Find the Area of a segment of a circle if the central angle of the segment is $105^\circ$ degrees and the radius is $70$.
Formulas I have:
- Area of a non-right angle triangle= $\frac{1}{2}a b \sin C$.
- Area of segment = ( area of sector ) $-$ (area of triangle).
Please, could you explain it step by step so I can understand, thanks
$\endgroup$23 Answers
$\begingroup$You can work out the area of the sector then subtract the area of the triangle.
The area of a sector is given by $\frac{1}{2}r^2\theta$ if $\theta$ is in radians or $\frac{1}{2}r^2\pi\frac{\theta}{180^\circ}$ if $\theta$ is in degrees.
The area of the triangle is give by $\frac{1}{2}r^2\sin\theta$.
Combining these two gives: $\frac{1}{2}\times70^2\times\pi\times\frac{105^\circ}{180^\circ}-\frac{1}{2}\times70^2\times\sin105^\circ \approx 2123.34$
$\endgroup$$\begingroup$Formulas that you need to know:-
(1) The area formula for $\triangle ABC$ $[A = \dfrac 12 ab \sin C]$. (Edited.)
(2) The area of a sector (OAB) formula $[A = \dfrac 12 r^2 \theta]$; where $\theta$ is the central angle and it should be in radian instead of degree.
Added. The conversion formula is $[\pi$ radians $= 180^0]$.
Additional info: The common naming convention and the derivation of the area formula:-
We already know that $[⊿ABC] = \dfrac {1}{2}b \times h$.
Since $h = a \sin C$, then $[⊿ABC] = \dfrac {1}{2}ab \times \sin C$.
$\endgroup$8$\begingroup$Forget about particular values and name the things to be measured.
Call the radius of the circle $r$ and the angle subtended by the sector $t$.
The area of the part of the circle containing the sector and internal triangle is $tr^2/2$.
The internal triangle has altitude $r\cos(t/2)$ and base $2r\sin(t/2)$, so its area is
$\begin{array}\\ (1/2)(2r\sin(t/2))(r\cos(t/2)) &=r^2(\sin(t/2)\cos(t/2))\\ &=r^2\sin(t)/2\\ \end{array} $
since $\sin(t) =2\sin(t/2)\cos(t/2) $.
The area of the sector is the difference of these expressions, which is $tr^2/2-r^2\sin(t)/2 =r^2(t-\sin(t))/2 $.
Now you can substitute the values of the radius and angle of the sector.
As a check, for $t = \pi$, this gives $\pi r^2/2$, which is area of the semicircle.
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