Question

this question........​

Answer 1

Answer:

• AB is the diameter of a circle
• The area of the circle is π
• Another circle is drawn with centre as C
• The circumference of the circle passes through A and B

We need to find the shaded portion.

Concept used:- The line joining two points, where circle interests, is perpendicular to the line joining the centres.

So, here we have AB as donated of one circle and C as centre of the second circle. By the concept, CO is the perpendicular to AB.

As CO is perpendicular to AB, hence, AO = BO (perpendicular bisects the diameter).

Now, we know that, πr² = π

Hence, r² = 1.

So, r = 1 units.

Now, we already have right angled triangles.

So, Finding the hypotenuse.

AO² + CO² = AC²

1² + 1² = AC²

AC = √2 units.

We know that, the area of the sector CAB = π/2 sq. units. ( because the sector is a semicircle )

Also, the area of triangle CAB = ½ * base * height

= ½ * 2 * 1

= 1 units.

So, the area of two segments formed in a circle with Centre C = π/2 - 1 units.

Now, finding the area of shaded portion.

Shaded portion = π/2 - (π/2 -1)

= π/2 - π/2 + 1

= 1 units. = Final Answer

Answer 2

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QUESTION :

this question........

SOLUTION :

From the question, we can see that the area of the circle is π sq units.

So the radius of the circle is 1 cm.

For the complete solution refer to the attachments.

The first attachment shows the figure.

The second attachment shows the calculation..

ANSWER :

The area of the required shaded region is 1 unit ^ 2.