radius of a special sector

[jschwa1]

I'm attempting to find the radius of this sector. It is given that the triangles are right triangles and the shaded area is 5670 square units. It just seems like there are too many missing variables. :? The solution is 146 units.

 

[mmm4444bot]

There's enough information to write expressions for both legs of each right-triangle (base and height) :idea: as the big triangle has base r and the small triangle has hypotenuse r.

These four expressions will all be in terms of r and trigonometric functions of 0.8 radians. Use them to write an expression for the area of each triangle; those two expressions differ by 5670, providing an equation in r.

Do they ask you to round the answer to the nearest unit ?

 

[galactus]

I got this one.

I just got some company so I will be back later to explain.

\(\displaystyle \underbrace{\frac{r^{2}tan(4/5)}{2}}_{\text{area of large triangle}}-\overbrace{\frac{r-r(1-cos(4/5))}{2}(rsin(4/5))}^{\text{area small triangle}}\)

This whittles down to the area of the shaded region=\(\displaystyle \frac{sin^{3}(4/5)r^{2}}{2cos(4/5)}=5670\)

Solving for r gives r=146.3

 

[Denis]

I just got some company so I will be back later to explain.

You got people visiting you?! They must wanna borrow money :idea:

 

[Deleted member 4993]

Or ride his ATV.......

 

[BigGlenntheHeavy]

\(\displaystyle Let \ A_2 \ = \ area \ of \ larger \ right \ triangle\)

\(\displaystyle Let \ A_1 \ = \ area \ of \ smaller \ right \ triangle.\)

\(\displaystyle Then \ A_2-A_1 \ = \ 5670 \ sq. \ units, \ can \ you \ take \ it \ from \ here?\)

 

[Deleted member 4993]

I got this one.

I just got some company so I will be back later to explain.

\(\displaystyle \underbrace{\frac{r^{2}tan(4/5)}{2}}_{\text{area of large triangle}}-\overbrace{\frac{r-r(1-cos(4/5))}{2}(rsin(4/5))}^{\text{area small triangle}}\)

This whittles down to the area of the shaded region=\(\displaystyle \frac{sin^{3}(4/5)r^{2}}{2cos(4/5)}=5670\)

Solving for r gives r=146.3

I agree ? r = 146.2949512 = 146.3 units

 

[soroban]

Hello, jschwa1!

It is given that the triangles are right triangles and the shaded area \(\displaystyle BDEC\) is 5670 square units.
Find the radius \(\displaystyle R.\)

                        D
                        o
                      * |
                    *   |
                B *     |
                o       | Rtan@
              * |   *   |
         R  *   |Rsin@  |
          *     |      *|
        * 0.8   |       |
      o - - - - o - - - o
      A  Rcos@  C       E
      : - - - - R - - - :

\(\displaystyle \text{In right triangle }ABC\!:\:\begin{Bmatrix}AC &=& R\cos0.8 \\ BC &=& R\sin0.8 \end{Bmatrix}\)

. . \(\displaystyle \text{Hence, area of }\Delta ABC \;=\;\tfrac{1}{2}(R\cos0.8)(R\sin0.8) \;=\;\tfrac{1}{2}R^2\sin0.8\cos0.8\;=\;A_1\)


\(\displaystyle \text{In right triangle }ADE\!:\;\begin{Bmatrix}AE &=& R \\ DE &=& R\tan0.8 \end{Bmatrix}\)

. . \(\displaystyle \text{Hence, area of }\Delta ADE \;=\;\tfrac{1}{2}(R)(R\tan0.8) \;=\;\tfrac{1}{2}R^2\tan0.8\;=\;A_2\)


\(\displaystyle \text{Since }\,A_2 - A_1 \:=\:5670,\:\text{ we have: }\;\tfrac{1}{2}R^2\tan0.8 - \tfrac{1}{2}R^2\sin0.8\cos0.8 \:=\:5670\)

. . \(\displaystyle \tfrac{1}{2}R^2(\tan0.8 - \sin0.8\cos0.8) \:=\:5670 \quad\Rightarrow\quad R^2 \;=\;\frac{11,\!340}{\tan0.8 - \sin0.8\cos0.8}\)


\(\displaystyle \text{Hence: }\;R^2 \;=\; 21,\!402.21275\)

. . \(\displaystyle \text{Therefore: }\;R \;=\;146.2949512 \;\approx\;146.3\)