GreenlightTestPrep wrote:
In the above figure, the small circle and big circle have diameters of 3 and 6 respectively. If AB||CD, and both circles share the same center, what is the area of the shaded region?
A) 9/2 + 2π
B) 9/2 + 3π
C) 3√3 + 3π
D) 6√3 + 2π
E) (9/2)√3 + 3π
Here,
plz refer to the diagram attached,
First of all let us joined the AO , CO , BO, DO as they are all radii of the bigger circle = 3 (since radius = dia/2 = 6/2 = 3)
Let us join AC and BD, which are = 3 (since the diameter of the smaller circle = 3)
we find that △ AOC and △BOD are equilateral △ as all sides are equal and so all angles in the △'s are 60°
Now let us consider the sector AOC, we have the central angle = 60°
We know,
\(\frac{Central angle}{360} = \frac{Sector Area AOC}{Circle Area}\)
or \(\frac{60}{360}= \frac{(Sector Area)}{(9\pi)}\)
or Sector Area AOC = \(\frac{(3\pi)}{2}\)Similarly
Sector Area BOD = \(\frac{(3\pi)}{2}\)Now we need the area of the △ AOB and △COD, in which we need the base of both the △'s i.e. AB and CD
Now let us consider the right angled △ BCD
we have BD =3 , CB = 6 , angle BDC =90° , angle DBC = 60° and angle BCD = 30° (angle BDC = 90° because the line AB is parallel to line CD and both are inscribed within a circle)
SO the triangle is a 30-60-90 triangle and the sides are in the ratio of \(1: \sqrt{3} : 2\)
or we can write as \(3 : 3\sqrt{3} : 6\) ; such that the side \(CD = 3\sqrt{3}\)
Now the
Area of the △COD,
\(\frac{1}{2} * base * altitude = \frac{1}{2} * CD * \frac{3}{2}\) (Altitude = 1.5 i.e. the radius of the smaller circle)
= \({\frac{1}{2} * 3\sqrt{3} * \frac{3}{2}}\)
i.e. Area of the △COD = \(\frac{9\sqrt3}{4}\)Similarly in △ AOB the area = \(\frac{9\sqrt3}{4}\)Now the area of the shaded region = Area of the sector AOC + Area of the sector BOD + Area of the △ AOB + Area of the △ COD = \(\frac{(3\pi)}{2} + \frac{(3\pi)}{2} + \frac{9\sqrt3}{4} + \frac{9\sqrt3}{4}\) = \(3\pi + \frac{9\sqrt3}{2}\)
Attachments
FIG 1jpg.jpg [ 21.5 KiB | Viewed 1767 times ]
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