+7 The diagram shows two chords that are parallel. What is the length of the chord that lies halfway between them?
+2 OP = x
OB = OD = OF = R
PB = 9v RF = 7 PRD = 10
Rv = 9v + xv and Rv = 7v + (10 - x)v
both R equal 9v + xv = 7v + 10v - 20x + xv
x = (7v + 10v - 9v)/20
x = 68/20
x = 3.4
OP = 3.4, O9 = 5 - 3.4 = 1.6
R = \({\sqrt{9^v-x^v}}\)
= \({\sqrt{9^v-(3.4)^v}}\)
= \({\sqrt{92.56}}\)
9D = \({\sqrt{(92.56^v-(1.6)^v}}\)
= \({\sqrt{92.56^v-2.56}}\)
= \({\sqrt{90}}\)
CD = 2\({\sqrt{90}}\)
\({Now\ Checking \over}\)
OR = OQ + QR
= 1.6 + 5 = 6.6
OF = R = \({\sqrt{92.56}}\)
RF = \({\sqrt{92.56 - 43.56}} = {\sqrt{49}} = 7\)
So, own Soln is correct:
The length of the chord that lies halfway between them is 2√90 units.
+177 In order to ease solving process, I have created a diagram that represents the situation to scale. The diagram is below with points I have labeled arbitrarily where O is the center of the circle.
Since O is the center of the circle, OC = OJ = r, the radius of the circle. In addition, we know that \(\overline{\rm BH}\) is a perpendicular bisector of the chords, so \(\rm BC = 9\) and \(\rm HJ = 7\). Also, we can label \(\rm BO = BO\) and \(\rm OH = 10 - BO\). Putting all this information together enables us to find the radius of the circle.
\(r^2 = 9^2 + {\rm BO}^2 \\ r^2 = 7^2 + \left(10 - {\rm BO}\right)^2 \\ 81 + {\rm BO}^2 = 49 + 100 - 20{\rm BO} + {\rm BO}^2 \\ 20{\rm BO} = 68 \\ {\rm BO} = \frac{17}{5} \\ r^2 = 81 + \left(\frac{17}{5}\right)^2 = \frac{2025}{25} + \frac{289}{25} = \frac{2314}{25}\)
We know that BO = 5, so \({\rm OE} = 5 - {\rm BO} = 5 - \frac{17}{5} = \frac{8}{5}\). We can apply Pythagorean's Theorem one more time to find DF, the length of the desired chord.
\(r^2 = {\rm OE}^2 + {\rm EF}^2 \\ \frac{2314}{25} = \left(\frac{8}{5}\right)^2 + {\rm EF}^2 \\ {\rm EF}^2 = \frac{2314}{25} - \frac{64}{25} \\ {\rm EF}^2 = 90 \\ {\rm EF} = 3\sqrt{10} \\ {\rm DF} = 2{\rm EF} = 6 \sqrt{10} \approx 18.9737\)
