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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