Subject: Compulsory Maths

The plane surface enclosed by the boundary line of a plane closed figure is known as its area . The area is always measured in a square unit. For example,cm^{2} , mm^{2},m^{2}etc.

**Area of rectangles :**

In the adjoining graph , the area of each square room is 1 cm^{2}. So , the surface enclosed by the rectangle is 10 cm^{2.}

\(\therefore\) Area of rectangle = 10 cm^{2}

i.e . 5 rooms along length x 2 rooms breadth = 10cm^{2}

\(\therefore\) Area of rectangle = length x breadth = l x b

**Area of triangles**

Here , area of the triangle EFG= length x breadth

=FG xGH

=base x height

=b x\(\frac{1}{2}\)h

\(\frac{1}{2}\)bh

So the area of triangle =\(\frac{1}{2}\)base x height = \(\frac{1}{2}\)b xh

**Area of trianglesgle **

Here , area of triangle = area of rectangle FGHI

= length x breadth

=FG XGH

=base x hieight

=b x\(\frac{1}{2}\)

=\(\frac{1}{2}\)bh

So , the area of tringle =\(\frac{1}{2}\)base x height = \(\frac{1}{2}\) b xh

**Area of Parallelogrms**

Here , area of the parallelogram EFGH =Area of the rectangle of EFGH

= length x breadth

=EF x FI

= base x height

= b x h

So, the area of parallelogram = base x height

= b x h

**Area of circles :**

The length of rectangle EFGH (l) =\(\frac{1}{2}\) x circumference = \(\frac{1}{2}\) x2 πr

The breadth of rectangle EFGH (b) = r

area of the circle = Area of rectangle EFGH

= length x breadth

=πr x r

=πr^{2}

So , the area of circle =**πr ^{2}**

- Area of rectangle = length x breadth = l x b.
- The area of parallelogram = base x height.
- The area of circle =
**πr**^{2}

- 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 area of the given figure.

Solution:

Here,

area of two rectangles = l × b

= 20 cm × 15 cm

= 300 cm^{2}

Also,

area of two triangles = 2 (\(\frac{1}{2}\) ×b × h)

= 2 × \(\frac{1}{2}\) × 15 × 10 cm^{2}= 150 cm^{2}

Again,

∴ Area of the figure = 300 cm^{2} + 150 cm^{2}

= 450 cm^{2}

Find the area of the given figure.

Solution:

Area of the rectangle = l × b

= 20 cm × 15 cm

= 300 cm^{2}

Radius of each semi-circle = \(\frac{14}{2}\)

= 7 cm

Area of two semi-circles = 2(\(\frac{1}{2}\)πr^{2})

= \(\frac{22}{7}\) × 7 × 7 cm^{2}

= 154 cm^{2}

∴ Area of the figure = 350 cm^{2} + 154 cm^{2 }= 504 cm^{2}

Find the area of the shaded region in the given figure.

Solution:

Here,

Area of parallelogram = b × h

= 16 cm × 10cm

= 160 cm^{2}

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

= \(\frac{1}{2}\) × 16 cm × 10 cm

= 80 cm^{2}

∴ Area of the shaded region = Area of parallelogram -Area of triangle

= 160 cm^{2} - 80 cm^{2}

= 80 cm^{2}

Find the area of the shaded regions in the given figures.

Solution:

Area of rectangle = l × b

= 12 cm × 9 cm

= 108 cm^{2}

Area of parallelogram = b × h

= 6 cm × 4 cm

= 24 cm^{2}

∴ Area of the shaded region = Area of rectangle - Area of parallelogram

= 108 cm^{2} - 24 cm^{2}

= 84 cm^{2}

Find the area of shaded regions in the given figure.

Solution:

Here,

Area of bigger rectangle = l × b

= 20 cm × 16 cm

= 320 cm^{2}

Area of smaller rectangle = l × b

= 14 cm × 10 cm

= 140cm^{2}

∴ Area of the shaded region = Area of bigger recatngle - Area of smaller rectangle

= 320 cm2 - 140 cm^{2}

= 180 cm^{2}

The perimeter of a square field is 96 m. Find its length.

Solution:

The perimeter of the square field = 96 m

or, 4 l = 96 m

or, l = \(\frac{96}{4}\)

or, l = 24 m

∴ The length of the field (l) = 24m

A rectangular room is twice as long as its breadth and its perimeter is 48 m. Find its length and breadth and area.

Solution:

Let, the breadth (b) pf the room be x m.

So, the length (l) of the room will be 2x m.

Now,

The perimeter of the rectangular room = 48 m

or, 2 (l + b) = 48 m

or, 2 (2x + x) = 48 m

or, 6x = 48 m

or, x = \(\frac{48}{6}\) m

or, x = 8 m

∴ The breadth of the room (b) = x = 8 m

The length of the room (l) = 2x

= 2 × 8m

= 16 m

Again.

Area of the room = l × b

= 16 m × 8 m

= 128 m^{2}^{}^{∴ Thw area of a room is 128 m2}

If the perimeter of a circular ground is 220 m, find its radius and area.

Solution:

Here,

The perimeter of circular ground = 220 m

or, 2πr = 220 m

or, 2 × \(\frac{22}{7}\) × r = 220 m

or, r = \(\frac{220 × 7}{2 × 22}\) m

or, r = 35 m

Now,

Area of the circular ground = πr^{2}

= \(\frac{22}{7}\) × 35 m × 35 m

= 3850 m^{2}

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