Question

[Tex]\rm Question [\tex]

The radii of two concentric circles is 13 cm and 8 cm respectively. PQ is diameter of bigger circle and QR is tangent to a smaller circle touching it at R. Find the length of PR​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

• The radii of two concentric circles having centre O is 13 cm and 8 cm respectively.

• PQ is diameter of bigger circle and QR is tangent to a smaller circle touching it at R.

Construction :- Join OR and produce QR to meet the bigger circle in S. Join PS.

Now, As OR is radius and QR is tangent.

So, OR is perpendicular to QR.

Now, In right-angle $\triangle$OQR

By using Pythagoras theorem, we have

$\rm \: {OQ}^{2} = {OR}^{2} + {QR}^{2}$

$\rm \: {13}^{2} = {8}^{2} + {QR}^{2}$

$\rm \: 169 = 64+ {QR}^{2}$

$\rm \: 169 - 64 = {QR}^{2}$

$\rm \: 105 = {QR}^{2}$

$\bf\implies \:QR = \sqrt{105} \: cm$

Now, we know that, perpendicular drawn from centre bisects the chord. So, QR = RS

$\bf\implies \:QR \: = \: RS \: = \: \sqrt{105} \: cm$

Now, O is the midpoint of PQ and R is the midpoint of QS.

So, By using Midpoint Theorem, we have

$\bf\implies \:PS = 2 \times OR = 2 \times 8 = 16 \: cm$

Now, PQ is diameter. $\rm\implies \:\angle PSQ = 90\degree$

[$\because$ Angle in semi-circle is right angle. ]

Now, In right-angle $\triangle$ PSR

By using Pythagoras Theorem, we have

$\rm \: {PR}^{2} = {PS}^{2} + {SR}^{2}$

$\rm \: {PR}^{2} = {16}^{2} + {( \sqrt{105} )}^{2}$

$\rm \: {PR}^{2} = 256 + 105$

$\rm \: {PR}^{2} = 256 + 105$

$\rm \: {PR}^{2} = 361$

$\bf\implies \:PR \: = \: 19 \: cm$