In the adjoining figure, seg QR is a chord of the circle with centre O. P is the midpoint of the chord QR. If QR = 24, OP = 10, find radius of the circle. To find solution of the problem, write the theorems that are useful. Using them, solve the problem. (Textbook pg. no. 48)
Best answer
Theorems which are useful to find solution:
i. The segment joining the centre of a circle and the midpoint of a chord is perpendicular to the chord.
ii. In a right angled triangle, sum of the squares of the perpendicular sides is equal to square of its hypotenuse.
QP = \(\frac { 1 }{ 2 } \) (QR) [P is the midpoint of chord QR]
\(\frac { 1 }{ 2 } \) × 24 = 12 units
Also, seg OP ⊥ chord QR [The segment joining centre of a circle and midpoint of a chord is perpendicular to the chord]
In ∆OPQ, ∠OPQ = 90°
∴ OQ2 = OP2 + QP2 [Pythagoras theorem]
= 102 + 122
= 100 + 144
= 244
∴ OQ = \(\sqrt { 244 }\) = 2\(\sqrt { 61 }\) units.
∴ The radius of the circle is 2\(\sqrt { 61 }\) units.
Log in or create a free account to answer this question.
