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# HALP PLSSS

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The diagram shows two chords that are parallel. What is the length of the chord that lies halfway between them?

Aug 10, 2023

### 3+0 Answers

#2
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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 + x= 7v + 10v - 20x + xv

x = (7v + 10- 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.

Aug 11, 2023
#3
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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.

$$r^2 = {\rm OE}^2 + {\rm EF}^2 \\ \frac{2314}{25} = \left(\frac{8}{5}\right)^2 + {\rm EF}^2 \\ {\rm EF}^2 = \frac{2314}{25} - \frac{64}{25} \\ {\rm EF}^2 = 90 \\ {\rm EF} = 3\sqrt{10} \\ {\rm DF} = 2{\rm EF} = 6 \sqrt{10} \approx 18.9737$$

Aug 11, 2023
#5
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Thank you so much for your help!

ImAMathKid  Aug 11, 2023