Asked by Candice
What is the area of the minor segment cut off a circle of radius 10 cm by a chord of length 12 cm?
Could you please show me the working out for this question?
The answer in the textbook is 16 sq cm.
Thanks!!!
Could you please show me the working out for this question?
The answer in the textbook is 16 sq cm.
Thanks!!!
Answers
Answered by
Reiny
Make a sketch showing the chord of 12 and the the two radii of 10.
I see an isosceles triangle. Draw an altitude from the centre to that chord, making two congruent right-angled triangles.
let the height be x, then x^2 + 6^2 = 10^2
x^2 = 100-36 = 64
x = √64 = 8
So the area of the large triangle, 10,10,12 is
(1/2) (12)(8) = 48 cm^2
We have to find the central angle of the sector.
Let each angle at the centre of the right-angled triangles be Ø
sinØ = 6/10 = .6
Ø = 36.87‡
and the central angle is 2Ø = 73.74°
area of whole circle = π(10)^2 = 100π
area of sector/100π = 73.74/360
area of sector = 64.35 cm^2
sooo, the segment is 64.35 - 48 = 16.35 cm^2
(I carried all decimals my calculator could hold and only rounded off the final answer.)
I see an isosceles triangle. Draw an altitude from the centre to that chord, making two congruent right-angled triangles.
let the height be x, then x^2 + 6^2 = 10^2
x^2 = 100-36 = 64
x = √64 = 8
So the area of the large triangle, 10,10,12 is
(1/2) (12)(8) = 48 cm^2
We have to find the central angle of the sector.
Let each angle at the centre of the right-angled triangles be Ø
sinØ = 6/10 = .6
Ø = 36.87‡
and the central angle is 2Ø = 73.74°
area of whole circle = π(10)^2 = 100π
area of sector/100π = 73.74/360
area of sector = 64.35 cm^2
sooo, the segment is 64.35 - 48 = 16.35 cm^2
(I carried all decimals my calculator could hold and only rounded off the final answer.)
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