Question

Question no 31 plz tell fast

Answer 1

It is given that the triangle ABC is a right angled triangle.
And, length of two sides ( which are not hypotenuse ) are 3 cm ( AB ) and 4 cm ( AC )



By Pythagoras theorem :

= > BC^2 = AB^2 + AC^2

= > BC^2 = ( 3 cm )^2 + ( 4 cm )^2

= > BC^2 = 9 cm^2 + 16 cm^2

= > BC^2 = 25 cm^2

= > BC = 5 cm [ ignoring the negative value, since side can't be negative ]


Thus,
Length of BC is 5 cm.



On observing the diagram, we get :

BC is a diameter of a semi circle which have A , B and C on its circumference.
Name this semi circle as C1.

So, radius of C1 = Diameter / 2 = BC / 2 = 5 cm / 2 = 2.5 cm


Then,
= > Area of C1 = 1 / 2 πr^2
= > Area of C1 = 1 / 2 x 22 / 7 x ( 2.5 cm )^2
= > Area of C1 = 137.5 / 14 cm^2


Also,
= > Area of the triangle = 1 / 2 x base x height
= > Area of triangle = 1 / 2 x 4 cm x 3 cm
= > Area of triangle = 6 cm^2


= > Area of C1 which is not in triangle = Area of C1 - Area of triangle
= > Area of C1 which is not in triangle = 137.5 / 14 - 6 cm^2 = ( 137.5 - 84 cm^2 ) / 14 = 53.5 / 14 cm^2.



Now,
As we can see, AC is also a diameter of an another circle.
Name this circle as C2 .

So, radius of C2 = AC / 2 = 4 cm / 2 = 2cm.


Then,
= > Area of C2 = 1 / 2 πr^2
= > Area of C2 = 1 / 2 x 22 / 7 x ( 2 cm )^2
= > Area of C2 = 88 / 14 cm ^2


And,
AB is also a diameter of a circle.
Name this circle as C3.

So, radius of C3 = AB / 2 = 3 cm / 2 = 1.5 cm

= > Area of C3 = 1 / 2 πr^2
= > Area of C3 = 1 / 2 x 22 / 7 x ( 1.5 cm )^2
= > Area of C3 = 49.5 / 14 cm^2


Therefore,
= > Area of the shaded region = Area of C2 + Area of C3 - Area C1 which is not in triangle .

= > Area of the shaded region = 88 / 14 + 49.5 / 14 - 53.5 / 14 cm^2

= > Area of the shaded region = 84 / 14 cm^2

= > Area of the shaded region = 6 cm^2



Hence,
The required area of the shaded region is 6 cm^2.
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