rocko911 wrote:
umg wrote:
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O is the Center of the circle with Diameter 12 and length of arc BOD is \(4\pi\). What is the area of quadrilateral AOBC if AD is parallel to CB and BE is parallel to CA. (Figure not drawn to scale)
A. \(12\sqrt{3}\)
B. \(18\sqrt{3}\)
C. \(36\sqrt{3}\)
D. \(72\sqrt{3}\)
E. \(144\sqrt{3}\)
How do we know angle C is 60 degrees?
and can u help me evaluating the outer equilateral triangle?
Here is the explanation..
In Quad AOBC,
AO // BC
BO // CA
So, this is a Parallelogram. Hence, it can be a square, rectangle, or rhombus. We know that
Angle AOB = 60
Hence, it is not a Square or Rectangle. Moreover, to prove that this is a Rhombus,
OA = OB
Radii of Circle.
Because Adjacent sides of a Rhombus are Equal, Quad AOBC is definitely a Rhombus.
Opposite angles of Rhombus are equal. Hence,
Angle AOB = Angle ACB = 60
Join AB and we have 2 equilateral Triangles. Find the area of one and multiply it by 2.
BTW, this is a special kind of Rhombus for which we do not need the lengths of diagonals to calculate the area.
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