Question

d) Let's Construct a quadrilateral PQRS in which PQ = 7cm, QR = 6.5 cm, RS
-5.2 cm, SP = 4.4 cm, ZPQR = 60°​

Answer 1

Answer:

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