Question

what is the length of the longer of the two chords shown?
length of longer chord = $square$ units
the circle has two intersecting chords, with segment lengths: 4, $x-7$, 12, $x+3$

Explanation:

Step1: Apply chord segment theorem

When two chords intersect in a circle, the products of the lengths of their segments are equal:
$$12 \times 4 = (x-7)(x+3)$$

Step2: Simplify and solve for $x$

First calculate the left side, then expand and rearrange into quadratic form:
$$48 = x^2 - 4x - 21$$
$$x^2 - 4x - 69 = 0$$
Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ where $a=1$, $b=-4$, $c=-69$:
$$x = \frac{4 \pm \sqrt{16 + 276}}{2} = \frac{4 \pm \sqrt{292}}{2} = \frac{4 \pm 2\sqrt{73}}{2} = 2 \pm \sqrt{73}$$
Since length cannot be negative, we take $x = 2 + \sqrt{73} \approx 2 + 8.544 = 10.544$ (the other solution would make $x-7$ negative).

Step3: Calculate chord lengths

First chord length: $12 + 4 = 16$ units
Second chord length: $(x-7)+(x+3) = 2x - 4$
Substitute $x = 2 + \sqrt{73}$:
$$2(2 + \sqrt{73}) - 4 = 4 + 2\sqrt{73} - 4 = 2\sqrt{73} \approx 17.088$$

Step4: Compare lengths

$2\sqrt{73} \approx 17.088 > 16$, so the second chord is longer.

Answer:

$2\sqrt{73}$ (or approximately 17.09) units