Question

E (4+ √5) cm BC = (2 + √5) cm ED = (4 + √5) cm DC = 2√√5 cm Show that the length of AB is (p√√5+q)cm, where p and q are intergerts.​

Answer 1

Proof:

Let
`x` be the length of segment AB. By the power of a point formulas,

`CB \cdot CA = CD \cdot CE\\\implies (2+√(5))(2+√(5)+x)=(2√(5))(4+3√(5)).`

Expanding both sides, we get:

`4+2√(5)+2x+2√(5)+5+x√(5)=8√(5)+30\\\implies 9+4√(5)+2x+x√(5) = 8√(5)+30\\\implies x(2+√(5))=21+4√(5)\\\implies x = (21+4√(5))/(2+√(5)) = ((21+4√(5))(2-√(5)))/((2+√(5))(2-√(5))) = (42-21√(5)+8√(5)-20)/(-1) = -(22-13√(5)) = 13√(5)-22.`

Hence, it is possible for the length of AB to be in the form
`p√(5)+q` for integers
`p` and
`q.`