O is the Center of the circle with Diameter 12

Question

Untitled.png 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} Self Made

rohit8865 · Accepted Answer

umg

can u please confirm the answer!!!

Below is my take...

umg · Answer

Ops! Yes. Sorry. Marked the wrong OA. Fixed it.

Leo8 · Answer

FullSizeRender (13).jpg +B

rocko911 · Answer

How do we know angle C is 60 degrees?

and can u help me evaluating the outer equilateral triangle?

Umang bhatia · Answer

in Quad AOBC

as we found angle BOD to 120
therefore, Angle AOB is 60

also as CB|| AD, therfore angle CBO = angle BOD

and similarly ang(CAO)= angle AOE= 120 degress

now in quad you have 3 angles so forth comes out be 60 by summing four angles to 36p

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umg · Answer

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 :arrow:

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.

sahilvijay · Answer

Answer is B.

as it is clearly visible from pic, the area is 18√3

Fdambro294 · Answer

1st thing that needs to be realized is that the question never says that Lines CB and CA are TANGENT to the Circle. This means the 90 degree/Perpendicular Rule with the Radius does NOT apply.

(2nd)
We know that Quadrilateral ACBO must be a Parallelogram because:

CB is PARALLEL to Opposite Side AO (which travels across the Circle to Point D)

and

CA is PARALLEL to Opposite Side BO (which travels across the Circle to Point E)

(3rd)
Sides AO and BO are EQUAL Adjacent Sides of the Parallelogram. They are both Radii.

Given:

-1- Opposite Sides are Parallel

and

-2- Adjacent Sides are Equal

We are dealing with a Square or a Rhombus

(3rd)
Given that Minor Arc BOD = 4(pi) and the Radius of the Circle = 6:

The Central Angle that is Subtended by Arc BOD = 4(pi) / Circumference = X / 360 deg

Since the Circumference = 12(pi), the Central Angle <BOD = (1/3) * 360 deg = 120 deg

Vertically Opposite Angle <AOE = 120 deg. also

Completing the Circle around the Center O:

Angle <AOB = 60 deg

Lastly, Given the Parallel Lines of AO and CB ---- <CBO is an Alternating Interior Angle that is EQUAL to <BOD = 120 deg

Given the Parallel Lines of CA and BO ---- <CAO is also an Alternating Interior Angle that is EQUAL to <EOA

In Summary, the Angles of the Paralleogram are given as follows:

Angle at Vertex O = 60

Angle at Vertex A = 120

Angle at Vertex B = 120

Angle at Vertex C must = 60 in order to complete the Quadrilateral and because Opposite Angles are Equal in a Paralleogram

Angle <AOB = 60 deg

Since Adjacent Angles are NOT 90 Degrees and Equal, we are dealing with a RHOMBUS

(4th)
2 Concepts apply to a Rhombus:

Concept 1: Diagonals are Perpendicular Bisectors and Angle Bisectors

Concept 2: You can Find the Area of a Rhombus Multiplying the Diagonal Lengths and Dividing by 2

Diagonal AB and Diagonal CO create FOUR 30-60-90 Right Triangles

Using the Ratio of Sides and Given that in the Right 2 Triangles we have the Sides AO = BO = 6 to use to find the Actual Values

Diagonal AB is Bisected into 2 Lines = 3

Diagonal CO is Bisected into 2 Lines each = 3*sqrt(3)

Finally, AREA of Rhombus CAOB =

(6) * (6sqrt(3)) * (1/2) =

18sqrt(3)

-B-

O is the Center of the circle with Diameter 12

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O is the Center of the circle with Diameter 12

Prep Toolkit

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