Question

pls ans the above mentioned question​

Answer 1

ᴀɴsᴡᴇʀ

★ given →

• QR is a chord of the a circle and P is the mid point of the chord QR
• P bisects the chord QR
• QR = 24cm
• OP = 10cm
• QP = RP = 24/2 = 12cm

★ to find →

radius of the circle?

★ solution →

☆ 1st Theorem : "A straight line passing through the centre of a circle to bisect a chord is perpendicular to the chord."

and it is given that P bisects the chord QR

➪ P ⊥ QR

➪ ∠OPQ = ∠OPR = 90°

so, ∆OPQ is right angle triangle

☆ Pythagoras theorem : "In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides."

➪ OP² + PQ² = OQ²

➪ 10² + 12² = OQ²

➪ 100 + 144 = OQ²

➪ 244 = OQ²

➪ OQ = √244

➪ OQ = 2√61cm

➪ OQ = 2 × 7.81

➪ OQ = 15.62cm

so, radius of the circle is 2√61cm or 15.62cm

✩ theorems which have been used to solve this question;

• "A straight line passing through the centre of a circle to bisect a chord is perpendicular to the chord."
• "In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides" (Pythagoras theorem)