Advertisements
Advertisements
Question
Using the following figure, show that BD = `sqrtx`.
Advertisements
Solution
AB = x and BC = 1
AC = AB + BC
= x + 1
diameter = x + 1
radius OA = OD = OC = OB = OC - BC
`= (x + 1)/2 - 1 = (x + 1 -2)/2 = (x - 1)/2`
Using pythagoras in ΔBOD
P2 + B2 = H2
`"P"^2 + ((x - 1)/2)^2 = ((x + 1)/2)^2`
`"P"^2 = ((x + 1)/2)^2 - ((x - 1)/2)^2`
= `((x + 1)^2 - (x - 1)^2)/4`
`= ((x^2 + 1 + 2x) - (x^2 + 1 - 2x))/4`
`= (x^2 + 1 + 2x - x^2 - 1 + 2x)/4`
`"P"^2 = (4x)/4`
P2 = x
P = `sqrtx`
APPEARS IN
RELATED QUESTIONS
Simplify by rationalising the denominator in the following.
`(3 - sqrt(3))/(2 + sqrt(2)`
Simplify by rationalising the denominator in the following.
`(sqrt(15) + 3)/(sqrt(15) - 3)`
Simplify by rationalising the denominator in the following.
`(sqrt(7) - sqrt(5))/(sqrt(7) + sqrt(5)`
Simplify by rationalising the denominator in the following.
`(sqrt(12) + sqrt(18))/(sqrt(75) - sqrt(50)`
In the following, find the values of a and b:
`(3 + sqrt(7))/(3 - sqrt(7)) = "a" + "b"sqrt(7)`
In the following, find the values of a and b:
`(5 + 2sqrt(3))/(7 + 4sqrt(3)) = "a" + "b"sqrt(3)`
If x = `(4 - sqrt(15))`, find the values of
`x^2 + (1)/x^2`
If x = `(4 - sqrt(15))`, find the values of:
`(x + (1)/x)^2`
If x = `((sqrt(3) + 1))/((sqrt(3) - 1)` and y = `((sqrt(3) - 1))/((sqrt(3) + 1)`, find the values of
x2 + y2
Show that: `(3sqrt2 - 2sqrt3)/(3sqrt2 + 2sqrt3) + (2 sqrt3)/(sqrt3 - sqrt2) = 11`
