Advertisements
Advertisements
प्रश्न
Draw a line segment of length `sqrt5` cm.
Advertisements
उत्तर
Construct a right-angled triangel OAB with

OA = 2 cm,
∠OAB = 90° and
AB = 1 cm
Using OB2 = OA2 + AB2
OB2 = 22 + 12
OB2 = 4 + 1
OB2 = 5
OB = `sqrt5`

APPEARS IN
संबंधित प्रश्न
Rationalize the denominator.
`1/(3 sqrt 5 + 2 sqrt 2)`
Simplify : `sqrt18/[ 5sqrt18 + 3sqrt72 - 2sqrt162]`
Simplify by rationalising the denominator in the following.
`(1)/(5 + sqrt(2))`
Simplify by rationalising the denominator in the following.
`(5 + sqrt(6))/(5 - sqrt(6)`
Simplify by rationalising the denominator in the following.
`(4 + sqrt(8))/(4 - sqrt(8)`
Simplify by rationalising the denominator in the following.
`(7sqrt(3) - 5sqrt(2))/(sqrt(48) + sqrt(18)`
If x = `(7 + 4sqrt(3))`, find the values of :
`(x + (1)/x)^2`
If x = `sqrt3 - sqrt2`, find the value of:
(i) `x + 1/x`
(ii) `x^2 + 1/x^2`
(iii) `x^3 + 1/x^3`
(iv) `x^3 + 1/x^3 - 3(x^2 + 1/x^2) + x + 1/x`
Show that: `x^2 + 1/x^2 = 34,` if x = 3 + `2sqrt2`
Using the following figure, show that BD = `sqrtx`.

