Question

What is the area of a segment that has a height of 0.15 and a base length of 2.4m

Answer 1

Answer:

  about 0.240748 m²

Step-by-step explanation:

Let's call the height of the segment h, and the chord length c. The radius of the circle can be r, and the distance from the circle center to the chord midpoint can be t.

Then t = r - h, and the Pythagorean theorem tells us ...

  t² + (c/2)² = r²

  (r -h)² + (c/2)² = r² . . . . . . substitute for t

  r² -2rh +h² +c²/4 = r² . . . eliminate parentheses

  h² +c²/4 = 2rh . . . . . . . . . subtract r²-2rh

  r = (4h² +c²)/(8h) . . . . . . . divide by 2h to solve for r

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The central angle α subtended by this segment is then ...

  α = 2·arcsin((c/2)/r) = 2·arcsin(c/(2r))

and the area of the segment is the difference between the area of the sector and the area of the triangle between the segment and the circle center:

  A = (1/2)r²·α - (1/2)(c)(t)

  = r²·arcsin(c/(2r)) -(c/2)(r - h)

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Computing r and filling in the values, we get ...

  r = (4·0.15² +2.4²)/(8·0.15) = 4.875

  A = 4.875²·arcsin(2.4/9.75) -(2.4/2)(4.875 -0.15)

  = 23.765625·arcsin(16/65) -5.67 . . . . . "exact" value

  A ≈ 0.24074833 . . . . square meters