Question

in the figure AB is the diameter PC is perpendicular to AB. PC=3cm. find the radius of semicircle ?​

Answer 1

Answer:

Hey mate here is your answer

Step-by-step explanation:

Let AB be the chord of the given circle with centre O and a radius of 10 cm.

Then AB =16 cm and OB = 10 cm

From O, draw OM perpendicular to AB.

We know that the perpendicular from the centre of a circle to a chord bisects the chord.

∴ BM = (162) cm=8 cm

In the right  ΔOMB, we have:

OB2 = OM2 + MB2   (Pythagoras theorem)

⇒ 102 = OM2 + 82

⇒ 100 = OM2 + 64

⇒ OM2 = (100 - 64) = 36

⇒ OM=36−−√ cm=6 cm

Hence, the distance of the chord from the centre is 6 cm.

Answer 2

Step-by-step explanation:

(1) It is given that line AB is tangent to the circle at A.

∴ ∠CAB = 90º (Tangent at any point of a circle is perpendicular to the radius throught the point of contact)

Thus, the measure of ∠CAB is 90º.

(2) Distance of point C from AB = 6 cm (Radius of the circle)

(3) ∆ABC is a right triangle.

CA = 6 cm and AB = 6 cm

Using Pythagoras theorem, we have

BC2=AB2+CA2⇒BC=62+62−−−−−−√ ⇒BC=62–√ cm

Thus, d(B, C) = 62–√ cm

(4) In right ∆ABC, AB = CA = 6 cm

∴ ∠ACB = ∠ABC (Equal sides have equal angles opposite to them)

Also, ∠ACB + ∠ABC = 90º (Using angle sum property of triangle)

∴ 2∠ABC = 90º

⇒ ∠ABC = 90°2 = 45º

Thus, the measure of ∠ABC is 45º.