Sachin

Area and Perimeter is an important and basic topic in the Mensuration of 2-D or Planar Figures. The area is used to measure the space occupied by the planar figures. The perimeter is used to measure the boundaries of the closed figures. In Mathematics, these are two major formulas to solve the problems in the 2-dimensional shapes.

Each and every shape has two properties that are Area and Perimeter. Students can find the area and perimeter of different shapes like Circle, Rectangle, Square, Parallelogram, Rhombus, Trapezium, Quadrilateral, Pentagon, Hexagon, and Octagon. The properties of the figures will vary based on their structures, angles, and size. Scroll down this page to learn deeply about the area and perimeter of all the two-dimensional shapes.

Area and Perimeter Definition

Area: Area is defined as the measure of the space enclosed by the planar figure or shape. The Units to measure the area of the closed figure is square centimeters or meters.

Perimeter: Perimeter is defined as the measure of the length of the boundary of the two-dimensional planar figure. The units to measure the perimeter of the closed figures is centimeters or meters.

Formulas for Area and Perimeter of 2-D Shapes

1. Area and Perimeter of Rectangle:

• Area = l × b
• Perimeter = 2 (l + b)
• Diagnol = √l² + b²

Where, l = length
b = breadth

2. Area and Perimeter of Square:

• Area = s × s
• Perimeter = 4s

Where s = side of the square

3. Area and Perimeter of Parallelogram:

• Area = bh
• Perimeter = 2( b + h)

Where, b = base
h = height

4. Area and Perimeter of Trapezoid:

• Area = 1/2 × h (a + b)
• Perimeter = a + b + c + d

Where, a, b, c, d are the sides of the trapezoid
h is the height of the trapezoid

5. Area and Perimeter of Triangle:

• Area = 1/2 × b × h
• Perimeter = a + b + c

Where, b = base
h = height
a, b, c are the sides of the triangle

6. Area and Perimeter of Pentagon:

• Area = (5/2) s × a
• Perimeter = 5s

Where s is the side of the pentagon
a is the length

7. Area and Perimeter of Hexagon:

• Area = 1/2 × P × a
• Perimeter = s + s + s + s + s + s = 6s

Where s is the side of the hexagon.

8. Area and Perimeter of Rhombus:

• Area = 1/2 (d1 + d2)
• Perimeter = 4a

Where d1 and d2 are the diagonals of the rhombus
a is the side of the rhombus

9. Area and Perimeter of Circle:

• Area = Πr²
• Circumference of the circle = 2Πr

Where r is the radius of the circle
Π = 3.14 or 22/7

10. Area and Perimeter of Octagon:

• Area = 2(1 + √2) s²
• Perimeter = 8s

Where s is the side of the octagon.

Solved Examples on Area and Perimeter

Here are some of the examples of the area and perimeter of the geometric figures. Students can easily understand the concept of the area and perimeter with the help of these problems.

1. Find the area and perimeter of the rectangle whose length is 8m and breadth is 4m?

Solution:

Given,
l = 8m
b = 4m
Area of the rectangle = l × b
A = 8m × 4m
A = 32 sq. meters
The perimeter of the rectangle = 2(l + b)
P = 2(8m + 4m)
P = 2(12m)
P = 24 meters
Therefore the area and perimeter of the rectangle is 32 sq. m and 24 meters.

2. Calculate the area of the rhombus whose diagonals are 6 cm and 5 cm?

Solution:

Given,
d1 = 6cm
d2 = 5 cm
Area = 1/2 (d1 + d2)
A = 1/2 (6 cm + 5cm)
A = 1/2 × 11 cm
A = 5.5 sq. cm
Thus the area of the rhombus is 5.5 sq. cm

3. Find the area of the triangle whose base and height are 11 cm and 7 cm?

Solution:

Given,
Base = 11 cm
Height = 7 cm
We know that
Area of the triangle = 1/2 × b × h
A = 1/2 × 11 cm × 7 cm
A = 1/2 × 77 sq. cm
A = 38.5 sq. cm
Thus the area of the triangle is 38.5 sq. cm.

4. Find the area of the circle whose radius is 7 cm?

Solution:

Given,
Radius = 7 cm
We know that,
Area of the circle = Πr²
Π = 3.14
A = 3.14 × 7 cm × 7 cm
A = 3.14 × 49 sq. cm
A = 153.86 sq. cm
Therefore the area of the circle is 153.86 sq. cm.

5. Find the area of the trapezoid if the length, breadth, and height is 8 cm, 4 cm, and 5 cm?

Solution:

Given,
a = 8 cm
b = 4 cm
h = 5 cm
We know that,
Area of the trapezoid = 1/2 × h(a + b)
A = 1/2 × (8 + 4)5
A = 1/2 × 12 × 5
A = 6 cm× 5 cm
A = 30 sq. cm
Therefore the area of the trapezoid is 30 sq. cm.

6. Find the perimeter of the pentagon whose side is 5 meters?

Solution:

Given that,
Side = 5 m
The perimeter of the pentagon = 5s
P = 5 × 5 m
P = 25 meters
Therefore the perimeter of the pentagon is 25 meters.

FAQs on Area and Perimeter

1. How does Perimeter relate to Area?

The perimeter is the boundary of the closed figure whereas the area is the space occupied by the planar.

2. How to calculate the perimeter?

The perimeter can be calculated by adding the lengths of all the sides of the figure.

3. What is the formula for perimeter?

The formula for perimeter is the sum of all the sides.

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In Mensuration the circumference and area of a circle are defined as the length of the boundary of the circle and region occupied by the circle in 2-D Geometry. Let us discuss in detail the area and circumference of the circle using the formulas and solved example problems. We provide a detailed explanation of how to calculate the circumference and area of the circle.

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What is Circumference and Area of Circle?

Circumference of Circle:

The circumference of the circle is the measure of the boundary of the circle. The circumference of the circle is also known as the perimeter of the circle. The perimeter or circumference of the circle is measured in units.
C = Πd or 2Πr

Area of Circle:

The area of the circle is the region covered by the circle or sphere in two-dimensional mensuration. The units to measure the area of the circle is square units.
A = Πr²
Where,
A is the area of the circle
r is the radius of the circle

What is the radius of the circle?

The radius of the circle is the distance from the center to the outline of the circle. Radius plays an important role in calculating the area and perimeter of the circle.

Properties of Circle

The properties of the circle are given below,

• The diameter of the circle is the longest chord of the circle.
• The circle is said to be congruent if it has the same radii.
• A circle can confine rectangle, square, trapezium, etc.
• If the tangents are drawn at the end of the diameter they are parallel to each other.

Area and Circumference of Circle Formula

Circumference:
The circumference of the circle is the measure of the boundary of the circle. The formula for the circumference of the circle is given below,
C = Πd
Where,
C is the circumference of the circle
Π is the mathematical constant
The approximate value of pi is 3.14 or 22/7
d is the diameter of the circle
C = 2Πr
Where,
C is the circumference of the circle
Π is the mathematical constant
The approximate value of pi is 3.14 or 22/7
r is the radius of the circle
Area: 
The formula for the area of the circle is as follows,
A = Πr²
A is an area of the circle
Π is the mathematical constant
The approximate value of pi is 3.14 or 22/7
r is the radius of the circle
Area of Semi-Circle:
The area of the semicircle is the region covered by the 2D figure. The formula for the area of the semi-circle is as follows,
A = Πr²/2
Perimeter of the Semi-circle:
The formula for the perimeter of the semi-circle is given below,
P = 2Πr/2 = Πr

Solved Examples on Circumference and Area of a Circle?

Get the step by step explanation on the formula of Area and Circumference of Circle here.

1. What is the circumference of the circle with a radius of 7cm?

Solution:

Given,
r = 7cm
We know that,
Circumference of the circle = 2Πr
C = 2 × 22/7 × 7 cm
C = 44 cm
Thus the circumference of the circle is 44 cm.

2. What is the circumference of the circle with a diameter of 21 cm?

Solution:

Given,
d = 21 cm
We know that,
Circumference of the circle = 2Πr
C = Πd
C = 22/7 × 21cm
C = 22 × 3cm
C = 66 cm
Therefore the circumference of the circle is 66 cm.

3. Find the area of the circle with a radius of 14m?

Solution:

Given,
r = 14m
We know that,
Area of Circle = Πr²
A = 22/7 × 14 × 14 sq.m
A = 22 × 2 × 14 m²
Thus the area of the circle is 616m²

4. Find the area of the circle if its circumference is 124m?

Solution:

Given,
Circumference of the circle = 124m
We know that,
Circumference of the circle = 2Πr
124m = 2Πr
Πr = 124/2
Πr = 62
r = 62 × 7/22
r = 19.72 m
Now find the area of the circle using the radius.
Area of Circle = Πr²
A = 3.14 × (19.72)²
A = 1221 sq. m
Therefore the area of the circle is 1221 sq. meters.

5. Find the area and circumference of the circle of radius 14m?

Solution:

Given,
radius = 14m
We know that,
Circumference of the circle = 2Πr
C = 2 × 22/7 × 14m
C = 2 × 22 × 2m
C = 88m
Now find the area of the circle using the radius.
Area of Circle = Πr²
A = Π(14)²
A = 22/7 × 14 × 14 sq.m
A = 22 × 2 × 14
A = 44 × 14 sq.m
A = 616 sq.m
Thus the area of the circle is 616 sq.m

FAQs on Circumference and Area of Circle

1. How to calculate the area of a circle?

The area of the circle can be calculated by the product of pi and radius squared.

2. What is the diameter of the circle?

The diameter of the circle is 2r.

3. How to calculate the circumference of the circle?

The circumference of the circle can be calculated by multiplying the diameter with pi.

The Perimeter and Area of the Square are used to measure the length of the boundary and space occupied by the square. These are two important formulas used in Mensuration. Perimeter and Area of the Square formulas are used in the 2-D geometry.

Square is a regular quadrilateral where are the sides and angles are equal. The concepts of the Perimeter and Area Square formula, Derivation, Properties, are explained here. The solved examples with clear cut explanations are provided in this article. Students can understand how and where to use the formulas of Area and Perimeter of Square.

What is the Area and Perimeter of the Square?

Area of a square: The area of the square is defined as the region covered by the two-dimensional shape. The units of the area of the square are measured in square units i.e., sq. cm or sq. m.

Perimeter of a square: The perimeter of the square is a measure of the length of the boundaries of the square. The units of the perimeter are measured in cm or m.

Area of Square Formula

The area of the square is equal to the product of the side and side.
Area = Side ×  Side sq. units
A = s² sq. units

Perimeter of Square Formula

The perimeter of the square is the sum of the lengths.
P = s + s + s +s
P = 4s units
Where s is the side of the square.

Diagonal of Square Formula

The square has two diagonals with equal lengths. The diagonal of the square is greater than the sides of the square.

• The relationship between d and s is d = a√2
• The relationship between d and Area is d = √2A

What is Square?

A square is a regular polygon in which all four sides are equal. The measurement of the angles of the square is also equal.

Properties of Squares

The properties of the square are similar to the properties of the rectangle. Go through the properties of squares from the below section.

• All sides of the squares are equal.
• It has 4 sides and 4 vertices.
• The interior angles of the square are equal to 90º
• The diagonlas of square bisect at 90º
• The diagonals of the square are divided into two isosceles triangles.
• The opposite sides of the squares are parallel to each other.
• Each half of the square is equal to two rectangles.

Solved Problems on Perimeter and Area of Square

Below we have provided the solved examples of perimeter and area of a square with a brief explanation. Scroll down this page to check out the formulas of Area and Perimeter of Square.

1. What is the Area and Perimeter of the square if one of its sides is 4 meters?

Solution:

Given the side of the square is 4 meters.
Area of the square = s × s
A = 4 m × 4 m
A = 16 sq. meters
The perimeter of the square = 4s
P = 4 × 4 m
P = 16 meters.
Therefore the area and perimeter of the square are 16 sq. meters and 16 meters.

2. Find the area of the square if the side is 10 cm?

Solution:

Given,
s = 10 cm
Area of the square = s × s
A = 10 cm × 10 cm
A = 100 sq. cm
Therefore the area of the square is 100 sq. cm

3. The perimeter of the square is 64 cm. Find the area of the square?

Solution:

Given,
The perimeter of the square is 64 cm
P = 4s
64 cm = 4s
s = 64/4 = 16 cm
Thus the side of the square is 16 cm.
Now the find the area of the square.
Area of the square = s × s
A = 16 cm × 16 cm
A = 256 sq. cm
Therefore the area of the square is 256 sq. cm.

4. If the area of the square is 81 cm², then what is the length of the square?

Solution:

Given,
A = 81 cm²
Area of the square = s × s
81 sq. cm = s²
s² = 81 sq. cm
s = √81 sq. cm
s = 9 cm
Thus the length of the square is 9 cm.

5. The length of the square is 25 cm. What is the area of the square?

Solution:

Given,
The length of the square is 25 cm
Area of the square = s × s
A = 25 × 25
A = 625 sq. cm
Therefore the area of the square is 625 sq. cm.

FAQs on Perimeter and Area of Square

1. How to find the perimeter of the square?

Add all the sides of the square to find the perimeter of the square.

2. What is the formula for the perimeter of a square?

The Perimeter of Square formula is sum of the lengths i.e, side + side + side + side = 4s

3. What is the formula for the area of the square?

The area of the square formula is the product of side and side. A = s × s.

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Solve different types of problems on calculating speed and get acquainted with various models of questions asked in your exams. Be aware of the Formula to Calculate and Relationship between Speed Time and Distance. Practice Speed Problems on a regular basis so that you can be confident while attempting the exams. We even provided solutions for all the Questions provided and explained everything in detail for better understanding. Try to solve the Speed Questions on your own and then cross-check where you are lagging.

We know the Speed of the Object is nothing but the distance traveled by the object in unit time.

Formula to find out Speed is given by Speed = Distance/Time

Word Problems on Calculating Speed

1.  A man walks 25 km in 6 hours. Find the speed of the man?

Solution:
Distance traveled = 25 km
Time taken to travel = 6 hours
Speed of Man = Distance traveled/Time taken
= 25km/6hr
= 4.16 km/hr
Therefore, a man travels at a speed of 4.16 km/hr

2. A car covers a distance of 420 m in 1 minute whereas a train covers 70 km in 30 minutes. Find the ratio of their speeds?

Solution:
Speed of the Car = Distance Traveled/Time Taken
= 420m/60 sec
= 7 m/sec

Speed of the Train = Distance Traveled/Time Taken
= 70 km/1/2 hr
= 140 km/hr

To convert it into m/sec multiply with 5/18
= 140*5/18
= 38.8 m/sec
= 39 m/sec (Approx)
Ratio of Speeds = 7:39

3. A car moves from A to B at a speed of 70 km/hr and comes back from B to A at a speed of 40 km/hr. Find its average speed during the journey?

Solution:
Since the distance traveled is the same the Average Speed= (x+y)/2 where x, y are two different speeds
Substitute the Speeds in the given formula
Average Speed = (70+40)/2
= 110/2
= 55 km/hr
The Average Speed of the Car is 55 km/hr

4. A bus covers a certain distance in 45 minutes if it runs at a speed of 50 km/hr. What must be the speed of the bus in order to reduce the time of journey by 20 minutes?

Solution:
Speed = Distance/Time
50 = x/3/4
50 = 4x/3
4x = 150
x = 150/4
= 37.5 km

Now by applying the same formula we can find the speed

Now, time = 40 mins or 0.66 hr since the journey is reduced by 20 mins

S = Distance/Time
= 37.5/0.66
= 56.81 km/hr

5. Ram traveled 200 km in 3 hours by train and then traveled 140 km in 3 hours by car and 5 km in 1/2 hour by cycle. What is the average speed during the whole journey?

Solution:
Distance traveled by Train is 200 km in 3 hours
Distance Traveled by Car is 140 km in 3 hours
Distance Traveled by Cycle is 5 km in 1/2 hour
Average Speed = Total Distance/Total Time
= (200+140+5)/(3+3+1/2)
= 345/6 1/2
= 345/(13/2)”
= 345*2/13
= 53.07 km/hr

6. A train covers 150 km in 3 hours. Find its speed?

Solution:
Speed = Distance/Time
= 150 km/3 hr
= 50 km/hr
Therefore, Speed of the Train is 50 km/hr.

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Find the Distance Problems in which an object will travel for a certain distance in a given period of time. Learn the Formula to Calculated Distance When Time and Speed are given. Refer to Solved Problems on Calculating Distance and understand the logic behind them. You can see out Time and Distance concept to know everything in detail. Practice the questions on finding distance and get the answers too from our page. Detailed Solutions makes it easy for you to grasp the concept.

Formula to Calculate the Distance = Speed * Time

Distance Word Problems Examples

1. A train moves at a speed of 45 km/hr. How far will it travel in 30 minutes?

Solution:

Speed of the Train = 45 kmph

Time = 30 min = 1/2 hr

Distance = Speed * Time

= 45Kmph*1/2 hr

= 22.5 Km

Therefore, the train travels a distance of 22.5 km

2. If a motorist moves with a speed of 40 km/hr and covers the distance from place A to B in 2 hours, find the distance between places A and B?

Solution:

Speed = 40 km/hr

Time taken to travel from Place A to Place B = 2 hrs

Distance = Speed*Time

= 40kmph*2 hr

= 80 km

Therefore, the distance between places A and B is 80 km.

3. How much father can an interstate bus go traveling 80 km/hr rather than 50 km/hr in 3 hours?

Solution:

Distance = Speed *Time

If Speed = 80 km/hr

Time = 3 hrs

Distance = 80*3

= 240 km

If Speed = 50 km/hr

Time = 3 hrs

Distance = Speed * Time

= 50*3

= 150 km

Difference between Distances = 240 km – 150 km

= 90 km

4. Sound travels at a speed of 1100 km in one hour. How many meters will it travel in one second?

Solution:

Speed = 1100 kmph

To convert kmph to m/sec multiply with 5/18

= 1100*5/18

= 305.55 m/sec

5. A car travels at a speed of 72 km/hr. How many meters will it travel in 1 second?

Solution:

Speed = 72 km/hr

To convert kmph to m/sec multiply with 5/18

= 72*5/18

= 20 m/sec

6. Mohan drives a car at a uniform speed of 40 km/hr, find how much distance is covered in 120 minutes?

Solution:

Speed = 40 km/hr

Time = 120 minutes = 2 hrs

Distance = Speed*Time

= 40 km/hr*2 hr

= 80 km

7. A car takes 2 hours to cover a distance if it travels at a speed of 30 kmph. What should be its speed to cover the same distance in 1 hour?

Solution:

Speed = 30 kmph

Time = 2hrs

Distance = Speed*TIme

= 30 kmph*2 hr

= 60 Km

Speed = ?

Distance = 60 km

Time = 1 hr

Speed = Distance/Time

= 60 km/1 hr

= 60 kmph

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Work with the Problems on Calculating Time and learn how to find Time when Speed and Distance are given. Know the Relationship between Speed Time and Distance with the formula provided. Check Worked out Examples for Distance with Solutions and cross-check your solutions while practicing. Check different types of Time Questions followed by illustrations for a better understanding of the concepts. Practice each of the Time Word Problems provided and score better grades in the exam.

Solved Time Questions with Answers

1. A car travel 70 km in 30 minutes. In how much time will it cover 120 km?

Solution:

Speed = Distance/Time

= 70 km/1/2 hr

= 140 kmph

Speed = Distance/Time

140 kmph = 120 km/Time

Time = 120 km/140 kmph

= 0.85 hr

car takes 0.85 hr to cover the distance 120 km.

2. Vinay covers 180 km by car at a speed of 60 km/hr. find the time taken to cover this distance?

Solution:

Speed = 60 km/hr

Distance = 180 km

Speed = Distance/Time

60 km/hr = 180 km/Time

Time = 180 km/60 km/hr

= 3 hr

Vinay takes 3 hrs to cover the distance 180 km at a speed of 60 km/hr.

3. A train covers a distance of 45 km in 20 minutes. Find the time taken by it to cover the same distance if its speed is decreased by 12 km/hr?

Solution:

Distance covered by train = 45 km

Time taken = 20 min = 20/60 = 1/3 hr

Speed of the train = Distance covered/Time taken

= 45km/1/3 hr

= 135 km/hr

Reduced Speed = 135 km/hr – 12 km/hr

= 123 km/hr

Time = Distance Traveled/Speed

= 45/123

= 0.365hr

= 0.365*60 min

= 21.95 min

4. A man is walking at a speed of 5 km per hour. After every km, he takes a rest for 3 minutes. How much time will it take to cover a distance of 6 km?

Solution:

Rest time = Number of rests * time of each rest

= 5*3 minutes

= 15 minutes

Total Time Taken = Distance/Speed +Rest Time

= (6/5)*60+15 minutes

= 72+15

= 87 minutes

5. A car takes 3 hours to cover a distance if it travels at a speed of 45 mph. What should be its speed to cover the same distance in 2 hours?

Solution:

Distance = Speed *Time

= 45mph*3 hr= 135 miles

Speed = Distance/Time

= 135 miles/2 hrs

= 67.5 miles per hour

Therefore, car needs to travel at a speed of 67.5 miles per hour in order to travel a distance of 135 miles in a time of 2 hrs.

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The Perimeter and Area of Rectangle are two important formulas in Mensuration. It calculates the space occupied by the rectangle and the length of boundaries of the rectangle. In this article, students can come across the concept of area and perimeter of rectangle deeply.

A rectangle is a quadrilateral with two equal sides and two parallel lines and four right angles. The concept of area and perimeter of rectangle formulas are explained with examples. Some of the examples of different shapes are given below.

What is Perimeter and Area of a Rectangle?

Perimeter: Perimeter of the rectangle is the sum of all the sides of the rectangle. The rectangle has two lengths and two breadths. To find the perimeter of the rectangle we have to add the length and breadth. It is measured in units. It is denoted by P.

Area: The area of the rectangle formula helps to calculate the length and breadth of the two-dimensional closed figure. To find the area of the rectangle we have to multiply the length and breadth of the rectangle. It is measured in square units. It is denoted by A.

Properties of Rectangle

• A Rectangle has two equal sides
• The rectangle is a quadrilateral
• The diagonals of the rectangle have the same length.
• The diagonal of the rectangle bisect each other.
• Sum of all the four angles is 360º
• Each angle of the rectangle is 90º
• If the sides of the rectangle are l and b then the diagonal of the rectangle d = √l² + b²
• The opposite sides of the rectangle are parallel.

Derivation of Perimeter of a Rectangle

The perimeter of the rectangle is the sum of all four sides.
P = l + l + b + b
P = 2l + 2b
P = 2(l + b)
Thus the perimeter of the rectangle is = 2(l + b)

Area of the Rectangle

The area of the rectangle is the product of length and breadth.
A = l × b

Perimeter and Area of Rectangle Formula

• Area of Rectangle = l × b
• Perimeter of Rectangle = 2(l + b)
• Length of the rectangle = A/b
• The breadth of the Rectangle = A/l
• Diagonal of the Rectangle = √l² + b²

Solved Problems on Area and Perimeter of a Rectangle

The formula of Perimeter and Area of Rectangle is explained step by step here with examples. Go through the below questions and solve the problems using the Area and Perimeter of the Rectangle formula.

1. Find the length of the rectangular plot whose breadth is 11 cm and the area is 165 cm². Also, find the perimeter of the rectangle?

Solution:

Given,
Breadth = 11 cm
Area = 165 cm²
Area of the rectangle = l × b
165 sq. cm = l × 11 cm
l = 165/11
l = 15 cm
Thus the length of the rectangle is 15 cm.
We know that,
Perimeter of the rectangle = 2(l + b)
P = 2(15 + 11)
P = 2(26)
P = 52 cm
Thus the perimeter of the rectangle is 52 cm.

2. Find the Perimeter of the Rectangle whose length is 10 cm and breadth is 8 cm?

Solution:

Given, Length = 10 cm
Breadth = 8 cm
We know that,
Perimeter of the rectangle = 2(l + b)
P = 2(10 cm + 8 cm)
P = 2(18 cm)
P = 36 cm
Therefore the perimeter of the rectangle is 36 cm.

3. Find the area of the rectangle whose length is 14 meter and width is 10 meters?

Solution:

Given,
Length = 14 meter
Width = 10 meter
We know that,
Area of the rectangle = l × w
A = 14 m × 10 m
A = 140 sq. meters
Therefore the area of the rectangle is 140 square meters.

4. A rectangular plot has its length of 16 cm and a perimeter of 60 cm. Find the width of the rectangular plot?

Solution:

Given,
Length = 16 cm
Perimeter = 60 cm
Width = ?
We know that
The perimeter of the rectangle = 2(l + w)
60 cm = 2(16 cm + w)
16 + w = 60/2
16 + w = 30
w = 30 – 16
w = 14 cm
Thus the width of the rectangular plot is 14 cm.

5. Find the area and perimeter of the rectangle whose length and breadth are 12 m and 6 m?

Solution:
Given,
length = 12 m
breadth = 6 m
We know that,
Area of the rectangle = l × b
A = 12 m × 6 m
A = 72 sq. meters
Now find the perimeter of the rectangle
P = l + l + b + b
P = 12 m + 12 m + 6 m + 6 m
P = 24 m + 12 m
P = 36 m
Therefore the area and perimeter of the rectangle is 72 square meters and 36 meters.

FAQs on Area and Perimeter of Rectangle

1. How to find the perimeter of a rectangle?

The perimeter of the rectangle can be calculated by adding all the sides of the rectangle.

2. What is the area of the rectangle?

The area of the rectangle is defined as the space occupied by the rectangle or closed figure. It is the product of length and breadth.

3. What is the unit for the perimeter of a rectangle?

The unit for the perimeter of the rectangle is cm or meters.

Wanna learn all the basics of Integers concepts? Then start with the absolute value of an integer. Integer Absolute Value is an important and basic concept to learn other concepts of Integers. Refer to all the rules, definitions, solved examples, types, etc to understand the concept more clearly. Follow the below sections to get more idea on how to find the Absolute Value of an Integer in detail and Solved Problems for finding the Integer Absolute Value using different approaches.

Absolute Value of an Integer – Introduction

Integers Absolute Value is the distance of that integer value from zero irrespective of (positive or negative) direction. While considering the absolute value, its numerical value is taken without taking the sign into consideration.

An integer is any positive or negative whole number. Therefore, the positive integer has a negative sign and vice versa.

A positive number is a number that is greater than zero. It is represented by a “+” symbol and can be written with or without a symbol in front of it. The gain in some value or something is written with a positive number. A negative number is a number that is less than zero. It is represented by a “-” symbol and can be written with a “-” symbol in front of it. The loss in some value or something is written with a negative number.

A number line is a kind of diagram on which the numbers are marked at intervals. These are used to illustrate simple and easy numerical operations. Using the number line allows seeing a number is in relationship to other numbers and from zero. Zero, is present in the middle and separates positive and negative numbers. On the right side of zero, we can find numbers that are greater than zero (positive numbers) and on the left side, we can find numbers that are less than zero (negative numbers). The absolute value of the integer is the same as the distance from zero to a specific number.

Representation of Absolute Value of an Integer

Integer Absolute Value is represented with two vertical lines. i.e., | |, one on either side of the integer.

|a| = a, when a is the positive integer

|a| = -a, when a is the negative integer

Examples:

1. Absolute integer value of -15 is written as |-15| = 15 {here mod of -15 = 15}
2. Absolute integer value of 8 is written as |8| = 8 {here mod of 8 = 8}

Adding and Subtracting Absolute Value of Integers

To add 2 integers with the same (positive or negative) sign, add absolute values and assign the sum with the same sign as same both values.

Example:

(-7) + (-4) = -(7 + 4)= – 11.

To add 2 integers with different signs, find the difference in absolute integer values and give that product the same sign of the largest absolute value.

Example:

(-7)+(2)= -5

How to find the absolute value?

1. First of all, find the absolute values of (7 and 2)
2. Find the difference of numbers between 7 and 2 (7-2=5)
3. Find the sign of the largest absolute value i.e., negative, as 7 is of (-)sign
4. Add the sign to the difference we got in Step 2.
5. Hence the final solution is -5

When an integer is subtracted or added by another integer, the result will be an integer.

Multiplication of Absolute Value of Integers

To multiply the integer values, we have to multiply the absolute values. If the integers that are to be multiplied have the same sign, then the result will be positive. If both the integers have different signs, then the result will be negative.

When an integer is multiplied by another integer, then the result will be an integer itself.

Absolute Value of Integers Examples

Question 1:

Write the absolute value of each of the following?

(i) 15

(ii) -24

(iii) -375

(iv) 0

(v) +7

(vi) +123

Solution: 

(i) 15

(ii) 24

(iii) 375

(iv) 0

(v) 7

(vi) 123

Question 2: 

Evaluate the following integers

(i) |-7| + |+5| + |0|

(ii) |10| – |-15| + |+12|

(iii) – |+3| – |-3| + |-6|

(iv) |-8| – |17| + |-12|

Solution: 

(i) 12

(ii) 7

(iii) 0

(iv) 3

Question 3:

State whether the statements are true or false?

(i) The absolute value of -3 is 3.

(ii) The absolute value of an integer is always greater than the integer.

(iii) |+5| = +5

(iv) |-5| = -5

(v) – |+5| = 5

(vi) – |-5| = -5

Solution:

(i) True

(ii) False

(iii) True

(iv) False

(v) False

(vi) True

Methods to Compute the Absolute Value of Integer

There are 3 approaches to find the Integers Absolute value. These approaches are helpful while solving problems. Finding the absolute value is the first step in solving the problems.

Method 1: 

As the absolute value of the integer is always positive. For the positive integer, the absolute number is the number itself. For the negative number, the absolute number is multiplied by other negative numbers.

Method 2:

Negative numbers are saved in 2’s complement form. To get the absolute number, toggle bits of the number, then 1 to the result.

Method 3:

The built-in function library finds the absolute value of an integer.

To solve most of the problems, knowing the absolute value of the integer is important. This method is used in the real world and it has all sorts of math clues. Absolute values are often used in distance problems and also sometimes used for inequalities. Integers Absolute values are really helpful to get clarity on many things.

We have mentioned the important concept of Integers Absolute Value in the above article. Hope you got a clear idea of definitions, rules, and how to solve the problems. If you have any doubts, you can contact us from the below comment box. You can also directly ping us to know other information. Stay tuned to our page to get the latest and important information about all mathematical concepts.

Lines and Angles concepts are explained by experts in order to help students to get excellent marks in their examination. The line is a combination of infinite points that has both directions infinitely. There are different lines available such as intersecting lines, perpendicular lines, transversal lines, etc. An angle is nothing but a combination of two rays with a common endpoint.

Based on the operations performed on a line, then the lines are classified as Parallel Lines, Perpendicular Lines, Transversal, etc. In the same way, the angles are classified as Complementary Angles, Supplementary Angles, Adjacent Angles, Vertically Opposite Angles based on their operation.

Quick Links of Lines and Angles

Working on Lines and Angles problems will help students to get an idea about how to solve the problems. All the concept links are given below. Just check out the concepts and open the topic that is difficult for you. After that solve all the problems available on that topic to get a grip on that entire topic.

• Basic Geometrical Concepts
• Angles
• Classification of Angles
• Related Angles
• Some Geometric Terms and Results
• Complementary Angles
• Supplementary Angles
• Complementary and Supplementary Angles
• Adjacent Angles
• Linear Pair of Angles
• Vertically Opposite Angles
• Parallel Lines
• Transversal Line
• Parallel and Transversal Lines

Line Segment

The line segment is a line that has two endpoints with a fixed length.

Ray

A ray is a line that has only one endpoint and extends infinitely in another direction. From the below figure, point A is the and point.

Perpendicular Lines

Two lines are said to be Perpendicular Lines when they meet at a single point and forms a right angle. From the below figure, PQ and RS are two lines and they intersect at a point x. Also, the two lines PQ and RS form a right angle 90º.

Parallel Lines

Two lines are said to be Parallel Lines when they do not intersect each other and they have the same distance throughout the plane. From the below figure, A and B are two parallel lines.

Transversal Line

A transversal Line is a line that intersects two lines at distinct points. From the below figure, C is the Transversal Line, that intersects the lines A and B at two points D and E.

Properties of Lines

• The points those are not lie on the same line are called non-collinear points.
• The points that lie on the same line are called Collinear points

Acute Angle

The angle that is less than the right angle is called Acute Angle.

Obtuse Angle

The angle that is more than the right angle is called Acute Angle.

Right Angle

The angle with 90 degrees is called as Right Angle.

Straight Angle

The angle that is equal to 180 degrees is called Straight Angle.

Complementary Angles

If the sum of the two angles is 90 degrees, then they are called Complementary Angles.

Supplementary Angles

If the sum of the angles is 180 degrees, then those angles are called Supplementary Angles.

Properties of Angles

• An angle is formed at a point where two rays intersecting.
• If the angle is more than 180 degrees and less than 360 degrees, then it is called a reflex angle.
• If the two angles are adjacent angles and add up to 180 degrees then they form linear pair.

Quick Access to Lines and Angles

We have given solutions for all questions with step by step explanation in understandable language. Therefore, students can get great knowledge on the entire concepts of Lines and Angles. All the questions and answers that are present in the updated syllabus are included on our website with individual concept explanation. So, students who want to learn a single topic can go through the required link available below and learn it from the concerned link.

Are you searching for Lines and Angles Material? You have free access to Lines and Angles Material on our website. You can read and practice all problems either online or offline. So, immediately bookmark our website and practice all the math problems at your comfortable times. Do hard work and get complete knowledge on the Lines and Angles concept. Clear your doubts by contacting us through the comment section.

 

When two triangles lie on the same base and between the same parallels, then those triangles measure equal area. Check the proof for Triangles on the same base and between the same parallel are equal in Area. Get the relationship between the areas of triangles having the same base in the following sections. For the reference of students, we have also included the solved examples on Triangle on the Same Base and between Same Parallels.

Proof for Area of Triangles are Equal Having the Same Base

Get a detailed explanation for how the areas of triangles are equal when their bases are the same and between the same parallels.

In the following figure, ∆ABD and ∆DEF having equal base length and between the same parallels BF and AD.

In this case, Area of △ ABC = Area of △ DEF

To prove that Triangle on the Same Base and between Same Parallels have the same area, follow these instructions.

Let us take △ ABC and △ ABD on the same base AB and between the same parallels AB and CD.

Draw a parallelogram ABPQ with AB base and lying between the same parallels AB and Q.

The relationship between the area of the triangle and parallelogram says that when they lie on the same base and between the same parallels, then the triangle is half of the parallelogram area.

So, Area of ∆ ABC = 1/2 x Area of parallelogram ABPQ

In the same way, Area of ∆ ABD = 1/2 x Area of parallelogram ABPQ

Therefore, Area of ∆ ABC = Area of ∆ ABD.

Hence proved.

Questions on Triangles on the Same Base and between Same Parallels

Example 1.

AD is the median of ∆ABC and ∆ADC. E is any point on AD. Show that area of ∆ABE = area of ∆ACE.

Solution:

Here, AD is the median for △ ABC.

So, BD = CD

In the above figure, △ ABD and △ ADC are having equal bases BD and CD and they are between the same parallels BC and U.

So, Area of △ ADC = Area of △ ABD —– (i)

Since E lies on AD

Therefore, Ed is the median for △ BCE

Now, BDE and CDE are having the equal base length BD = CD and between the same paralleks BC and m, then

Area of △ CDE = Area of △ BDE —— (ii)

Subtracting equation (i) and equation (ii), we get

Area of △ ADC – Area of △ CDE = Area of ∆ABD – Area of ∆BED

Area of △ ABE = Area of △ ACE

Hence, shown.

Example 2.

Prove that the medians of the triangle divide it into two triangles having equal area?

Solution:

AD is the median of the ∆ BCA and AE is the altitude of ∆ BCA and also ∆ ADC.

And AE is perpendicular to BC

Here, AD is the median of ∆ ABC

Therefore, BD = CD

On multiplying both sides of the equation by AE, we get

BD x AE = CD x AE

Again multiply both sides by 1/2

1/2 x BD x AE = 1/2 x CD x AE

Area of ∆ABD = Area of ∆ADC

Hence shown.

If a triangle and parallelogram having the same base and between the same parallels, then the area of the triangle is equal to half the area of the parallelogram. Get the example problems and solutions on Triangle and Parallelogram on the Same Base and between Same Parallels. Also, find the relationship between them in the following sections explained step by step.

Proof for Area of Triangle Equal to Half of the Parallelogram having the Same Base

In the following figure, parallelogram ABCD and ∆ABD are on the same base AB and between the same parallels AF and DC. Get the area of the triangle ABD and area of the parallelogram ABCD and check their areas.

Area of the parallelogram ABCD = base x height

= AB x CF

Here, CF and DE area parallel lines CF = DE.

Area of the parallelogram = AB x DE = AB x CF.

Area of △ ABD = 1/2 x base x height

= 1/2 x AB x DE

= 1/2 x Area of the parallelogram ABCD

Therefore, Area of the parallelogram ABCD = 1/2 x Area of △ ABD

Hence proved.

Here are various case studies for the area of the parallelogram is half of the area of the triangle when they lie on the same base.

1. When a triangle and a parallelogram lies on the same base and have the same altitude, then the area of the triangle is equal to half of the parallelogram.
2. If they lie between the same parallels and have the same altitude, then the area of the triangle is equal to half of the area of the parallelogram.
3. When the triangle and rectangle have the same base and between the same parallels, then the area of the triangle is half of the area of the rectangle.

Example Questions for Triangle and Parallelogram on Same Base and between Same Parallels

Example 1:

∆ ABD and parallelogram ABCD are on the same base AB. If the base and altitude of the parallelogram are 10 cm and 6 cm, find the area of the triangle.

Solution:

Given that,

Base of the parallelogram = 10 cm

Altitude of the parallelogram = 6 cm

Area of the parallelogram = base x altitude

= 10 x 6 = 60 cm²

We know that when a triangle and parallelogram having the same base and between the same parallels then the area of the triangle is equal to half of the area of the parallelogram.

So, area of △ ABD = 1/2x Area of the parallelogram ABCD

= 1/2 x 60 = 30 cm²

Therefore, triangle area is 30 cm².

Example 2:

∆ ABD and parallelogram ABCD are on the same base AB. If the base and height of the triangle is 16 cm and 12 cm, find the area of the parallelogram.

Solution:

Given that,

The base length of the triangle = 16 cm

Height of the triangle = 12 cm

Area of ∆ ABD = 1/2 x base x height

= 1/2 x 16 x 12

= 1/2 x 192 = 96 cm²

We know that when a triangle and parallelogram having the same base and between the same parallels then the area of the parallelogram is equal to double the area of the parallelogram.

Area of the parallelogram ABCD = 2 x Area of ∆ ABD

= 2 x 96 = 192 cm²

Therefore, parallelogram area is 192 cm².

Example 3:

In the adjacent figure, BCDE is a rectangle and ABC is a triangle show that area of the triangle ABC is equal to half of the area of the rectangle BCDE.

Solution:

Given that,

Triangle ABC and rectangle BCDE lies on the same base BC and between the same parallels BC and ED>

Area of triangle ABC = 1/2 x base x height

= 1/2 x BC x AP

AP and Cd are parallel lines and having the same length. So, AP = CD.

△ ABC area = 1/2 x BC x AP = 1/2 x BC x CD

Area of the rectangle = length x breadth

= BC x CD

= 1/2 x Area of △ ABC.

Therefore, the Area of the rectangle BCDE = 1/2 x Area of △ ABC

Hence, shown.

When one parallelogram and a rectangle lies on the same base and between the same parallels then they measure equal area. Here we are providing the example questions, answers for Parallelograms, and Rectangles on Same Base and between the Same Parallels. We are also offering proof that shows parallelogram and rectangle on same between same parallels have the same area.

Proof for Rectangles and Parallelograms on the Same Base have Equal Area

Below-mentioned is the steps for the parallelogram and rectangles having the same base and between the same parallels having equal area.

In the following figure, ABCD is a parallelogram and ABEF is the rectangle that lies on the same base AB and the same parallels AB and DF.

Area of parallelogram ABCD = base x altitude

AF is the altitude of the parallelogram.

So, Parallelogram area = AB x AF.

Area of the rectangle ABEF = base x height

= AB x AF

Therefore, the area of the parallelogram ABCD = Area of the rectangle ABEF.

Solved Examples for Parallelograms and Rectangles on Same Base and between Same Parallels

Example 1:

Parallelogram ABCD and rectangle ABFE have the same AB and the length and breadth of the rectangle are 15 cm and 9 cm. Find the area of the parallelogram.

Solution:

Given that,

Length of the rectangle l = 15 cm

The breadth of the rectangle b = 9 cm

Parallelogram ABCD and rectangle ABFE have the same AB.

So, area of the parallelogram = Area of the rectangle.

Area of the rectangle = length x breadth

= 15 x 9 = 135 cm²

Therefore, area of the parallelogram is 135 cm².

Example 2:

In the adjacent figure, ABCD is a parallelogram and EFCD is a rectangle. Also, AL ⊥ DC. Prove that

(a) Area (ABCD) = Area (EFCD)

(b) Area (ABCD) = DC × AL

Solution:

(a)

Rectangle is CDEF.

Parallelogram is ABCD.

In the given figure, AL and DE are parallel and have the same length. So, DE = AL.

Area of ABCD = CD x AL

= CD x DE square units.

Area of EFCD = CD x DE

So, Area of EFCD = Area of ABCD

Hence proved.

(b)

As ABCD is a parallelogram.

Area of ABCD = base x altitude

The base is CD and the altitude is AL.

Therefore, Area of ABCD = CD x AL

Hence proved.

Example 3:

Parallelogram ABCD and rectangle ABFE have the same AB and the base and height of the parallelogram are 28 cm and 16 cm. Find the area of the rectangle.

Solution:

Given that,

Base of the parallelogram = 28 cm

Height of the parallelogram = 16 cm

Area of the parallelogram = base x height

= 28 x 16 = 448 cm²

We know, that when Parallelograms and Rectangles on Same Base and between Same Parallels, then the area of the parallelogram = area of the rectangle.

So, the area of the rectangle = 448 cm².

Students can check how parallelograms having the same base and between the same parallels are equal in area and proof for that. When two or more parallelogram lies on the same base and between same parallels then they have the same area. Also, find the solved example questions here explaining the concept and learn the concept behind them.

Proof for Theorem: Parallelogram on the Same Base & Between Same Parallels are Equal in Area

Let us draw a parallelogram ABCD on a cardboard sheet or thick sheet of paper.

Now, draw a line segment DE as shown in the figure.

Now, cut the triangle ADE congruent to triangle ADE in a seperate sheet with the help of a tracing paper and place the triangle ADE in such a way that AD coincides BC. The entire process is as shown below.

Note that there are two parallelograms ABCD and EE’CD on the same base CD and between the same parallel AE’ and CD.

∆ADE ≅ ∆ A’ D’ E’

Therefore, Area of ADE = Area of A’D’E’

Also, Area of ABCD = Area of ADE + Area of EBCE

= Area of A’D’E’ + Area of EBCD

= Area of EE’CD

So, the two parallelograms are in an equal area.

Parallelograms on Same Base and Between Same Parallels Examples

Example 1:

A farmer was having a field in the form of a parallelogram ABCD. She took any point P on AB and joined it to points C and D. In how many parts the fields are divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?

Solution:

If DP and CP are joined parallelogram ABCD is divided into three triangles.

They are △ APD, △ CDP, △ BCP

△ CDP and parallelogram ABCD are on the same base i.e CD and between parallel sides CD ∥ AC

∴ Area of △ CDP = 1/2 x area of parallelogram ABCD —- (i)

∴ Area of △ APD + Area of △ BCP = 1/2 x area of parallelogram ABCD —- (ii)

From equation (i) and equation (ii)

Area of △ CDP = Area of △ APD + Area of △ BCP

∴ The farmer can use the part of the field to sow wheat, i.e. Area of △ APD + Area of △ BCP, and in the same area he can use the Area of △ CDP to sow pulses.

Example 2:

Parallelograms ABCD and ABEF are situated on the opposite sides of AB in such a way that D, A, F are not collinear. Prove that DCEF is a parallelogram, and parallelogram ABCD + parallelogram ABEF = parallelogram DCEF.

Solution:

AB and DC are two opposite sides of parallelogram ABCD,

Therefore, AB ∥ DC and AB = DC

Again, AB and EF are two opposite sides of parallelogram ABEF

Therefore, AB ∥ EF and AB ∥ EF

Therefore, DC ∥ EF and DC = EF

Therefore, DCEF is a parallelogram.

Therefore, ∆ADF and ∆BCE, we get

AD = BC (opposite sides of parallelogram ABCD)

AF = BE (opposite sides of parallelogram ABEF)

And DF = CE (opposite sides of parallelogram CDEF)

Therefore, ∆ADF ≅ ∆BCE (side – side – side)

Therefore, ∆ADF = ∆BCE

Therefore, polygon AFECD – ∆BCE = polygon AFCED – ∆ADF

Parallelogram ABCD + Parallelogram ABEF = Parallelogram DCEF

Two geometric shapes are said to be on the same base and between the same parallels, if they have a common side called the base and vertices opposite to the common base on the line parallel to the base. In the below sections, you can learn about the How to Identify figure on the same base and between the same parallels. We know the measure of the plane region enclosed by a closed figure called the area.

The area is measured in cm², m², and other square units. We also know how to calculate the area of different figures using various formulas. Here you will use the formulas to study the relationship between the area of figures when they lie on the same base and between the same parallels.

Some Geometric Figures on the Same Base and between Same Parallels

Below-mentioned is the geometric figures for which you can change the same base and between the same parallels.

The common base for triangles ABC, BCD is BC. And both triangles lie on the same base.

Parallelograms ABCD and triangle CDE lie on the same base CD.

Parallelograms ABCD and EFCD are on the same base DC.

Trapezium ABCD and parallelogram EFCD have a common side DC. We say that trapezium ABCD and parallelogram EFCD are on the same base DC.

Example Questions on Figures that Lie on the Same Base and Between Same Parallels

Example 1.

ABC is right-angled at B with BC = 10 cm and AC = 15 cm. also ∆ABC and ∆BCD are on the same base BC. Find the area of ∆BCD.

Solution:

In right angled ∆ ABC, AC = 15 cm and BC = 10 cm. using Pythagoras theorem, we get

AC² = AB² + BC²

15² = x² + 10²

x² = 15² – 10²

x² = 225 – 100

x² = 125

x = √125

x = 11.18 cm

Now, since ∆ ABC and ∆BCD are on the same base BC.

Therefore, area of ∆ ABC = Area of ∆BCD

1/2 × base × height = Area of ∆BCD

1/2 × 10 × 11.18 = Area of ∆BCD

Area of ∆BCD = 5 x 11.18 = 55.9 cm²

Therefore, area of ∆BCD is 55.9 cm².

Example 2.

AD is the median of ∆ABC. E is any point on AD. Show that area of ∆ABE = area of ∆ACE

Solution:

The common base for ∆ABE, ∆ACE is AE.

ABC is a triangle with AD as the median.

i.e BD = CD

E is any point on AD.

In ∆ABC,

Given that,

D is the midpoint of BC.

∴ AD is median

Since median divides the triangle into two triangles of equal area.

So, Area of ∆ABD = Area of ∆ABC —– (i)

In ∆EBC

Given that, D is the midpoint of BC.

∴ ED is median.

Since median divides the triangle into two triangles of equal area.

So, Area of ∆EBD = Area of ∆EDC —— (ii)

Subtracting equation (1) from equation (2)

Area of ∆ABD – Area of ∆EBD = Area of ∆ADC – Area of ∆EDC

Area of ∆ABE = Area of ∆ACE

Hence, proved.

Example 3.

Parallelogram PQRS and PQTU are on the same base PQ and between the same parallels PQ and UR. Area of parallelogram PQRS = 56 cm² and the altitude of the parallelogram PQTU = 7 cm. Find the length of the common side of two parallelograms.

Solution:

Given that,

Area of parallelogram PQRS = 56 cm²

The altitude of the parallelogram PQTU = 7 cm

Area of parallelogram PQRS = base x height

56 = PQ x 7

PQ = 56/7

PQ = 8 cm

Therefore, the length of the common side of the two parallelograms is 8 cm.