Question

Study the diagram, where AE is tangent to the circle at point A, and DE is secant to the circle at points C' and D.

Answer 1

25

Step-by-step explanation

to solve this we need to apply the secant-tan rule, it says

if a secant and tangent are drawn to a circle form the same external point, the product of the lengths of the secant and its external segement equals the square of the length of the tangent segment

`OM\cdot\text{ON=(OQ})^2`

so,

Step 1

indentify

OM= DE (unknown value)=DC+CE=DC+9

ON=CE=9

OQ=AE=15

replace

`\begin{gathered} OM\cdot\text{ON=(OQ})^2 \\ (DC+9)(CE)=(AE)^2 \\ \text{replace} \\ (DC+9)(9)=(15)^2 \\ 9DC+81=225 \\ \text{subtract 81 in both sides} \\ 9DC+81-81=225-81 \\ 9DC=144 \\ \text{divide both sides by 9} \\ (9DC)/(9)=(144)/(9) \\ DC=16 \end{gathered}`

so, we get that

DC=16

Step 2

Also, we know

`\begin{gathered} DE=DC+CE \\ \text{replacing} \\ DE=16+9 \\ DE=25 \end{gathered}`

therefore, the answer is

25

I hope this helps you