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.
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