The rectangle is inscribed in a circle with Dia = 10 cm. The area of the segment that is formed by the shorter side of the rectangle and the circumference of a circle is 4 cm2. Find the area of that rectangle.
Have fun.
The rectangle is inscribed in a circle with Dia = 10 cm. The area of the segment that is formed by the shorter side of the rectangle and the circumference of a circle is 4 cm2. Find the area of that rectangle.
Hello Guest!
\(\alpha\) = center angle
r = radius
s = chord
b = long side of the rectangle
A = segment area
\(A_R\) = rectangle area
\(\color{blue}A=\frac{r^2}{2}(\alpha -sin \alpha )\\ \alpha -sin \alpha=\frac{2\cdot A}{r^2}=\frac{8cm^2}{(5cm)^2}\\ \alpha -sin\ \alpha=0.32\\ \alpha -sin\ \alpha -0.32=0 \)
http://www.arndt-bruenner.de/mathe/scripts/gleichungssysteme2.htm
Solution found in 1st run after 3 iterations
\(\alpha = 1,27721309204\)
\(s=2r\cdot sin(\frac{\alpha}{2})\\ s=2\cdot 5cm\cdot sin(\frac{1.27721309204}{2})\\ \color{blue}s=5.961cm\)
\(cos(\frac{\alpha}{2})=\frac{\frac{b}{2}}{r}\\ b=2r\cdot cos(\frac{\alpha}{2})=10cm\cdot cos(\frac{1,27721309204}{2})\)
\(b=8.029cm\)
\(A_R=s\cdot b=5.961cm\cdot 8.029cm\)
\(A_R=47.862\ cm^2\)
!
asinus
Let θ be the central angle whose endpoints are on the shorter side of the rectangle
The area of the sector formed by this angle = ( r^2/2 ) ( θ)
And the area of the isosceles triangle formed by the two radii and the shorter side of the rectangle =
(r^2/2) sin ( θ)
So....the area between the shorter side of the rectangle and the circumference is given by
4 = (r/2/2) ( θ - sin θ)
4 = (5^2/2) ( θ -sin θ)
4 = (25/2) ( θ - sin θ)
θ - sin θ = 8/25
Using a little technology to solve this for θ ≈ 1.27721 rads ≈ 73.18°
From the Law of Cosines we can find the shorter side of the rectangle as
√ [ 5^2 + 5^2 - 2(5)(5)cos(73.18°) ] = √[50 -50cos(73.18°)] ≈ 5.96 cm
And using the Pythagorean Theorem, we can find the longer side of the rectangle as
√[10^2 - 5.96^2 ] ≈ 8.03 cm
So.....the area of the rectangle will be the product of these two sides ≈ 5.96 * 8.03 ≈ 47.86 cm^2
Here's a pic :