Question

part 9: trapezoid
abcd is an isosceles trapezoid with bases ab and cd, and median ef. use the given
information to solve each problem.

1. if dc = 30 and ab = 42, find ef.
2. if ∠a = 5x and ∠d = 4x, find the value of x.
3. if ef = x + 5 and ab + cd = 4x + 6, find ef.

part 10: coordinate geometry

1. determine what kind of quadrilateral pqrs is based on the following vertices. justify

your answer. p(2,3), q(5,9), r(11,6), s(8,0)

Explanation:

Step 1: Apply trapezoid median formula

The median of a trapezoid is the average of its two bases, so $EF = \frac{AB + DC}{2}$.
Substitute $AB=42$, $DC=30$:
$EF = \frac{42 + 30}{2}$

Step 2: Calculate the value

$EF = \frac{72}{2} = 36$

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Step 1: Use consecutive angle property

In a trapezoid, consecutive angles between the bases are supplementary: $\angle A + \angle D = 180^\circ$.
Substitute $\angle A=5x$, $\angle D=4x$:
$5x + 4x = 180^\circ$

Step 2: Simplify and solve for $x$

$9x = 180^\circ$
$x = \frac{180^\circ}{9} = 20^\circ$

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Step 1: Relate median to sum of bases

The median $EF = \frac{AB + CD}{2}$. Substitute $EF=x+5$, $AB+CD=4x+6$:
$x + 5 = \frac{4x + 6}{2}$

Step 2: Simplify the equation

Multiply both sides by 2:
$2(x + 5) = 4x + 6$
$2x + 10 = 4x + 6$

Step 3: Solve for $x$

$10 - 6 = 4x - 2x$
$4 = 2x$
$x = 2$

Step 4: Calculate $EF$

Substitute $x=2$ into $EF=x+5$:
$EF = 2 + 5 = 7$

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Step 1: Calculate side lengths (distance formula)

Distance formula: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

• $PQ = \sqrt{(5-2)^2 + (9-3)^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}$
• $QR = \sqrt{(11-5)^2 + (6-9)^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5}$
• $RS = \sqrt{(8-11)^2 + (0-6)^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}$
• $SP = \sqrt{(2-8)^2 + (3-0)^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5}$

Step 2: Calculate slope of sides

Slope formula: $m = \frac{y_2-y_1}{x_2-x_1}$

• Slope of $PQ$: $\frac{9-3}{5-2} = 2$
• Slope of $QR$: $\frac{6-9}{11-5} = -\frac{1}{2}$
• Slope of $RS$: $\frac{0-6}{8-11} = 2$
• Slope of $SP$: $\frac{3-0}{2-8} = -\frac{1}{2}$

Step 3: Analyze slopes for perpendicularity

Check product of adjacent slopes: $2 \times (-\frac{1}{2}) = -1$, so adjacent sides are perpendicular.

Answer:

1. $EF = 36$
2. $x = 20$
3. $EF = 7$
4. Quadrilateral PQRS is a square. All four sides are congruent (length $3\sqrt{5}$), and adjacent sides are perpendicular (product of slopes is -1), so it meets the criteria of a square.