Find the area of the shaded region AB-Diameter-7cm
Anyone answer the question plz
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Find the area of the shaded region AB-Diameter-7cm
Anyone answer the question plz
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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 :)