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}
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}
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}
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}
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}
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}
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}
The formula of area of rectangle is ______ .
The formula of area of triangle is ______ .
The formula of area of parallelograms is ______.
If the perimeter of a circular ground is 220 m, then find its radius.
Find the area of triangle in which base = 8 cm, height = 5.5 cm.
The perimeter of a square ground is 128 m, find its area.
Find the area of a square field is 625 m^{2}, find its perimeter.
If the perimeter of a circle is 44cm, find its area.
A square room is 15 m long, what will be its area of carpet required to cover its floor.
A rectangular field is 40 m long and its area is 1120 m^{2}, find its breadth.
ASK ANY QUESTION ON Area of Plane Figures
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Ask any queries on this note.enlargement construction of plane figure
Jan 25, 2017
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