Question

Find the area of the shaded region AB-Diameter-7cm



Anyone answer the question plz​

Answer 1

Step-by-step explanation:

Here, diameter of the circle (d) = 7 cm (given)

So, Radius of the circle = d/2

= 7/2 cm

Also,

Area of the circle = πr^2

= 22/7 × 7/2 × 7/2

= 11 × 7/2

= 77/2 cm^2

Let O be mid-point of AB.

As we know diameter always passes through the centre of the circle.

And, since AB is the diameter.

So, O will be the centre of the circle. (Since, O is the mid-point of diameter AB)

Also, Join OC

Now, OC = 7/2 (radius of the circle)

Also, OC = OB = 7/2 (radii of the circle)

In ∆ABC,

Angle A = 30° (given)

Angle C = 90° (given)

Angle B = ?

As we know that,

Angle A + Angle B + Angle C = 180° (A.S.P of a ∆)

=> 30° + Angle B + 90° = 180°

=> 120° + Angle B = 180°

Therefore, Angle B = 180° - 120°

= 60°

And, since OB = OC

Therefore, Angle OBC = Angle OCB = 60° (Since, angles opposite to equal sides are equal)

Now, In ∆OBC,

Angle OBC = Angle OCB = 60°

Angle COB = ?

As we know that,

Angle OBC + Angle OCB + Angle COB = 180° (A.S.P of a ∆)

=> 60° + 60° + Angle COB = 180°

=> 120° + Angle COB = 180°

Therefore, Angle COB = 180° - 120°

= 60°

Since, all angles of ∆ OBC are equal. Therefore, ∆OBC is an equilateral ∆.

And, OB = OC = BC = 7/2 cm........(i)

So, In ∆ABC,

AB = hypotenuse = 7cm (given)

BC = base = 7/2 cm [ using (i) ]

CA = perpendicular = ?

Also, By Pythagoras' theorem, we get :-

(hypotenuse)^2 = (perpendicular)^2 + (base)^2

=> (perpendicular)^2 = (hypotenuse)^2 - (base)^2

=> (CA)^2 = (AB)^2 - (BC)^2

= (7)^2 - (7/2)^2

= 49 - 49/4

= 147/4

Therefore, CA = √147/4

= (7√3)/2 cm

Now, Area of ∆ABC = 1/2 × base × height

= 1/2 × 7/2 × (7√3)/2

= (49√3)/8 cm^2

So, Area of shaded region = Area of circle - Area of ∆ABC

= 77/2 - (49√3)/8

= (308-49√3)/8

= 27.89 cm^2 (approx.)

= 28 cm^2

Hope it helps :)