Question

OABC is a square. A circle drawn with a centre O cuts the square in P and Q prove that BP = PQ

Answer 1

Answer:

Using Pythagoras theorem prove BP = PQ

Step-by-step explanation:

Let 'a' be the side of the square.

From the figure we get,OA = AB = BC = OC = a

And OP = OQ = r radius of circle

Consider the triangles  ΔOAP and ΔOCQ, both are right angled triangle

PA =$\sqrt{OP^2-OA^2}$

PA =$\sqrt{r^2- a^2}$    ----(1)

Similarly,

QC = $\sqrt{OQ^2-OC^2}$

QC = $\sqrt{r^2 - a^2}$   -----(2)

From (1) and (2) we get,

PA = QC we have AB = BC

Therefore BP = BQ

Answer 2

Answer:

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