Question

h.w due 2/10/2026

1. find x

$\frac{7×3}{4}= x$

1. find x
2. $ac= 11$

$cm=6$
$dm=10$
find $mb$
$\frac{6×5}{10}= 3$
$(dm)(mb)=(cm)(ma)$

1. find x

$\frac{8×7}{14}= x$
$x=4$
$(do)(oc)=(ao)(ob)$

Explanation:

Step1: Apply chord segment theorem

For two intersecting chords, the products of segment lengths are equal: $7 \times 3 = x \times (x+4)$

Step2: Simplify to quadratic equation

$21 = x^2 + 4x$
$x^2 + 4x - 21 = 0$

Step3: Factor the quadratic

$(x+7)(x-3) = 0$

Step4: Solve for valid x

$x = 3$ (discard $x=-7$, length can't be negative)

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Step1: Apply chord segment theorem

$3x^2 \times 3 = x \times (x-1)$

Step2: Simplify to quadratic equation

$9x^2 = x^2 - x$
$8x^2 + x = 0$

Step3: Factor and solve

$x(8x + 1) = 0$
$x = 0$ (discard $x=-\frac{1}{8}$, length can't be negative/zero, so no valid positive solution; if assuming non-zero, $x=-\frac{1}{8}$ is extraneous)

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Step1: Calculate length of MA

$MA = AC - CM = 11 - 6 = 5$

Step2: Apply chord segment theorem

$CM \times MA = DM \times MB$
$6 \times 5 = 10 \times MB$

Step3: Solve for MB

$30 = 10MB$
$MB = \frac{30}{10} = 3$

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Step1: Apply chord segment theorem

$AD \times DB = AO \times OC$
$14 \times x = 8 \times 7$

Step2: Solve for x

$14x = 56$
$x = \frac{56}{14} = 4$

Answer:

1. $x=3$
2. No valid positive solution (or $x=0$ if allowing zero length)
3. $MB=3$
4. $x=4$