MathRevolution wrote:
If O is a centre of a circle as above figure, what is the area of the region shaded?
1) ∠AOB=30 degrees.
2) The length of arc CA is 2π (pi)
*An answer will be posted in 2 days.
Area of shaded region = (Area of Triangle AOC + Area of minor Sector OAB) - Area of Triangle COB
Triangle AOC is a right angled triangle at O and notice that OA and OC are radii of circle O. Let radius of circle O be 'r'
So, Area of Triangle AOC = 1/2(OA)(OC) = r^2/2
Area of minor sector OAB = ∠AOB/360(πr^2)
To find the Area of Triangle COB draw an altitude/ perpendicular OD from point O on base BC. Note that OC and OB are radii of Circle O. Hence, Triangle COB is an isosceles triangle. So the perpendicular OD drawn on base BC will bisect the BC and ∠BOC. Area of Triangle COB = 1/2(OD)(CB)
Area of shaded region = r^2/2 + ∠AOB/360(πr^2) - 1/2(OD)(CB)
So we need to know r, ∠AOB and length of OD and CB
Statement 1
∠AOB=30 degrees
This statement does not provide any information about r, OD and CB. Hence, this is insufficient.
We can eliminate options A and D
Statement 2
The length of arc CA is 2π
Since ∠AOC = 90, we know that length of arc CA = 1/4 Circumference of Circle O.
We therefore, can find the value of r.
2π = 1/4(2πr)
r = 4.
But this statement does not provide us information about ∠AOB and length of OD and CB.
Hence, this statement is insufficient. We can eliminate option B.
Combining statement 1 and statement 2
We know r and ∠AOB. We need to find length of OD and CB
Look at Isosceles triangle BOC. ∠COB = 120 (as ∠AOC = 90 and ∠AOB = 30). Hence, ∠CBO = ∠BCO = 30
Note that perpendicular bisector OD bisects the isosceles triangle BOC in two right angled triangles at D, i.e. Triangle BOD and Triangle COD.
Both these triangles are 30-60-90 triangles. We can use the property of 30-60-90 triangles to find the length of OD, BD and DC.
So, combining the statements gives us all the unknowns required to answer the question. Both statements combined are sufficient. We can eliminate option E.
The correct answer is C.
Hope this helps!!
Nalin
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