in this picture a length of square edge is 8 cm. I want to calculate the radius of circle. i try to calculate it, but i don't know how.
I calculate this:
Asked
Active
Viewed 1,679 times
0
-
Hint: The "subtriangles" in the red triangle are similar. – David Mitra Feb 07 '16 at 13:09
-
It is known that ratio of radius and side is $5/8$. Look at this answer. – Alistair Feb 07 '16 at 13:38
3 Answers
1
Mark the center of circle as point $O$.
Mark the intersection of the red line and the square edge $BC$ as $F$.
Draw a line from $O$ to $B$ in your diagram. The length of this line must be equal to the radius of the circle, $r$.
The length of the line from $O$ to $F$ must then be equal to $8-r$ as $EO=r$ and $EF=8$.
Now use pythagorus in triangle $OBF$ to get: $$r^2=4^2+(8-r)^2$$and solve for $r$
Mufasa
- 5,434
1
Let M be the center of the circle and let M' be the (orthogonal) projection of M on AB. Let r denote the radius of the circle. Then $r=|EM|=|AM'|, |AM'|+|M'B|=8, |MM'|^2+|M'B|^2=|MB|^2=r^2$. So we get $(8-r)^2+4^2=r^2$ which imlpies $r=5$
Takirion
- 1,520
0
First of all, the cathetus you labeled with 2*sqrt(10) actually has length $\sqrt{80}=2\sqrt{20}$. But you don't actually need this.
Let's call the third corner of the red triangle F. Let M be the intersection of EF and BC. Then $\angle MEB=\angle MBF$. Therefore, the triangles $\triangle MEB$ and $\triangle MBF$ are alike. So $MF$ has length 2, $EF$ has length 10 and the radius is 5.
Xaver
- 586