+0  
 
0
686
4
avatar

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.smiley

 Aug 30, 2019
edited by Guest  Aug 30, 2019
edited by Guest  Aug 30, 2019
edited by Guest  Aug 31, 2019
 #1
avatar
+1

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

\(\alpha = 1,27721309204\)

 

\(s=2r\cdot sin(\frac{\alpha}{2})\\ s=2\cdot 5cm\cdot sin(\frac{1.27721309204}{2})\\ \color{blue}s=5.961cm\)

 

\(cos(\frac{\alpha}{2})=\frac{\frac{b}{2}}{r}\\ b=2r\cdot cos(\frac{\alpha}{2})=10cm\cdot cos(\frac{1,27721309204}{2})\)

\(b=8.029cm\)

 

\(A_R=s\cdot b=5.961cm\cdot 8.029cm\)

 \(A_R=47.862\ cm^2\)

laugh  !

asinus

 Aug 31, 2019
edited by asinus  Aug 31, 2019
edited by asinus  Sep 1, 2019
 #2
avatar
+1

My numbers are somewhat different>>

D = 10 cm

r  = 5 cm

(Central angle)  α = 73.168º

(Chord)   c = 5.96 cm

(Segment area)  A= 4 cm²

(Rec short side)   a = 5.960 cm

(Rec  long side)   b = 8.030 cm

(Rectangle area)  Ar = 47.859 cm²

smiley  

Guest Aug 31, 2019
 #3
avatar+128079 
+2

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 :

 

 

cool cool cool

 Aug 31, 2019
 #4
avatar
+1

Hi, Phill!  You did an exellent job again.

I created and posted this math problem on this forum, I couldn't solve it using formulas because I got stuck at this:   A = r2 / 2 ( θ - sin θ )

So, I've decided to use the old fashioned method - guesstimation. smiley

Guest Aug 31, 2019

4 Online Users

avatar
avatar
avatar
avatar