Question

In a circle two chords AB & CD, both subtend 50° at the centre O. If AB = 6cm, then find a CD. ​

Answer 1

Answer: Let AB and CD be two parallel chords of a circle with centre O such that AB = 6 cm and CD = 12 cm.

Construction:

Join OB and OD

Draw perpendicular bisectors OL of AB and OM of CD. Since

AB ∥ CD the points O, M and L are collinear.

Clearly, LM = 3 cm. Let OM = x cm and let OB = OD = r.

From right triangle OLB, we get:

Using Pythagoras Theorem:

OB2=OL2+BL2or r2=(x+3)2+32

(i)(∵BL=12AB=3cm)From right triangle ODM, we get:

Using Pythagoras Theorem:

OD2=OM2+MD2or r2=x2+62(ii)(∵MD=12CD=6cm)

Thus, from (i) and (ii), we get:

(x+3)2+32=x2+62 or 6x=18 or x=3cm

∴r2=x2+62=32+62=45or r=√45=6.7cm

Hence, the radius of the circle is 6.7cm.

Explations:  

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

∴ ∠CAB = 50º     (Tangent at any point of a circle is perpendicular to the radius throught the point of contact)

Thus, the measure of ∠CAB is 50º.

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–√  cmThus, 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

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