Question

Problem: Report Error 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 $a$.

Answer 1

The value of $\sqrt{a} = 2\sqrt{46}$ and a = 184

We let O be the centre, A₁ A A₂ , B₁ B B₂ represent the chords with length 10, 14 respectively.

Connecting the endpoints of the chords with the centre, we have several right triangles. However, we do not know whether the two chords are on the same side or different sides of the centre of the circle.

By the Pythagorean Theorem on Δ OBB₁,

we get x^2 + 7^2 = r^2

⇒ x = $\sqrt{r^{2} - 49 }$, where x is the length of the other leg. Now the length of the leg of Δ OAA₁ is either 6 + x or 6 - x depending whether or not A₁A₂, B₁B₂ are on the same side of the center of the circle:

                               $(6$ ± $\sqrt{r^{2} - 49 }$ ) + 5²  =  r²

                                 12 ± $12 \sqrt{r^{2} - 49}$ = 0

Only the negative works here (thus the two chords are on opposite sides of the center), and solving we get x=1, r =$5\sqrt{2}$. The leg formed in the right triangle with the third chord is 3 - x = 2, and by the Pythagorean Theorem again,

$(\frac{a}{2} )^{2} + 2^{2} = (5\sqrt{2} )^{2}$

⇒ $\sqrt{a} = 2\sqrt{46}$

a = 184

To learn more about Pythagoras theorem from the given link

brainly.com/question/343682

#SPJ4