Question

two concentric circles of radii 10cm and 6cm are drawn. find the length of the chord of the larger circle which touches the small circle

Answer 1

➩Concentric circles

•The circles having common centre are called concentric circles. (refer to the figure)

➩Given:

▪There are 2 Concentric circles , one circle is having radius-10 cm while the other one (smaller or inner circle) is having radius-6cm

➩Find:

▪Length of the chord of the larger circle which touches the small circle i.e AB

➩Solution

OP = 6cm (radius of smaller or inner circle)

OA = OB = 10cm (radius of bigger or outer circle)

• AB is tangent to smaller circle

• OP is perpendicular on AB

∴ ∠OPA = ∠OPB = 90°

If angles are 90° , then we can simply go for Pythagoras theorem.

In △OPB

${(OB)}^{2} = {(PB)}^{2} + {(OP)}^{2}$

${(10)}^{2} = {(PB)}^{2} + {(6)}^{2}$

${(PB)}^{2} = {(10)}^{2} - {(6)}^{2}$

${(PB)}^{2} = 100 - 36$

${(PB)}^{2} = 64$

${PB} = \sqrt{64}$

${PB} = 8 cm$

⭐We know that AP = PB as the perpendicular OP bisects the line into 2 equal parts where in falls.

So AP = PB = 8cm

AP + PB = AB

8 + 8 = 16cm

So , the length of the chord of the larger circle which touches the small circle is 16cm