Question

a sector with acute central angle theta is cut out fron a circle of radius 7cm area ofcircle circumscribing the sector is

Answer 1

Answer:

area of circle circumscribing the sector = 22/7x7x7-thita/360x22/7x7x7

                                                                 =154-thita/360x22x7

Step-by-step explanation:

Answer 2

Answer:

Suppose the vertices of the sector are labelled A,B and C with angle BAC =θ. 

Given diameter =14cm ⟹ Radius (R)=7cm

Then ∣AB∣=∣AC∣=R=7. 

From A, draw the bisector m of θ in the sector. 

This line m goes through the centre of the circumscribing circle and intersects that circle at point S(opposite to vertex A). 

The length AS=2r, where r is the radius that we need to determine.

Draw from S the line segment SB. 

Since AS goes through the centre of the circumscribing circle and B is a point on that same circle, 

△ASB is a right triangle and ∠SBA=90o. 

Furthermore, angle BAS=θ/2 

So ∣AS∣=cos(θ/2)∣AB∣

Since ∣AS∣=2r and ∣AB∣=R=7, we have 

2r=cos(θ/2)R=cos(θ/2)7

⟹ r=21×cos(θ/2)7=27sec(θ/2)

∴ Area of the circle circumscribing the sector ABC=πr2=722×472sec2(2

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