Question

can anyone solve this worksheet​

Answer 1

Step-by-step explanation:

Let PQRS be a quadrilateral circumscribing a circle with centre O.

Join OA, OB, OC and OD.

Since the two tangents drawn from an external point to a circle subtend equal angles at the centre.

∴ ∠1 = ∠8

Similarly, ∠2 = ∠3, ∠4 = ∠5 and ∠6 = ∠7 ........ (1)

Now,

Sum of the angles at the centre = 360°

∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°

⇒ 2(∠1 + ∠2 + ∠5 + ∠6) = 360° [From equation (1)]

and 2(∠7 + ∠8 + ∠3 + ∠4) = 360°

⇒ ∠1 + ∠2 + ∠5 + ∠6 = 180°

and ∠3 + ∠4 + ∠7 + ∠8) = 180°

Thus, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles of the centre of the circle.

Answer 2

Sorry , we can solve only one or two question at a time but send your worksheet......