Question

Let C be a circle of radius 5 meters having center at O. Let PQ be a chord of C that passes through points A and B where A is located 4 meters north of O and B is located 3 meters east of O. Then, the length of PQ, in meters, is nearest to

• A. 7.2
• B. 7.8
• C. 6.6
• D. 8.8

Answer 1

The Correct Option is D

The correct answer is (D): \(8.8\)

\(AB = \sqrt{3^2+4^2} = 5\)
\(OT = \frac{3×4}{5} = \frac{12}{5}\)
\(OT = 2.4M\)
In \(ΔOTQ\)
\(OQ^2 = OT^2+TQ^2 ⇒ OQ^2-OT^2\)
= \((5)^2-(2.4)^2\)
= \(19.24\)
\(⇒ TQ = \sqrt{19.24} ≈ 4.4\)
\(∴ PQ=2TQ = 8.8m\)

Answer 2

As , we have OA = 4m and OB = 3m and AB = 5m, so OR bisects the chord into QC and PC
Since AB = 5m, we have a + b = 5 …. (i)
4² - k² = a² …. (ii)
3² - k² = b² ….. (iii)
By subtracting eqn (iii) from (ii), we get :
a² - b² = 7 …. (iv)
Now , substituting eqn (i) in (iv), we get :
a - b = 1.4 ….. (v)  ∵ [ (a -b)(a-b) = 7 ; ∴ (a-b) = \(\frac{7}{5}\)]
Now , by solving eqn (i) and (v), we have the value of a = 3.2 and b = 1.8
Therefore, k² = 5.76
In △ POC , we have 5² - k² = (x + a)² 
By Substituting the values we have , we get :
(25 - 5.76) = ( x + 3.2)²
\(\sqrt{19.24}\) = (x + 3.2) or x = 1.19 m
Same way , solve y using △ QOC, we get y = 2.59 m
∴ PQ = 5 + 2.59 + 1.19
= 8.78
≈ 8.8 m
Therefore, the correct option is (D) : 8.8