Subject: Compulsory Maths

A part of the plane enclosed by a simple closed figure is called a plane region and the measurement of it, is called its area. The area is measured in square units. The length of the boundary of a closed figure is called the perimeter. The units of the perimeter are same as that of length, i.e., m, cm, mm, etc.

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

- Area of a square= length x length = (length)
^{2 }=*l*^{2} - Perimeter of a square = 4a
- Area of rectangle = l x b
- Perimeter of a rectangle = 2(l+b).
- Area of a parallelogram = base x height
- Area of a rhombus = \(\frac{1}{2}\) x d1 x d2.
- Area of a trapezium = \(\frac{1}{2}\) x (sum of parallel sides) x h
- Area of the circle = \(\frac{1}{2}\) circumference

- It includes every relationship which established among the people.
- There can be more than one community in a society. Community smaller than society.
- It is a network of social relationships which cannot see or touched.
- common interests and common objectives are not necessary for society.

Find the perimeter of a square flower-bed of side 20 m.

Solution:

Given, side(a) = 20m

Perimeter of square flower-bed = 4a

=4 x 20

=80

\(\therefore\) perimeter of square flower-bed = 80 m

Find the area and perimeter of a square room 35m.

Solution:

Given, side(a) = 35m

Now,

Area of square= a^{2}

= (35m)^{2}

= 1225m^{2}

\(\therefore\) area of square = 1225

Also,

Primeter of square = 4a

= 4 x 35m

= 140m

\(\therefore\) Perimeter of square = 140m

Find the area and perimeter of a rectangle field 12m long and 20 m wide.

Solution:

Given, length (l) =12 m and wide (b)=20 m

Area of rectangle field = l x b

=12m x 20m

= 240m^{2}

Perimeter of rectangle field = 2 (l+b)

= 2 (12m + 20m)

= 2 x 32m

= 64m

Find the area and perimeter of a rectanguler room 25m long and 30m wide.

Solution:

Given, length (l) = 25m and wide (b) = 30m

Area of a rectanguler room = lxb

= 25m x 30m

= 750m^{2}

Perimeter of a rectanguler room = 2(l+b)

= 2(25m+30m)

= 2 x 55m

= 110m

One side of a parallelogram is 14 cm. Its distance from the opposite side is 16.5 cm. Find the area of the parallelogram.

Solution:

Here, base = 14 cm and height = 16.5 cm

Area of the parallelogram = base x height

= 14 cm x 16.5cm

= 231cm^{2}

Find the area of the parallelogram having distance from the opposite side is 12.5 cm and one side of a parallelogram is 10 cm.

Solution:

Here, base = 10cm and height = 12.5cm

Area of a parallelogram = base x height

= 10cm x 12.5cm

=120cm^{2}

Find the area of a trapezium whose parallel sides are 14 cm and 26 cm and the distance between them is 10 cm.

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}

Find the area of a trapezium whose parallel sides are 10 cm and 30 cm and the distance between them is 10 cm.

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}

Find the area of a rhombus, the length of whose diagonals are 24cm and 16.5cm.

Solution:

Area of the rhombus = \(\frac{1}{2}\) xd_{1}d_{2}

= \(\frac{1}{2}\) x 24 x 16.5

=198cm^{2}

Find the area of a rhombus, whose length are 12cm and 14.5cm.

Solution:

Area of the rhombus = \(\frac{1}{2}\) x d_{1}d_{2}

= \(\frac{1}{2}\) x 12cm x 14.5cm

= 87 cm^{2}

Find the circumference of a circle of radius 10.5 cm.

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

Find the diameter of a circle whose circumference is 88 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

The diameter of a wheel of a van is 63 cm. Find the distancetravelled by the van during the period, the wheel makes revolutions.

Solutions:

Note that, in 1 revolution the car covers a distance equal to the circumference of the wheel.

Now, the diameter of the wheel=63 cm

Therefore, radius(r)=\(\frac{63}{2}\)cm

Circumference of the wheel= 2\(\pi\)r

=2\(\times\)\(\frac{22}{7}\)\(\times\)\(\frac{63}{2}\)

=198 cm

=1.98 m

Here, the distance covered in 1 revolution=1.98 m

Distance covered in 1000 revolutions=1.98\(\times\)1000

=1980 m

The circumference of a circle is 44 cm.Find its area.

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}

The area of a circle is 164 cm^{2}. Find its circumference.

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

One side of a parallelogram is 14 cm.Its distance from the opposite side is 16.5 cm.Find the area of the parallelogram.

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}

Find the area of the given triangle.

Solution:

Base(b)=10 cm, height(h)=6 cm

Area of \(\triangle\)PQR=\(\frac{1}{2}\)b\(\times\)h

=\(\frac{1}{2}\)QR\(\times\)PQ

=\(\frac{1}{2}\)10cm\(\times\)6cm

=30cm^{2}

Therefore, area of \(\triangle\)PQR=30cm^{2}

Find the perimeter of the given triangle.

Solution:

Given, AB=6cm, BC=5cm and CA=7cm

Perimeter of \(\triangle\)ABC = AB + BC + CA

= 6cm + 5cm + 7cm

= 18cm

Find the area of the following diagram

Solution:

Length of the rectangle(l)=5cm

Breadth of the rectangle(b)=3.6cm

Area of the rectangle=?

By using formula,

A= l\(\times\)b

=5cm\(\times\)3.6cm

=18.0cm^{2}

Find the perimeter and area of the rectangle of length 17 cm and breadth 13 cm.

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}

Find the breadth of the rectangular plot of land whose area is 660m^{2} and whose length is 33m. Find its perimeter.

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

Find the area of the rectangle if its perimeter is 48cm and its breadth is 6cm.

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}

A wire in the shape of rectangle of length 25 cm and breadth 17 cm is rebent to form a square. What will be the measure of each side?

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.

Find the circumference and area of radius 7 cm.

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}

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