Answer :

● To see how right triangle makes a circle?

1. Draw a line of 5 cm.

2. Using Line with given length, make a slider from 0 to 180° with increment 5.

3. Draw an angle ‘a’ at one end and draw ‘90° - a’ clockwise at the other end to make a triangle.

4. Draw several such triangles.

5. Apply trace on to the top vertex and sides of a triangle to get a circle.

● To see angle in a semicircle is right.

1. To show ∠APB = 90° . Firstly join P to the centre of a circle O which splits the angle at P into two parts.

2. Δ AOP and Δ BOP are isosceles triangle.

Thus, ∠A = x° and ∠B = y°

3. In ΔAPB, the sum of angle is 180°

⇒ x + y + (x + y) = 180°

⇒ 2(x + y) = 180°

⇒ x + y = 90°

⇒ ∠APB = x° + y° = 90°

Thus, angle in a semicircle is right angle.

● To see for any point inside the circle, such an angle is larger than right angle.

1. To show ∠APB > 90° . Extend one of the lines to meet the circle. Join this point to the other end of the diameter.

2. APB is the exterior angle at P of triangle AQP.

∠APB = sum of interior angle at Q and B

3. The angle at Q is right angle. So, ∠APB is larger than 90°

● To see for any point outside the circle, such an angle is smaller than right angle.

1. To show ∠APB < 90° . APB is the interior angle at P of triangle PQB. The right angle AQB is an exterior angle.

2. So, ∠APB is smaller than 90°

Now, in our given question we can see that,

As seen above, point on a circle, such an angle is a right angle.

Point outside the circle , such an angle is larger than 90°

Point inside the circle, such an angle is smaller than 90°

Thus, as ∠ACB = 110° > 90° . Therefore point C lies inside the circle.

As, ∠ADB = 90° . Therefore point D lies on the circle.

As, ∠AEB = 70° . Therefore point E lies outside the circle.

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