Question

PLEASE solve these 3 question of 1 mark best ANSWER will be amrk as brainliest

Answer 1

Hey,

AB + CD = AD + BC

(To prove above, see the theorem that length of tangents drawn from an external points to a circle are equal)

Let AB touches circle at 'P" and BC at Q and CD at R and DA at S now write:

AP = AS

BP = BQ

CR = CQ

DR = DS

adding all 4 above we get

AP + BP + CR + DR = AS + DS + BQ + CQ ,

OR

AB + CD = AD + BC.

HOPE IT HELPS YOU:-))

Answer 2

Continuation to Ruchika08.

(1)

Given Equation is x^2 - x = α(2x - 1)

⇒ x^2 - x - α(2x - 1) = 0

⇒ x^2 - x - 2xα + α = 0

⇒ x^2 - x(2α + 1) + α = 0

On comparing with ax^2 + bx + c 0 , we get a = 1, b = -(2α + 1), c = α.

Given that sum of roots is 0.

We know that sum of roots = -b/a

⇒ 0 = 2α + 1/1

⇒ 0 = 2α + 1

⇒ 2α = -1

⇒ α = -1/2.

Therefore, the value of α = -1/2.

(2)

Given : (1/x + 2), (1/x + 3), (1/x + 5) are in AP.

We know that if a,b,c are in AP, then 2b = a + c.

⇒ 2(1/x + 3) = (1/x + 2) + (1/x + 5)

⇒(2)/x + 3 = (x + 5 + x + 2)/(x + 2)(x + 5)

⇒2(x + 2)(x + 5) = (x + 3)(x + 5 + x + 2)

⇒ 2(x + 2)(x + 5) = (x + 3)(2x + 7)

⇒ 2(x^2 + 5x + 2x + 10) = 2x^2 + 7x + 6x + 21

⇒ 2(x^2 + 7x + 10) = 2x^2 + 13x + 21

⇒ 2x^2 + 14x + 20 = 2x^2 + 13x + 21

⇒ 14x - 13x = 21 - 20

⇒ x = 1.

Therefore, the value of x = 1.

Hope it helps!