Notes on Area of Plane Figures | Grade 7 > Compulsory Maths > Perimeter, Area and Volume | KULLABS.COM

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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,cm2 , mm2,m2etc.

Area of rectangles :

In the adjoining graph , the area of each square room is 1 cm2. So , the surface enclosed by the rectangle is 10 cm2.

$$\therefore$$ Area of rectangle = 10 cm2

i.e . 5 rooms along length x 2 rooms breadth = 10cm2

$$\therefore$$ 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

=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

=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

=πr x r

=πr2

So , the area of circle =πr2

• Area of rectangle = length x breadth = l x b.
• The area of parallelogram = base x height.
• The area of circle =πr2
.

### Very Short Questions

Solution:

Here,
area of two rectangles = l × b
= 20 cm × 15 cm
= 300 cm2
Also,
area of two triangles = 2 ($$\frac{1}{2}$$ ×b × h)
= 2 × $$\frac{1}{2}$$ × 15 × 10 cm2
= 150 cm2
Again,
∴ Area of the figure = 300 cm2 + 150 cm2
= 450 cm2

Solution:

Area of the rectangle = l × b
= 20 cm × 15 cm
= 300 cm2
Radius of each semi-circle = $$\frac{14}{2}$$
= 7 cm
Area of two semi-circles = 2($$\frac{1}{2}$$πr2)
= $$\frac{22}{7}$$ × 7 × 7 cm2

= 154 cm2
∴ Area of the figure = 350 cm2 + 154 cm
= 504 cm2

Solution:

Here,
Area of parallelogram = b × h
= 16 cm × 10cm
= 160 cm2
Area of triangle = $$\frac{1}{2}$$ b × h
= $$\frac{1}{2}$$ × 16 cm × 10 cm
= 80 cm2
∴ Area of the shaded region = Area of parallelogram -Area of triangle
= 160 cm2 - 80 cm2
= 80 cm2

Solution:

Area of rectangle = l × b
= 12 cm × 9 cm
= 108 cm2
Area of parallelogram = b × h
= 6 cm × 4 cm
= 24 cm2
∴ Area of the shaded region = Area of rectangle - Area of parallelogram
= 108 cm2 - 24 cm2
= 84 cm2

Solution:

Here,
Area of bigger rectangle = l × b
= 20 cm × 16 cm
= 320 cm2
Area of smaller rectangle = l × b
= 14 cm × 10 cm
= 140cm2
∴ Area of the shaded region = Area of bigger recatngle - Area of smaller rectangle
= 320 cm2 - 140 cm2
= 180 cm2

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 = πr2
= $$\frac{22}{7}$$ × 35 m × 35 m
= 3850 m2

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l × b
l + b
l ÷ b
l - b
• ### The formula of area of triangle is ______ .

(frac{1}{2}) b × h
b × h × l
b × h
l × b

l × h
l × b × h
b × h
l × b

36 m
37 m
35 m
34 m

22 cm2
24 cm2
21 cm2
23 cm2

1036 m2
1024 m2
1012 m2
1042 m2

100m
70m
80m
90m

123 cm2
154 cm2
147 cm2
138 cm2

225 m2
235 m2
240 m2
210 m2

22 m
28 m
20 m
24 m