Advertisements
Advertisements
प्रश्न
Using the following figure, show that BD = `sqrtx`.
उत्तर
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
संबंधित प्रश्न
Rationalize the denominator.
`3/(2 sqrt 5 - 3 sqrt 2)`
Rationalise the denominators of : `[ √3 + 1 ]/[ √3 - 1 ]`
Rationalise the denominators of : `[ sqrt3 - sqrt2 ]/[ sqrt3 + sqrt2 ]`
Simplify by rationalising the denominator in the following.
`(3sqrt(2))/sqrt(5)`
Simplify by rationalising the denominator in the following.
`(2sqrt(3) - sqrt(6))/(2sqrt(3) + sqrt(6)`
Simplify the following :
`(7sqrt(3))/(sqrt(10) + sqrt(3)) - (2sqrt(5))/(sqrt(6) + sqrt(5)) - (3sqrt(2))/(sqrt(15) + 3sqrt(2)`
In the following, find the values of a and b.
`(sqrt(3) - 1)/(sqrt(3) + 1) = "a" + "b"sqrt(3)`
In the following, find the values of a and b:
`(5 + 2sqrt(3))/(7 + 4sqrt(3)) = "a" + "b"sqrt(3)`
In the following, find the values of a and b:
`(sqrt(2) + sqrt(3))/(3sqrt(2) - 2sqrt(3)) = "a" - "b"sqrt(6)`
Show that Negative of an irrational number is irrational.
