3-inch pizzas

⇒ Area of one 3-inch pizza = \(\frac{\pi \times 3^2}{4}\) = \(\frac{9\pi}{4}\) square inches

⇒ Total area = 7 × \(\frac{9\pi}{4}\) = \(\frac{63\pi}{4}\) = \(15.75\pi\) square inches

Option 4: Five 5-inch pizzas

⇒ Area of one 5-inch pizza = \(\frac{\pi \times 5^2}{4}\) = \(\frac{25\pi}{4}\) square inches

⇒ Total area = 5 × \(\frac{25\pi}{4}\) = \(\frac{125\pi}{4}\) = \(31.25\pi\) square inches

∴ The best value for her money is Four 6-inch pizzas.

Given:

Photo length = 50 cm

Photo width = 30 cm

Wooden strip width = 3 cm

Formula used:

Median length of frame = Photo length + Wooden strip width

Median width of frame = Photo width + Wooden strip width

Total length of wooden strip = Perimeter of outer frame = 2 x (Outer length + Outer width)

Calculations:

Outer length of frame = 50 + 3 = 53 cm

Median width of frame = 30 + 3 = 33 cm

Total length of wooden strip required = 2 x (Outer length + Outer width)

⇒ Total length = 2 x (53 + 33)

⇒ Total length = 2 x 86 = 172 cm

∴ The total length of the wooden strip required is 172 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:

∠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.