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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

#1
+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

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

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

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

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

Mar 19, 2020

#1
+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

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

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

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

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

heureka Mar 19, 2020
#2
+111321
+2

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

CPhill  Mar 19, 2020
#3
+24949
+1

Thank you, CPhill !

heureka  Mar 19, 2020
#4
0

Guest Mar 19, 2020
#5
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
+111321
+1

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

CPhill  Mar 19, 2020