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Given:

113°25'

Concept used:

∠A + ∠B = 180°

1° = 60'

Calculation:

(180° – 113°25')

⇒ (179°60' – 113°25')

⇒ 66°35'

∴ The supplementary angle of 113°25' is 66°35'

Note down some definitions:

• Line: When we speak of a line we mean a straight line.
• Line-segment (definition): The set of two distinct points A, B and the points between them is known as the line-segment determined by A and B.
• Length of a line segment (definition): The distance between the endpoints of a line segment is known as the length of the line segment.
• Collinear Points: If two or points are in the same line, they are said to be collinear.

Hence, the above statement is of line segment length.

Concept used:

Cyclic quadrilateral:- A quadrilateral that is circumscribed in a circle is called a cyclic quadrilateral. It means that all the four vertices of quadrilateral lie in the circumference of the circle

Property of the cyclic quadrilateral

The sum of the opposite angles of the cyclic quadrilateral is 180° 

Calculation:

From the above concept used we come to know that

The sum of the opposite angles of the cyclic quadrilateral is 180° 

Given:

Two statements are given

Concept Used:

Properties of trapezium

Calculation:

Isosceles trapezium is a trapezium having the non parallel sides equal and we know the property as well if non parallel sides are equal then th diagonals of the trapezium is also equal.

∴ Statement 1 is correct

Now, If the diagonals of trapezium are equal and bisect each other then it is a parallelogram this is not true because trapezium is having single pair of parallel line which is not necessarily equal but parallelogram have two pair of parallel line.

∴ Statement 2 is incorrect

Hence, option (1) is correct

Supplementary angles: Sum of two angles which are supplementary is 180⁰

⇒ (2/3) × 90° 

⇒ 60° 

⇒ 60° + x = 180° 

⇒ x = 120° 

Complimentary angles: When the sum of the measures of the two angles is 90°, the angles are called Complimentary angles.

Calculation:

Let the measure of one angle be x° 

Then other angle = 90° - x

⇒ x° = 90° - x°

2x° = 90°

∴ x = 45° 

Given:

The two given equations are 13x – 3y = 13 and 7x – 11y = 22.

Calculation:

The equation 13x – 3y = 13 intersect x-axis when y = 0

⇒ 13x – 0 = 13, x = 1

Point P will be (1, 0) = (a, b)     ----(1)

Now the equation 7x – 11y = 22 intersect y-axis when x = 0

⇒ 7× 0 – 11y = 22, y = -2

Point Q will be (0, -2) = (c, d)     ----(2)

From (1) & (2) we get a, b, c, d = 1, 0, 0, -2

⇒ a + b² + c³ + d⁴ = 1 + 0 + 0 + 2⁴ = 1 + 16

∴ The value is 17.

Given:

In triangle ABC,  AB = 20 cm, BC = 15 cm

Concept Used:

The circumcentre of a right angle triangle lies on the mid-point of the hypotenuse.

Formula Used:

Pythagoras theorem - In a Right angle triangle,

 Hypotenuse² = Perpendicular² + Base²

Calculation:

In ΔABC,

AC² = AB² + BC²

⇒ AC² = 20² + 15²

⇒ AC² = 625

⇒ AC = 25 cm

The distance of point A from the circumcentre = Circumradius of the triangle = 25/2 = 12.5 cm

∴ The distance of point A from the circumcentre is 12.5 cm

Given:

one side of triangle = 6 cm

second side of triangle = 10 cm

concept used:

the difference of two other sides < x < sum of two other side

calculation:

(10 - 6) < x < (10 + 6)

⇒ 4 < x < 16

the third side of the triangle should lie between 4 and 16

∴ minimum integral value is 5 and maximum integral value is 15

Hence, the answer is 5 cm

Given:

Point A = (5, a) and Point B = (17, 12)

The distance between the two points is 13 units.

Formula Used:

Distance between two points

= √[(x₂ – x₁)² + (y₂ – y₁)²]

Calculation:

Point A = (5, a)

⇒ x₁ = 5 and y₁ = a

Point B = (17, 12)

⇒ x₂ = 17 and y₂ = 12

Using the formula,

⇒ 13 = √(x2 – x1)2 + (y2 – y1)2

⇒ 13 = √(17 – 5)² + (12 – a)²

⇒ 13² = (12)² + (12 – a)²

⇒ 169 = 144 + (12 – a)²

⇒ (12 – a)² = 169 – 144

⇒ (12 – a)2 = 25

⇒ 12 – a = √25

⇒ 12 – a = 5

⇒ a = 12 – 5 

⇒ a = 7

∴ The value of a is 7 units.

Given:

Points A(1, 1) and B(4, 3)

Formula Used:

Equation of line containing two points (x₁, y₁) and (x₂, y₂) 

y = mx + C

m = (y₂ – y₁)/(x₂ – x₁)

Calculation:

m = (3 – 1)/(4 – 1)

⇒ m = 2/3

Equation of line y = (2/3)x + C

For value of C, point A(1, 1) will satisfy the equation

⇒ 1 = (2/3)(1) + C

⇒ C = 1 – 2/3

⇒ C = 1/3

Now, the equation of line is y = (2/3)x + (1/3)

⇒ 3y = 2x + 1

∴ The equation of line containing the points A(1, 1) and B(4, 3) is 3y = 2x + 1.

Given:

A line joining points P(3, 4) and Q(7, a) is parallel to 3x – y + 20 = 0.

Formula used:

If points are given (x₁, y₁) and (x₂, y₂)

Then slope = (y₂ – y₁)/(x₂ – x₁)

If a line equation is written in format y = mx +c 

Where m is the slope of the line.

The slope of parallel lines is equal.

Calculation:

Slope of the line 3x – y + 20 = 0 is 

⇒ y = 3x + 20

So, slope of the line is 3

And slope of point P(3, 4) and Q(7, a) is 3

⇒ slope = (y2 – y1)/(x2 – x1)

⇒ 3 = (a – 4)/(7 – 3)

⇒ 12 = a – 4

⇒ a = 12 + 4

⇒ a = 16

∴ The value of the a is 16.

Given:

ABC is a triangle in which E and F are the midpoint of AB and AC and EF ∥ BC and EF is 10.5 cm

Concept Used:

According to midpoint theorem If the line segment adjoins midpoints of any of the sides of a triangle, then the line segment is said to be parallel to all the remaining sides, and it measures about half of the remaining sides.

Calculation:

EF = 10.5 cm where E and F are the midpoint of AB and AC and EF ∥ BC the according to midpoint theorem

∴ EF = BC/2

⇒ BC = 2 × 10.5 = 21 cm

Given:

The ratio of the angles of a triangle = 3 : 5 : 7

Concept used:

The sum of the angles of the triangle is 180°.

Calculation:

Let the angles of the triangle be 3x, 5x, and 7x.

Sum of the angles, 3x + 5x + 7x = 180

15x = 180 

Or, x = 12

Value of the largest angle = 7x = 84° 

∴ Largest angle is 84°.

Given:

ΔABC and ΔPQR are similar to each other.

Ratio of area of ΔABC and ΔPQR = 1 : 16

Length of side AC = 28 cm

Concept:

Relation among areas, side of similar triangles ABC and PQR

(Area of ΔABC and ΔPQR) = (AB/PQ)² = (BC/QR)² = (AC/PR)²

Calculation:         

Let PR be x cm

(Area of ΔABC and ΔPQR) = (AC/PR)²

⇒ (1/16) = (28/x)²

Taking square root on both the sides

⇒ √(1/16) = √(28/x)²

⇒ (1/4) = (28/x)

⇒ x = 28 × 4

⇒ x = 112 cm

∴ The length of PR is 112 cm. 

Given:

O is the circumcentre inside the ΔDEF.

∠EOF = 150° 

Concepts used:

The measure of the angle formed with the given base at circumcentre = 2 × vertex angle formed opposite to the same base

Calculation:

⇒ ∠EOF = 2 × ∠D

⇒ 150° = 2 × ∠D

⇒ ∠D = 150°/2

⇒ ∠D = 75°.

∴ The measure of vertex ∠D is 75°.

Given:

Sides of triangles are 12 cm, 16 cm, and 22 cm

Formula used:

Area of triangle = \(\sqrt {s\left( {s\; - \;a} \right)\left( {s\; - \;b} \right)\left( {s\; - \;c} \right)} \)

Here, s is semi perimeter, and a, b, and c are sides of the triangle

Inradius = Area/s

Calculation:

Let a be 12 cm, b be 16 cm, and c be 22 cm

Semi perimeter (s) = (12 + 16 + 22)/2

⇒ s = 50/2 = 25

Area of triangle = \(\sqrt {s\left( {s\; - \;a} \right)\left( {s\; - \;b} \right)\left( {s\; - \;c} \right)} \)

⇒ Area of triangle = \(\sqrt {25\left( {25\; - \;12} \right)\left( {25\; - \;16} \right)\left( {25\; - \;22} \right)} \)

⇒ Area of triangle = \(\sqrt {25 \times 13 \times 9 \times 3} \)

⇒ Area of triangle = √8775 cm²

⇒ Area of triangle ≈ 93.67

Inradius = Area/s

⇒ Inradius = 93.67/25

∴ Inradius is 3.7 cm approximately

GIVEN:

Three statements.

CONCEPT:

Geometry

FORMULA USED:

Sum of interior angles of a polygon = (n - 2) × 180°

CALCULATION:

A:

We know that:

Sum of exterior angle of a polygon is always 360°.

So, sum of exterior angle of a triangle = Sum of exterior angle of a quadrilateral.

B:

Sum of interior angles of a polygon is given by (n - 2) × 180°

Where:

‘n’ = Number of sides

C:

We know that:

Sum of interior and exterior angle of a polygon is always 180°.

Sum of interior and exterior angle = 72° + 108° = 180°

Given:

Area of square = 400 sq. units

Formula Used:

Area of square = Side × Side

Calculation:

Area of square = Side × Side

⇒ 400 sq. units = Side × Side

⇒ √2 × 2 × 2 × 2 × 5 × 5 = Side

⇒ 20 units = side

∴ The length of each side is 20 units.

The correct option is 4 i.e. 20units

GIVEN:

Three statements.

CONCEPT:

Geometry

CALCULATION:

All the above given statements are theorems :

A:

If there is a line and a point not on the line, then there exists exactly one line though the point that is parallel to the given line.

B:

If a line is tangent to a circle, it is perpendicular to the radius of the circle drawn to the point of tangency.

C:

A straight line can be determined by two points.

Given:

2x + 5y - 12 = 0          ------(1)

 x + y = 3          ------(2)

y = 0          ------(3)

Concept used:

At x axis, y = 0

Area of triangle = (1/2) × Base × Height

Calculations:

⇒ 2x + 5y - 12 = 0          ------(1)

⇒ x + y = 3          ------(2)

By solving Equation (1) and (2)

⇒ x = 1 and y = 2

In equation (1) By putting y = 0

⇒ (x, y) = (6, 0)

In equation (2) by putting y = 0

⇒ (x, y) = (3, 0)

⇒ Vertices of triangle are (1, 2) , (6, 0) and (3, 0)

⇒ Area of triangle = (1/2) × 3 × 2

⇒ Area of triangle = 3 unit²

∴ The area of triangle is 3 unit2

Given:

Equation (1) = 3x - 2y = 8

Equation (2) = 4x + 3y = 5

Intersect point = P(∝, β)

Concept used:

We will solve the linera equations. The value of x and y will be the cordinates of intersenct point P.

Calculations:

⇒ 3x - 2y = 8         ------(1)

⇒ 4x + 3y = 5          ------(2)

By solving equation (1) and (2)

⇒ x = 2, y = -1

⇒ ∝ = 2, β = -1

⇒ (2 ∝ - β)

⇒ (2 × 2 + 1)

⇒ 5

∴ The value of (2 ∝ - β) is 5.

Complimentary angles: When the sum of the measure of the two angles is 90° , the angles are called Complimentary angles.

⇒ (1/3) × 90° 

⇒ 30° 

⇒ 30° + x = 90

∴ x = 60° is compliment of 1/3 of 90° 

Given: 

Radius of incircle (r) = √2 cm

AB = 5 cm, BC = 6 cm, AC = 9 cm.

Concepts used:

Area of ΔABC = 1/2 × r × (AB + BC + AC) 

Let the length of sides of triangle AB, BC and AC be a, b and c respectively.

Using heron's formula -

s = (a + b + c)/2

Area of ΔABC = √s(s – a)(s – b)(s – c)

Calculation:

Using heron's formula -

s = (5 + 6 + 9)/2 = 10 cm

Area of ΔABC = √s(s – a)(s – b)(s – c)

⇒ √10 × (10 – 5) × (10 – 6) × (10 – 9) cm² 

⇒ √(10 × 5 × 4 × 1) cm² 

⇒ √200 cm² 

⇒ 10√2 cm² 

∴ The area of ΔABC is 10√2 cm2.

Short trick:

Area of ΔABC = 1/2 × r × (AB + BC + AC) cm2 

⇒ 1/2 × √2 × (5 + 6 + 9) cm² 

⇒ 1/2 × √2 × 20 cm² 

⇒ 10√2 cm² 

∴ The area of ΔABC is 10√2 cm².

Given:

AB = 12 cm

AC = 15 cm

BC = 18 cm

Concept used:

The angle bisector of an angle of a triangle divides the opposite side into two parts that are proportional to the other two sides of the triangle.

Calculations:

The length of BC is, 

BC = BD + DC

⇒ DC = 18 – BD

Using the concept,

⇒ BD/DC = AB/AC

⇒ BD/(18 – BD) = 12/15

⇒ BD/(18 – BD) = 4/5

⇒ 5BD = 72 – 4BD

⇒ 9BD = 72 

⇒ BD = 8 cm

∴ The length of BD is 8 cm

Given:

Two factual statements are given

Concept Used:

Basic concept of circles

Calculation:

We know that the circle can only have maximum two points in common and minimum one point in common

∴ Statement 1 is incorrect

Now, to find the centre of circle we need at least two chords to find the exact centre of circle

∴ Statement 2 is incorrect

Hence, option (4) is correct

GIVEN:

The radii of the two circles R₁ = 6 cm & R₂ = 8 cm respectively and the distance between the center of the two circles(C₁C₂) = 15 cm 

FORMULA USED:

Length of direct common tangent = √(C₁C₂)² - (R₁ - R₂)² & Length of the transverse tangent = √(C₁C₂)² - (R₁ + R₂)²

CALCULATION:

R1 = 6 cm & R2 = 8 cm and C1C2 = 15 cm 

⇒ Length of direct common tangent = √(C1C2)2 - (R1 - R2)2 

⇒ √(C1C2)2 - (R1 - R2)2 = √(15)² - (8 - 6)²

⇒ √225 - 4

⇒ √221 cm

 Length of the transverse tangent = √(C1C2)2 - (R1 + R2)2

⇒ √(15)² - (8 + 6)²

⇒ √225 - 196

⇒ √29 cm

⇒  the ratio of the length of the direct common tangent to the length of the transverse common tangent =  √221 ∶ √29

∴ The ratio of the length of the direct common tangent to the length of the transverse common tangent =  √221 ∶  √29

Given:

Number of faces = 6

Number of vertices = 8

Formula used:

Euler's formula for polyhedron

F + V - E = 2

Where,

F = Number of Faces of Polyhedron

V = Number of Vertices of Polyhedron

E = Number of Edges of Polyhedron

Calculations:

Let the polyhedron has E edges,

⇒ 6 + 8 - E = 2

⇒ 14 - E = 2

⇒ E = 14 - 2

⇒ E = 12

∴ The Polyhedron has 12 edges

Given:

Number of faces, (F) = 7

Number of vertices, (V) = 10

Formula used:

From Euler's formula,

No. of edges, (E) = F + V − 2

Calculation:

No. of edges, (E) = F + V − 2

⇒ E = 7 + 10 − 2

⇒ E = 17 − 2

⇒ E = 15

∴ The number of edges of a polyhedron is 15

Given:

End points of the line segment are (–2, 2) and (10,–6)

Ratio = 3 : 1

Formula Used:

If end points of a line segment are (x₁, y₁) and (x_2, y₂) and point (x, y) divides the line segment internally in the ratio  m : n

\(x = \;\frac{{m{x_2} + n{x_1}}}{{m + n}}\;\;\;\;\;\;\;\;y = \;\frac{{m{y_2} + \;n{y_1}}}{{m + n}}\;\)

Calculation:

Let the coordinates of the point be (x, y) which divides the line segment in the ratio 3 : 1

\(x = \;\frac{{3 \times 10 + \;1 \times \left( { - 2} \right)}}{{3\; + \;1}}\;\;\)

\( ⇒ x = \;\frac{{30\; - \;2}}{4}\;\;\)

\(⇒ x = \;\frac{{28}}{4}\;\;\)

⇒ x = 7

\(\;y = \;\frac{{3 \times \left( { - 6} \right)\; + \;1 \times 2}}{{3\; + \;1}}\;\)

\(\; ⇒ y = \;\frac{{ - 18 + \;2}}{4}\;\)

\(\; ⇒ y = \;\frac{{ - 16}}{4}\;\)

⇒ y = –4

∴ The coordinates of the point are (7, –4)

Formula Used 

Sum of complementary angle is 90° 

Calculation 

Let the angle be x and y 

according to the question 

⇒ x = 8y 

⇒ x + y = 90° 

⇒ 9y = 90° 

⇒ y = 10° 

⇒ x = 8y = 8 × 10 = 80° 

∴ The measurement of angle is 80° 

Given:

An angle is eight times its complementary angle

Concept used:

Two angles are said to be complementary angles if their sum is 90º

Calculations:

Let  the required angle be y

Hence, its complementary angle is (90 − y)

⇒ y = 8 × (90 - Y)

⇒ y = 720 - 8y

⇒ 9y = 720

⇒ y = 720/9

⇒ y =  80º

∴  The measurement of the angle is 80º

Given 

The lines = x + 2y = 5 and 2x + 4y = 10

Calculation 

x + 2y = 5

2x + 4y = 10

First find the equation 1 

x + 2y = 5

x = 0 

0 + 2y = 5 

y = 5/2 

assume y = 0 

x + 2 (0) = 5 

x = 5 

then the 2 equation 

2x + 4y = 10

x = 0 

2(0) + 4y = 10 

y = 5/2

assume again y = 0 

2x + 4 (0) = 10 

2x = 10 

x = 5 

∴ These are co-incident line, because two line lie on same point.

Given

The ratio of two supplementary angles is 13 : 5

Concept used

Sum of two angles which are Supplementary angle is 180° 

Calculation

Let first angle be 13x then second angle be 5x

⇒ 13x + 5x = 180° 

⇒ 18x = 180° 

⇒ x = 10° 

⇒ First angle is 13 × 10 = 130° 

⇒ Second angle is 5 × 10 = 50° 

∴ Difference between angles is 130° - 50° = 80° 

Given:

Length of the side AB = 8√3 cm

Concept used:

Length of the centroid of equilateral triangle = length of the side/√3 

Calculation:

Length of the centroid = AB/√3

Lenght of the centroid = (8√3)/√3 = 8 cm

∴ The length of the centroid is 8 cm.  

Given:

The Sum of all interior angles of a polygon is 1400°

Formula used:

Sum of interior angles of a regular polygon of side 'n' = (n - 2) × 180° 

Calculation:

Let number of sides of the polygon is n

(n - 2) × 180 = 1440

⇒ n - 2 = 8

⇒ n = 10

∴ number of sides of the polygon is 10

Given:

Difference between exterior and interior angle a polygon is 120° (interior angle > exterior angle).

Formula Used:

1. Each exterior angle of the polygon = 180° - Each interior angle of the polygon

2. Number of diagonals of regular polygon = [n (n – 3)] / 2

3. An exterior angle of polygon = 360°/n

Where n = number of sides.

Calculation:

Let, the interior angle is x and exterior angle is y

Accordingly,

x - y = 120°      ----(1)

x + y = 180°      ----(2)

From (1) and (2) get,

y = 30° and x = 150°

The exterior of the given polygon is 30°

The number of sides of the polygon

= 360°/30° = 12

Number of diagonals of the polygon is {12 × (12 – 3)}/2

⇒ (12 × 9)/2 = 54

∴ The number of diagonals of the polygon is 54

Given:

Value of each external angle = 120°

Formula used:

θ = 360°/n

where,

θ = Each external angle

n = Number of sides 

Calculations:

θ = 360°/n

⇒ 120° = 360°/n

⇒ n = 3

∴ The number of sides is 3.

Given:

Two statements are given

Concept Used:

Basic concept of trapezium, rhombus and parallelogram

Calculation:

Let ABCD is a trapezium

∴ ∠A = ∠B and ∠C = ∠D

With this condition we are not reaching to any conclusion to prove that it is a parallelogram

∴ Statement 1 is incorrect

Now, If diagonals of rhombus are equal then by the help of congruency we can say that each angle is equal to 90°

∴ Statement 2 is correct

Given :

The ratio of the adjacent angle of the rhombus = 4 : 5

Concept used :

The sum of the adjacent angles of the rhombus is 180°

Calculation :

Let the smaller and larger angle be 4x and 5x respectively.

⇒ 4x + 5x = 180°

⇒ 9x = 180°

⇒ x = 20°

Difference between the larger and smaller angle = 5x - 4x

= x = 20°

∴ The difference between the larger and smaller angle is 20°.

Given

One of the internal angles of a regular polygon is 135°

Concept

Each interior angle of a regular polygon = [(n -2)/n] × 180° 

Number of diagonals = [n(n - 3)/2] 

Calculation

⇒ 135° = [(n -2)/n] × 180°  

⇒ (135°/180°) = [(n -2)/n]

⇒ (3/4) = [(n -2)/n]

⇒ 3n = 4n - 8

⇒ n = 8

Now, we get

⇒ Number of diagonals = 8(8 - 3)/2

⇒ Number of diagonals = 20 

∴ Number of diagonals is 20

Given:

If a regular polygon has 16 sides.

Concept used:

Polygon.

Calculation:

A polygon has 16 sides.

Each exterior angle = 360/n

Where n is the number of sides.

Exterior angle = 360/16

Exterior angle = 90/4

Sum of interior angle + Exterior angle = 180° 

interior angle = 180 - 90/4

interior angle = \(\frac{{720\ - \ 90}}{4}\)

interior angle = \(157\frac{1}{2}\)

Concept 

Measure of each interior angle of a regular polygon with 'n' sides = 180°(n - 2)/n

Calculation

Interior angle = 180°(n - 2)/n = 108º

⇒ 180ºn - 360º = 108ºn

⇒ (180º - 108º)n = 360º

⇒ 72º × n = 360º

∴ n = 360º/72º = 5

Ans: 1

Given:

The length and breadth of a rectangle are in the ratio 5 : 3.

the length is 8 m more than the breadth.

Formula used:

area of rectangle = length × breadth

Calculation:

The ratio of length and breadth of rectangle = 5 : 3

⇒ length of rectangle = 5x, breadth of rectangle = 3x

⇒ 5x - 3x = 8

⇒ 2x = 8

∴ x = 4

length of rectangle = 5x = 5 × 4 = 20 m

breadth of rectangle = 3x = 3 × 4 = 12 m

area of rectangle = length × breadth

⇒ 20 ×  12 = 240 m²

Formula used :

External angle = 180° - internal angle

Number of sides of polygon = 360/(external angle)

Calculation :

Internal angle = 156°

⇒ External angle = 180° - 156°

⇒ External angle = 24°

⇒ Number of sides of regular polygon = 360°/24°

⇒ Number of sides of regular polygon = 15

∴ The number of sides of the regular polygon is 15.

Given

The ratio between the interior angle and the exterior angle of a regular polygon is 2 : 1

Concept

Exterior angle + interior angle = 180° 

Side of polygon = (360°/Exterior angle)

Calculation

Let ratio of angles in x 

⇒ The ratio between the interior angle and the exterior angle of a regular polygon is = 2x : x

⇒ 2x + x = 180°

⇒ 3x = 180°

⇒ x = (180°/3)

⇒ x = 60°

⇒ Interior angle = 120°

⇒ Exterior angle = 60°

⇒ Side of polygon = (360°/60°)

⇒ Side of polygon = 6

∴ Side of polygon = 6

Given:

In ΔABC, AB = AC

∠A = (5x – 30)° 

∠C = x

Concept used:

If two sides of a triangle are equal their opposite angles will also be equal.

The sum of all angles of a triangle is 180^∘ .

Calculation:

AB = AC

⇒ ∠C = ∠B = x

According to question,

5x – 30° + x + x = 180° 

⇒ 7x – 30° = 180° 

⇒ 7x = 210° 

⇒ x = 30° 

∴ The value of ∠B is 30° 

Given: 

Length of the floor = 5 m 

Breadth of the floor = 4 m 

Side of square carpet = 3 m 

Formula used:

Area of the rectangle = (l × b)

Where, 

l → Lenght

b → Breadth 

Area of square = (side)² 

Calculations:

Area of the rectangle = (l × b)

⇒ (5 × 4) = 20 m²

Area of square carpet = (side)2 square unit

⇒ (3)² = 9 m²

Uncarpeted area = (20 – 9) = 11 m²

∴ The area of the floor that is not carpeted is 11 m²