Thread: stuck on these 3 problems

1. stuck on these 3 problems

solved.

2. Hello, oohhoohh!

2. Find the radius of a circle in which a 96 cm chord
is 16 cm closer to the center than an 80 cm chord.
Code:
             40       E
C * - - - - - - - o - - - - - - - . D
*             |             .
*           |16         .
R *        D|       48.
A . - - - * - - - o - - - . - - - * B
.     *     |     .     *
.   *   |x  .   * R
. * | . *
*
O

The center of the circle is $O.$
The radius is: . $R \:=\:OA = OB = OC = OD.$
The chords are: . $AB = 96,\;CD = 80$
Their distance is $DE = 16.$
Let $x = OD.$

In right triangle $ODB\!:\;\;x^2 + 48^2 \:=\:R^2$ .[1]

In right triangle $OEC\!:\;\;(x+16)^2 + 40^2 \:=\:R^2$ .[2]

Equate [1] and [2]: . $x^2 + 48^2 \:=\x+16)^2 + 40^2 \quad\Rightarrow\quad 32x \:=\:448 \quad\Rightarrow\quad x \:=\:14" alt="x^2 + 48^2 \:=\x+16)^2 + 40^2 \quad\Rightarrow\quad 32x \:=\:448 \quad\Rightarrow\quad x \:=\:14" />

Substitute into [1]: . $14^2 + 48^2 \:=\:R^2 \quad\Rightarrow\quad R^2 \:=\:2500$

. . Therefore: . $R \:=\:50\text{ cm}$