Advertisements
Advertisements
प्रश्न
Using the following figure, show that BD = `sqrtx`.
Advertisements
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.
`2/(3 sqrt 7)`
Rationalize the denominator.
`1/(3 sqrt 5 + 2 sqrt 2)`
Rationalise the denominators of : `[ √3 + 1 ]/[ √3 - 1 ]`
Simplify by rationalising the denominator in the following.
`(2sqrt(6) - sqrt(5))/(3sqrt(5) - 2sqrt(6)`
Simplify by rationalising the denominator in the following.
`(7sqrt(3) - 5sqrt(2))/(sqrt(48) + sqrt(18)`
Simplify the following
`(sqrt(5) - 2)/(sqrt(5) + 2) - (sqrt(5) + 2)/(sqrt(5) - 2)`
Simplify the following :
`(3sqrt(2))/(sqrt(6) - sqrt(3)) - (4sqrt(3))/(sqrt(6) - sqrt(2)) + (2sqrt(3))/(sqrt(6) + 2)`
If x = `(7 + 4sqrt(3))`, find the value of
`x^2 + (1)/x^2`
If x = `(4 - sqrt(15))`, find the values of
`(1)/x`
If x = `((2 + sqrt(5)))/((2 - sqrt(5))` and y = `((2 - sqrt(5)))/((2 + sqrt(5))`, show that (x2 - y2) = `144sqrt(5)`.
