Question

In the figure below Z is the center ofbthe circle. Suppose that QR= 10, ST = 10, UZ = 3X+4, VZ= 16. Find x and TV.

Answer 1

The line from the center perpendicular to the chord ST, bisects it.

ST = 10 ( Given)

Therefore,

`TV=(1)/(2)* ST=(1)/(2)*10=5`

Let the radius be r., then

`r^2=TV^2+VZ^2=5^2+16^2=25+256=281`

Also,

`\begin{gathered} UZ^2+5^2=r^2 \\ \text{then} \\ (3x+4)^2+25=281 \\ \text{ this implies that} \\ (3x+4)^2=281-25=256 \\ \text{ Therefore} \\ 3x+4=\sqrt[]{256} \end{gathered}`
`\begin{gathered} 3x+4=16 \\ \text{then} \\ 3x=16-4=12 \\ \text{therefore} \\ x=4 \end{gathered}`