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Maths / Mensuration / Area of Shaded Region
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Area of Shaded Region

Area of shaded region

We will calculate the area of different figures separately. Then we will calculate the area collectively. We come across these types of figures in our daily life and also in the form of various interesting designs. Flower beds, drains covers, window designs, designs on table covers,are some of such examples. We illustrate the process of calculating areas of figures through some illustration.

Illustration: Find the area of the shaded region in figure.

Solution  

  Area of semi circle is half of the area of a circle

   Area of shaded region = area of semicircle with radius 14 cm  + 2 [ area of semicircle with radius 7 cm]

              = frac{1}{2}pi (14)^{2}+2[frac{1}{2}pi (7)^{2}]

              = pi (14)(7)+pi (7)^{2}

              = pi (7)(14+7)

              = pi (7)(21)

              = frac{22}{7}times 7times 21 cm^{2}

              = 22times 21 cm^{2} = 462;cm^2

Illustration: Find the area of the shaded region

Solution:ABCD is a square with side 14 cm.

Area of the square =14 times 14

                           =196;cm^2

As there are two circles in one row, The diameter of two circles= 14 cm

Diameter of one circle = 7 cm

Radius; of; each; circle=frac {7}{2}cm

Area ;of; one; circle=pi r^2=frac {22}{7} times frac {7}{2} times frac {7}{2}cm^2

                                           =frac {154}{4}cm^2

Area; of; four; circles=4 times frac {154}{4}cm^2= 154 cm^2

Hence, the area of the shaded region in the given figure

                            = area of square - area of four circles

                           = 196 - 154 = 42 sq cm                      

 

Illustration:  In the figure, ABC is a quadrant of a circle of radius 7 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.

Area ;of ;quadrant =frac{1}{4}times frac{22}{7}times 7times 7= frac{77}{2}= 38.5; cm^2

Area ;of ;triangle = frac{1}{2}times Basetimes height=frac{1}{2}times 7times 7

                                              = frac{49}{2}= 24.5; cm^2

Using pythagoras theorem,

BC^2=AB^2+BC^2 = 7^2+ 7^2= 49 +49= 98

BC= 7sqrt2cm

BC is the diameter:, Therefore Radius = 3.5sqrt2cm

Area of semi circle with BC as diameter

=frac{1}{2}times frac{22}{7}times 3.5sqrt2times 3.5sqrt2=38.5 ;cm^2

Area of shaded region = area of triangle + area of semi-circle - area of quadrant

                                    = 24.5+ 38.5  -38.5 = 24.5 sq cm.

 

 

 


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