Please scroll down to get to the study materials.
Triangle, Rectangle, Square, Circle etc are the plane figures. The total length of the boundary lines of a plane figure is called its perimeter.
We can divide the square into small squares of 1 cm side length to find the area of a square by the method of counting squares.
Consider a square that has a side length of 4 cm using the method of counting squares, we find that the area of the square = 16 cm^{2}
Clearly, the square contains 4 rows of 4 squares. Therefore, Area = 4cm x 4cm = 16cm^{2}
This suggests that:
The area of a square is equal to its side-length multiplied by its side-length. That is
Area = length x length
= (length)^{2}
Using A for area and l for length, we can write it as:
A = l^{2}
\(\therefore\) Area of square = l^{2}
Also,Perimeter of a square = a + a + a + a
= 4a
To find the area of a rectangle by the method of counting squares, we divide the rectangle into small squares of 1 cm side length.
Consider a rectangle of length 5 cm and width 3 cm
Using the method of counting squares, we find that the area of the rectangle is 15cm^{2}.
Clearly, the rectangle contains 3 rows of 5 squares.
Therefore, area = 5cm x 3cm = 15 cm^{2}
This suggests that:
The area of a rectangle is equal to its length multiplied by its width. That is,
Area = Length x Width
Using A for area, l for length and b for width, we can write it simply as:
A = l x b
\(\therefore\) Area of rectangle = l x b
Also, perimeter of a rectangle = 2(l+b)
Let, ABCD be the parallelogram. Let DE? AB and BN? DC
Here, AB// DC, DE = BN
Area of Parallelogram = area of \(\triangle\) DAB + area of\(\triangle\)BCD
= \(\frac{1}{2}\) AB X DE + \(\frac{1}{2}\) DC x BN
=\(\frac{1}{2}\) AB X DE + \(\frac{1}{2}\) AB X DE ( AB = DC, DE =BN )
Thus, area of a parallelogram = base x height
We know that a trapezium is a quadrilateral whose one pair of opposite sides is parallel. If two non-parallel sides of a trapezium, it is called an isosceles trapezium.
Let, ABCD be a trapezium having parallel sides AB and DC. Draw DF? AB and CE? AB. Let DF = CE = h. Then area of the trapezium ABCD = area of \(\triangle\)AFD + area of rectangle FECD + area of \(\triangle\)EBC
=\(\frac{1}{2}\)AF x DF + FE x DF + \(\frac{1}{2}\) EB xCE
=\(\frac{1}{2}\)AF x h + FE x h+ \(\frac{1}{2}\) EB xh
=\(\frac{1}{2}\) h(AF + 2FE + EB)
= \(\frac{1}{2}\)h(AF + FE + EB + FE) [AF+FE+EB = AB and FE = DC]
= \(\frac{1}{2}\)h (AB+DC)
Thus, area of a trapezium = \(\frac{1}{2}\) x (sum of parallel sides) x distance between them.
We know that the rhombus is a parallelogram having sides equal. We also know that the diagonals of a rhombus bisect each other at right angles.
Consider a rhombus ABCD whose diagonals AC and BD bisect each other at right angles at a point O.
Let, AC =d_{1} and BD =d_{2}
Then, AO =\(\frac{1}{2}\)d_{1} and BO =\(\frac{1}{2}\)d_{2}
Area of \(\triangle\)AOB = \(\frac{1}{2}\) AO x BO = \(\frac{1}{2}\) x \(\frac{1}{2}\)d_{1}x \(\frac{1}{2}\)d_{2} = \(\frac{1}{8}\)d_{1}d_{2}
Since diagonals of a rhombus divide it in to four congruent right angled triangles,
Area of rhombus = 4 x area of \(\triangle\) AOB
=4 x \(\frac{1}{8}\)d_{1}d_{2}
=\(\frac{1}{2}\)d_{1}d_{2}
Thus, area of rhombus ABCD = \(\frac{1}{2}\) xproduct of diagonals.
Note: Since square is also a rhombus having equal diagonals, area of a square = \(\frac{1}{2}\)d^{2}
A triangle is a polygon with three vertices, and three sides or edges that are line segments. To find the area of a triangle by the method of counting squares, firstly we divide a rectangle into small squares of 1 cm side length. Secondly, we draw the largest triangle to divide the rectangle into three parts as shown below:
Finally, we estimate the area of a triangle by counting the squares. Area of rectangle = 7x 5 = 35cm^{2}
Area of Triangle = 17.5cm^{2}
This shows that area of triangle =\(\frac{1}{2}\) base x height =\(\frac{1}{2}\)bh
Also, If a, b and c denote three sides of a \(\triangle\)ABC, Perimeter of \(\triangle\)ABC = AB + BC + CA
= c + a + b
= a + b + c
Solution:
Given, side(a) = 20m
Perimeter of square flower-bed = 4a
=4 x 20
=80
\(\therefore\) perimeter of square flower-bed = 80 m
Solution:
Area of the trapezium = \(\frac{1}{2}\) x (sum of parallel sides)xh
= \(\frac{1}{2}\)x(14+26)x10
= \(\frac{1}{2}\)x40x10 = 200cm^{2}
Solution:
Area of the trapezium = \(\frac{1} {2}\) x (sum of parallel sides) x h
= \(\frac{1}{2}\) x (20+30) x 10
=\(\frac{1}{2}\) x 50 x 10
= 250cm^{2}
Solution:
Area of the rhombus = \(\frac{1}{2}\) xd_{1}d_{2}
= \(\frac{1}{2}\) x 24 x 16.5
=198cm^{2}
Solution:
Area of the rhombus = \(\frac{1}{2}\) x d_{1}d_{2}
= \(\frac{1}{2}\) x 12cm x 14.5cm
= 87 cm^{2}
Soution:
The circumference of the circle is given by
c =2\(\pi\)r
=2\(\times\)\(\frac{22}{7}\)\(\times\)10.5
=66 cm
Thus, circumference=66 cm
Solution:
We know that, c=2\(\pi\)r
or, 88=2\(\times\)\(\frac{22}{7}\)\(\times\)r
or, r=\(\frac{88\times7}{2\times22}\)\(\times\)r
or, r=\(\frac{88 x 7 }{2 x 22}\)
\(\therefore\) r=14 cm
Thus, radius=14 cm
Solution:
Circumference= 44 cm
So, 2\(\pi\)r=44
or, r=\(\frac{44}{2\pi}\)
or, r=\(\frac{44\times7}{2\times22}\)
\(\therefore\) r=7 cm
Area of the circle=\(\pi\)r^{2}
=\(\frac{22}{7}\)\(\times\)7 \(\times\)7
=154cm^{2}
Solution:
Area of a circle=154 cm^{2}
So, \(\pi\)r^{2}=154
or, r^{2}=\(\frac{154}{\pi}\)
or, r^{2}=\(\frac{154\times7}{22}\)
or, r^{2}=49
or, r=\(\sqrt{49}\)
\(\therefore\) r =7cm
Solution:
Here, base=14 cm and height= 16.5 cm
Area of the parellogram= base\(\times\)height
=14 cm\(\times\)16.5 cm
=231 cm^{2}
Solution:
Given, AB=6cm, BC=5cm and CA=7cm
Perimeter of \(\triangle\)ABC = AB + BC + CA
= 6cm + 5cm + 7cm
= 18cm
Solution:
Given: length = 17 cm, breadth = 13 cm
Perimeter of rectangle = 2 (length + breadth)
= 2 (17 + 13)cm
= 2 × 30cm
= 60cm
Also,
Area of rectangle = length × breadth
= 17cm × 13cm
= 221cm^{2}
Solution:
Given, Area of Rectangular plot = 660m^{2} and Length = 33m
Now, We know,
Area of rectangle = Length \(\times\) breadth
or, 660m^{2} = 33m \(\times\) breadth
or, breadth = \(\frac{660m^2}{33m}\)
\(\therefore\) breadth = 20m
Again,
Perimeter of the rectangular plot = 2 (length + breadth)
= 2 (33m + 20m)
= 2 × 53m
= 106m
Solution:
Given, Perimeter of rectangle = 48cm and Breadth = 6cm
We know, Perimeter of rectangle = 2(length + breadth)
So, 48cm = 2(length + 6cm)
or, \(\frac{48cm}{2} = length + 6cm
or, 24cm = length + 6cm
or, length = 24cm - 6cm
\(\therefore\) length = 18cm
Now,
Area of rectangle = length × breadth
= 18cm × 6cm
= 108cm^{2}
Solution:
Perimeter of rectangle = 2(length + breadth)
= 2(25cm + 17cm)
= 2 × 42cm
= 84cm
Let the side of square be x.
Then,
Perimeter of square = 4x
We know,
Perimeter of rectangle = Perimeter of Square
So, 84cm = 4x
or, x = \(\frac{84cm}{4}\)
\(\therefore\) x = 21cm
Therefore, sides of square = 21cm.
Solution:
Circumference of circle = 2πr
= 2 × 22/7 × 7
= 44 cm
Area of circle = πr^{2}
= 22/7 × 7 × 7 cm^{2}
= 154 cm^{2}
Find the area of a rectangular field 20m long and 10m wide.
Find the perimeter of a field 10m long and 12m wide.
Find the area of the Triangle, whose base is 10cm and height is 6cm.
The Perimeter of a Square is 36 cm, find its area.
The area of a square is 121cm^{2}, find its Perimeter.
Find the area and perimeter of a rectangle whose, length is 25.5cm and breadth is 14.6 cm.
The perimeter of a rectangle is 46m and its length is 15m. Find its breadth.
The perimeter of a rectangle is 46m and its length is 15m. Find its breadth.
The area of a rectangle is 8.4m^{2} and its length is 3m. Find its perimeter.
Find the area and perimeter of a rectangle whose length is 20cm and breadth is 12cm.
Find the area of a triangle whose, base is 24cm and altitude is 12cm.
Find the area of a trapezium whose parallel sides are 14cm and 26cm and the distance between them is 10cm.
Find the area of a rhombus, the length of whose diagonals are 24cm and 16.5cm.
Find the area of a right-angled triangle whose base is 8cm and hypotenuse is 10cm.
Find the area of aTriangle whose base is 1.5m and altitude is 0.8m.
ASK ANY QUESTION ON Area and Perimeter of Plain Figures
You must login to reply
Formula
Formula
Mar 25, 2017
0 Replies
Successfully Posted ...
Please Wait...
a field having length 120ft and width 90ft has squared pond with edge 20ft find the area of field excluding the pond
Jan 19, 2017
1 Replies
Successfully Posted ...
Please Wait...
area of shaded portion
How to solve
Jan 06, 2017
2 Replies
Successfully Posted ...
Please Wait...