Rectangle MCQ Quiz - Objective Question with Answer for Rectangle - Download Free PDF

Given:

A circle circumscribes a rectangle with sides of 16 cm and 12 cm. 

Concept used:

By Pythagoras theorem

AC² = AB² + BC²

The diagonal of the rectangle is the diameter of the circle.

Area of circle = πr²

Calculation:

The diagonal of the rectangle is the diameter of the circle.

Now, by Pythagoras' theorem,

⇒ (2r)² = (16² + 12²)

⇒ 4r² = (256 + 144)

⇒ 4r² = 400

⇒ r² = 100

⇒ r = 10 cm

Now, area of circle = π(10)² = 100 π square cm

∴ The required area is 100 π square cm.

Given:

The ratio between the length and the breadth of a rectangular park is 3 : 2

The perimeter of the park is 120 m

Formula used:

Perimeter of a rectangle = 2(length + breadth)

Area of a rectangle = length × breadth

Calculation:

Let the length be 3x and the breadth be 2x

Perimeter = 2(3x + 2x)

⇒ 120 = 2(5x)

⇒ x = 12

Length = 3x = 3 × 12 = 36 m

Breadth = 2x = 2 × 12 = 24 m

Area = length × breadth

⇒ Area = 36 × 24 = 864 m²

∴ The correct answer is option (4).

Given:

The area of the square formed on the diagonal of a rectangle as its side is 108¹/₃% more than the area of the rectangle.

The perimeter of the rectangle is 28 units.

Formula used:

Area of square = (1 + 108¹/₃% / 100) × Area of rectangle

Diagonal of rectangle = side of square

Diagonal of rectangle = √(l² + b²)

Perimeter of rectangle = 2(l + b)

Calculation:

Let the length and breadth of rectangle be l and b respectively.

Perimeter = 28

⇒ 2(l + b) = 28

⇒ l + b = 14

Now,

Area of rectangle = l × b

Area of square = (1 + 108¹/₃% / 100) × l × b

⇒ (1 + 325/300) × l × b

⇒ (425/300) × l × b

⇒ Area of square = (17/12) × l × b

Now, diagonal of rectangle = side of square

⇒ √(l² + b²) = √((17/12) × l × b)

⇒ l² + b² = (17/12) × l × b

From l + b = 14, Then, l = 14 - b

Substitute l into the equation:

(14 - b)² + b² = (17/12) × (14 - b) × b

⇒ 196 - 28b + b² + b² = (17/12) × (14b - b²)

⇒ 196 - 28b + 2b² = (238b/12 - 17b²/12)

⇒ 2b² + b² = (238b/12 - 28b + 196)

⇒ 2b² + 12(238b/12 - 28b + 196)/17 = 0

Solving this equation:

b = 6

l = 14 - b = 8

⇒ Difference = l - b = 8 - 6 = 2

∴ The correct answer is option (3).

Rectangle

• A rectangle is a quadrilateral.
• The opposite sides are parallel and equal to each other.
• Each interior angle is equal to 90 degrees.
• The sum of all the interior angles is equal to 360 degrees.
• The diagonals bisect each other
• Both diagonals have the same length.

Explanation:

According to the question, we draw a figure of a rectangle ABCD

AB = CD

BC = AD

AC = BD

It will be a rectangle and each angle will be a right angle.

Given:

A rectangular field has its length and breadth in the ratio of 8 ∶ 7 respectively.

A man riding a bicycle completes one lap of this field along its perimeter at the speed of 28.8 km/hr in 2.5 minutes.

Formula used:

Speed = Distance / Time

Perimeter (P) = 2(length + breadth)

Area = length × breadth

Calculation:

Speed = 28.8 km/hr

Time = 2.5 minutes = 2.5/60 hours = 1/24 hours

Distance (Perimeter) = Speed × Time

⇒ Distance = 28.8 × 1/24 km

⇒ Distance = 1.2 km = 1200 meters

Let length = 8x and breadth = 7x

Perimeter = 2(8x + 7x) = 2(15x) = 30x

30x = 1200

⇒ x = 1200 / 30

⇒ x = 40

Length = 8x = 8 × 40 = 320 meters

Breadth = 7x = 7 × 40 = 280 meters

Area = length × breadth

⇒ Area = 320 × 280

⇒ Area = 89600 sq.m

∴ The correct answer is 89600

Figure:

Calculation:

As the diagonals of a rectangle intersect each other,

⇒ AO = OB

⇒ ∠OBA = ∠OAB = 25° [∵ Angle opposite to equal side are equal]

By angle sum property in ΔAOB,

⇒ ∠AOB + ∠OAB + ∠OBA = 180°

⇒ ∠AOB + 25° + 25° = 180°

⇒ ∠AOB = 130°

By linear pair property,

⇒ ∠DOA + ∠AOB = 180°

⇒ ∠DOA + 130° = 180°

⇒ ∠DOA = 50°

Given:

Radius of semicircle = 10 cm

Formula Used:

Area of semicircle = (1/2)πr²

Area of rectangle = length × breadth

Calculation:

Area of semicircle = (1/2)πr²

⇒ Area of semicircle = (1/2) × (22/7) × (10)²

⇒ Area of semicircle = 157.14 cm²

∵ A is the mid point of OP

OA = 5 cm

In a ΔAOD,

OD² = OA² + AD²

⇒ (10)² = (5)² + AD²

⇒ AD² = 100 - 25

⇒ AD2 = 75

⇒ AD = 5√3

Area of rectangle ABCD = AB × AD

⇒ Area of rectangle ABCD = 10 × 5√3

⇒ Area of rectangle ABCD = 50√3

⇒ Area of rectangle ABCD = 86.60 cm²

∴ Area of shaded portion = Area of semicircle - Area of rectangle

⇒ Area of shaded portion = 157.14 - 86.60

⇒ Area of shaded portion = 70.54 cm²

∴ The area of shaded portion is 70.54 cm².

GIVEN:

Length of a rectangle is 17 m more than its breadth.

Length of its diagonal = 25 m

CONCEPT:

Mensuration

FORMULA USED:

Perimeter of rectangle = 2(l + b)

CALCULATION:

Let the breadth of rectangular field = ‘x’ m

So, the length of rectangular field = (x + 17) m

From the question:

(x + 17)² + x² = 25²

⇒ x² + 289 + 34x + x² = 625

⇒ x² + 17x – 168 = 0

⇒ x² + 24x – 7x – 168 = 0

⇒ x(x + 24) – 7(x + 24) = 0

⇒ x = 7

Hence,

Breadth of rectangle = 7 m

Length of rectangle = 7 + 17 = 24 m

Perimeter of rectangle = 2 × (7 + 24)

⇒ 62 m

Therefore the correct answer is 62m.

Calculations:

As AD = BC and DC is common

⇒ ΔADC and ΔBDC are congruent

⇒ ∠ACD = ∠BDC = 25°

In ΔDOC,

∠DOC = 180° - (25 + 25) = 130°

⇒ ∠DOA = 180° - 130° = 50°

∴ The  acute angle between the diagonals is 50°.

Given:

Area of rectangle = 300 cm²

Diagonal = 25 cm

FORMULA USED:

Area of a rectangle = length × breadth

Perimeter of a rectangle = 2 (length + breadth)

Diagonal = √(length2 + breadth2)

Calculation:

According to question,

⇒ length × breadth = 300

Again,

Diagonal = √(length2 + breadth2)

⇒ √(length2 + breadth2) = 25 

⇒ (length2 + breadth2) = 625

Let the length be a and breadth be b

Then,

⇒ (a + b)² = a² + b² + 2ab

⇒ (a + b)² = 625 + 2 × 300

⇒ (a + b)² = 1225

⇒ a + b = 35

Perimeter of rectangle = 2(a + b)

⇒ 2 × 35

⇒ 70

∴ Perimeter of rectangle is 70.

Given : 

A rectangle ABCD in which BC = 24, DP = 10 cm and CD = 15 cm 

CP = 15 - 10 = 5  

Concept used :

The similarity of a triangle  

If values of corresponding angles between two triangles are the same then triangles are said to be similar 

If two triangles are similar then the ratio of their corresponding sides are the same 

Calculations :

In ∆ADP and ∆QCP 

∠ADP = ∠QCP (both are 90°)

∠APD = ∠QPC (vertically opposite angles)

As AD||BQ, then 

∠PAD = ∠PQC (alternate angles) 

So, ∆ADP and ∆QCP are similar triangles 

Now,

AD/QC = DP/CP = AP/QP 

In ∆APD 

AD² + DP² = PA² 

24² + 10² = PA2   (AD = BC opposite side of a rectangle)

PA = 26 

⇒ 24/QC = 10/5 = 26/QP 

⇒ QP = 13 

Now, 

AQ = AP + PQ = 26 + 13  

⇒ 39 

∴ AQ will be 39 cm 

Given:

∠BOC = 44°

Concept used:

Diagonals of the rectangle bisect each other.

Calculation:

∠BOC = ∠AOD = 44° 

In ΔAOD, 

∠AOD + ∠DAO + ∠ADO = 180°

 ∠DAO + ∠ADO = 180 - 44 = 136° 

Since, DO = AO, ∠DAO = ∠ADO 

So, 2 × ∠DAO = 136

Or, ∠DAO = 68° 

Given:

l, b and p in GP where,

l = length, b = breadth and p = perimeter

Concept used:

(a - b)² = a² - 2ab + b²

(a + b)² = a² + 2ab + b²

Perimeter of rectangle = 2 × (l + b)

if a, b and c is in GP (in order) then,

b² = ac

Explanation:

According to the question 

b, l, p are in G.P

So, l² = b × p

⇒ l² = b × 2(l + b)

⇒ l² - 2lb = 2b²

⇒ l² - 2lb + b² = 2b² + b²

⇒ (l - b)² = 3b²

⇒ l - b = √3b

⇒ l = √3b + b

⇒ l = b (√3 + 1)

⇒ l/b = (√3 + 1)/1

⇒ (l/b)² = (4 + 2√3)/1

∴ The ratio of l² : b² is (4 + 2√3) : 1.

GIVEN:

Length of a rectangular field is 14 cm more than its width

Length of its diagonal = 34 cm

CONCEPT:

Mensuration

FORMULA USED:

Perimeter of rectangle = 2(l + b)

CALCULATION:

Let length and width of the field is ‘x + 14’ and ‘x’ respectively.

Diagonal = 34 = \(\sqrt {{{\left( {x + 14} \right)}^2} + {x^2}} \)

⇒ 1156 = x² + 196 + 28x + x²

⇒ x² + 14x – 480 = 0

⇒ x² + 30x - 16x - 480 = 0 

⇒ x(x + 30) - 16(x + 30) = 0

⇒ (x + 30) (x - 16) = 0

⇒ x = 16

So,

Perimeter of field = 2 [(x + 14) + x]

= 2 × 46

= 92 cm

∴ The perimeter of the field is 92 cm.

Calculation:

According to the question-

OT = OR

⇒ 3x + 1 = 2x + 4

⇒ 3x - 2x = 4 - 1

⇒ x = 3

∴ The correct answer is 3.