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two parallel chords in a circle have lengths 10 and 14, and the distance between them is 6. The chord parallel to these chords and midway between them is of

length sqrt(a). Find the value of a.

 

 Mar 18, 2020

Best Answer 

 #1
avatar+24949 
+1

two parallel chords in a circle have lengths 10 and 14, and the distance between them is 6.

The chord parallel to these chords and midway between them is of

length \(\sqrt{a}\). Find the value of a.

Phythagora's

\(\begin{array}{|lrcll|} \hline (1) & x^2+7^2 &=& r^2 \\ (2) & (3-x)^2 + \left(\dfrac{\sqrt{a}}{2}\right)^2 &=& r^2 \\ (3) & 5^2 + \Big(3+(3-x)\Big)^2 &=& r^2 \\ \hline \end{array} \)

 

\(\begin{array}{|lrcll|} \hline \mathbf{x=\ ?} \\ \hline (1)=(3): & r^2 = x^2+7^2 &=& 5^2 + \Big(3+(3-x)\Big)^2 \\ & x^2+7^2 &=& 5^2 + \Big(3+(3-x)\Big)^2 \\ & x^2+7^2 &=& 5^2 + (6-x)^2 \\ & x^2+7^2 &=& 5^2 + 6^2-12x+x^2 \\ & 7^2 &=& 5^2 + 6^2-12x \\ & 12x &=& 5^2 + 6^2 - 7^2 \\ & 12x &=& 12 \quad | \quad : 12 \\ & \mathbf{x}&=& \mathbf{1} \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline \mathbf{r^2=\ ?} \\ \hline (1): & r^2 &=& x^2+7^2 \quad | \quad \mathbf{x=1} \\ & r^2 &=& 1^2+7^2 \\ & \mathbf{r^2}&=& \mathbf{50} \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline \mathbf{a=\ ?} \\ \hline (2): & (3-x)^2 + \left(\dfrac{\sqrt{a}}{2}\right)^2 &=& r^2 \\ & (3-x)^2 + \dfrac{a}{4} &=& r^2 \quad | \quad \mathbf{x=1},\ \mathbf{r^2=50} \\ & (3-1)^2 + \dfrac{a}{4} &=& 50 \\ & 2^2 + \dfrac{a}{4} &=& 50 \\ & 4 + \dfrac{a}{4} &=& 50 \quad | \quad -4 \\ & \dfrac{a}{4} &=& 46 \quad | \quad * 4 \\ & \mathbf{a}&=& \mathbf{184} \\ \hline \end{array}\)

 

laugh

 Mar 19, 2020
 #1
avatar+24949 
+1
Best Answer

two parallel chords in a circle have lengths 10 and 14, and the distance between them is 6.

The chord parallel to these chords and midway between them is of

length \(\sqrt{a}\). Find the value of a.

Phythagora's

\(\begin{array}{|lrcll|} \hline (1) & x^2+7^2 &=& r^2 \\ (2) & (3-x)^2 + \left(\dfrac{\sqrt{a}}{2}\right)^2 &=& r^2 \\ (3) & 5^2 + \Big(3+(3-x)\Big)^2 &=& r^2 \\ \hline \end{array} \)

 

\(\begin{array}{|lrcll|} \hline \mathbf{x=\ ?} \\ \hline (1)=(3): & r^2 = x^2+7^2 &=& 5^2 + \Big(3+(3-x)\Big)^2 \\ & x^2+7^2 &=& 5^2 + \Big(3+(3-x)\Big)^2 \\ & x^2+7^2 &=& 5^2 + (6-x)^2 \\ & x^2+7^2 &=& 5^2 + 6^2-12x+x^2 \\ & 7^2 &=& 5^2 + 6^2-12x \\ & 12x &=& 5^2 + 6^2 - 7^2 \\ & 12x &=& 12 \quad | \quad : 12 \\ & \mathbf{x}&=& \mathbf{1} \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline \mathbf{r^2=\ ?} \\ \hline (1): & r^2 &=& x^2+7^2 \quad | \quad \mathbf{x=1} \\ & r^2 &=& 1^2+7^2 \\ & \mathbf{r^2}&=& \mathbf{50} \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline \mathbf{a=\ ?} \\ \hline (2): & (3-x)^2 + \left(\dfrac{\sqrt{a}}{2}\right)^2 &=& r^2 \\ & (3-x)^2 + \dfrac{a}{4} &=& r^2 \quad | \quad \mathbf{x=1},\ \mathbf{r^2=50} \\ & (3-1)^2 + \dfrac{a}{4} &=& 50 \\ & 2^2 + \dfrac{a}{4} &=& 50 \\ & 4 + \dfrac{a}{4} &=& 50 \quad | \quad -4 \\ & \dfrac{a}{4} &=& 46 \quad | \quad * 4 \\ & \mathbf{a}&=& \mathbf{184} \\ \hline \end{array}\)

 

laugh

heureka Mar 19, 2020
 #2
avatar+111321 
+2

Very impressive, heureka....I had no idea  how to approach this !!!

 

 

cool cool cool

CPhill  Mar 19, 2020
 #3
avatar+24949 
+1

Thank you, CPhill !

 

laugh

heureka  Mar 19, 2020
 #4
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0

thanks for your help

Guest Mar 19, 2020
 #5
avatar
0

I think you can use "Ptolemy's Theorem" to solve this question!
{[14^2 + 6^2] + [10^2 + 6^2]} / 2 = 184
The length of the midway chord is:
Sqrt(184).

 Mar 19, 2020
 #6
avatar+111321 
+1

Thx, Guest...I haven't heard of this before !!!!......I'm going to have to investigate "Ptolemy's Theorem".....

 

 

 

cool cool cool

CPhill  Mar 19, 2020

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