Question

Please help with this question.

Answer 1

Answer:

The radius of the circle is 10cm.

Step-by-step explanation:

Given:- A circle with centre O, TR is a diameter, PS = SQ = 8cm, SR = 4cm

To find:- Radius = (OR)

Construction:- Join OP and OQ making them the radius of the Circle with centre O

Proof:-

In ΔOPS and ΔOQS,

OP = OQ [Radii of the same circle]

PS = SQ [Given]

OS = OS [Common side]

∴ΔOPS ≅ ΔOQS

Then,

∠OSP = ∠OSQ [Corresponding parts of Congruent Triangles]

But,

PQ is a chord, which means PQ is a straight line.

Thus,

∠OSP + ∠OSQ = 180° [Linear Pair]

From above,

∠OSP + ∠OSP = 180°

2∠OSP = 180°

∠OSP = 180°/2

∠OSP = 90°

Then from above,

∠OSQ = 90°

Hence,

ΔOPS and ΔOQS are right triangles.

Thus,

Using Pythagoras theorem,

OS² + PS² = OP²

But,

we know that,

OP = OR = OQ [Radii of the same circle]

So, it becomes,

OS² + PS² = OR²

But again,

OR = OS + SR

OR = OS + 4

Then,

OS² + PS² = (OS + 4)²

Also,

PS = 8cm

Using the identity,

(x + y)² = x² + 2xy + y²

OS² + 8² = (OS² + 2(4)(OS) + 4²)

OS² + 64 = OS² + 8OS + 16

OS² - OS² + 64 = 8OS + 16

8OS + 16 = 64

8OS = 64 - 16

8OS = 48

OS = 48/8

OS = 6cm

Now,

OR = OS + SR

OR = 6 + 4

OR = 10cm

Hence,

The radius of the circle is 10cm.

Hope it helped and believing you understood it........All the best