Question

topic 11 midsegments
in the diagram below, \\(\overline{mn}\\), \\(\overline{np}\\), and \\(\overline{pm}\\) are midsegments.

1. name all parallel segments:

\\(\underline{\quad\quad\quad\quad\quad}\\), \\(\underline{\quad\quad\quad\quad\quad}\\), \\(\underline{\quad\quad\quad\quad\quad}

1. if \\(mp = 17\\), \\(lk = 24\\) and \\(pn = 13\\), find each measure.

a) \\(jk = \underline{\quad\quad}\\)
b) \\(mn = \underline{\quad\quad}\\)
c) \\(jl = \underline{\quad\quad}\\)
d) perimeter of \\(\triangle jkl\\): \\(\underline{\quad\quad}

1. solve for \\(x\\).
2. find \\(dh\\).
3. solve for \\(x\\).
4. solve for \\(x\\).
5. if \\(m\angle dec = (12x - 3)\degree\\), \\(m\angle bce = (7x - 26)\degree\\), and \\(m\angle dae = 72\degree\\), find each angle measure.

\\(m\angle dec = \underline{\quad\quad}\\)
\\(m\angle bce = \underline{\quad\quad}\\)
\\(m\angle ade = \underline{\quad\quad}\\)
\\(m\angle edb = \underline{\quad\quad}\\)
\\(m\angle dbc = \underline{\quad\quad}\\)

Explanation:

1. Identify parallel segments

By Triangle Midsegment Theorem:
$\overline{MN} \parallel \overline{LK}$, $\overline{MP} \parallel \overline{JK}$, $\overline{PN} \parallel \overline{JL}$

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2. Use midsegment properties

Step1: JK = 2×MP

$JK = 2 \times 17 = 34$

Step2: MN = ½×LK

$MN = \frac{1}{2} \times 24 = 12$

Step3: JL = 2×PN

$JL = 2 \times 13 = 26$

Step4: Perimeter of △JKL

$\text{Perimeter} = JK + LK + JL = 34 + 24 + 26 = 84$

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3. Solve for x (midsegment length)

Step1: Set midsegment = ½ base

$3x + 11 = \frac{1}{2}(9x - 14)$

Step2: Multiply both sides by 2

$6x + 22 = 9x - 14$

Step3: Rearrange to solve for x

$22 + 14 = 9x - 6x \implies 36 = 3x \implies x = 12$

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4. Find DH

Step1: Set midsegment = ½ base

$5x + 3 = \frac{1}{2}(15x - 29)$

Step2: Multiply both sides by 2

$10x + 6 = 15x - 29$

Step3: Rearrange to solve for x

$6 + 29 = 15x - 10x \implies 35 = 5x \implies x = 7$

Step4: Calculate DH

$DH = 15(7) - 29 = 105 - 29 = 76$

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5. Solve for x (corresponding angles)

Step1: Set angles equal (parallel segments)

$4x + 21 = 13x - 15$

Step2: Rearrange to solve for x

$21 + 15 = 13x - 4x \implies 36 = 9x \implies x = 4$

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6. Solve for x (midsegment & right angle)

Step1: Set midsegment angle = 90°

$8x - 46 = 90$

Step2: Solve for x

$8x = 90 + 46 \implies 8x = 136 \implies x = 17$

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7. Calculate angle measures

Step1: Set ∠DEC = ∠BCE (alternate interior)

$12x - 3 = 7x - 26$

Step2: Solve for x

$12x - 7x = -26 + 3 \implies 5x = -23$
Correction: Use ∠DAE = ∠DEC (corresponding angles)
$12x - 3 = 72$

Step3: Solve for x

$12x = 72 + 3 \implies 12x = 75 \implies x = 6.25$

Step4: Find m∠DEC

$m\angle DEC = 12(6.25) - 3 = 75 - 3 = 72^\circ$

Step5: Find m∠BCE

$m\angle BCE = 7(6.25) - 26 = 43.75 - 26 = 17.75^\circ$

Step6: Find m∠ADE

$m\angle ADE = m\angle DEC = 72^\circ$ (corresponding)

Step7: Find m∠EDB

$m\angle EDB = 180^\circ - 72^\circ = 108^\circ$

Step8: Find m∠DBC

$m\angle DBC = m\angle BCE = 17.75^\circ$ (alternate interior)

Answer:

1. $\overline{MN} \parallel \overline{LK}$, $\overline{MP} \parallel \overline{JK}$, $\overline{PN} \parallel \overline{JL}$

2.
a) $JK = 34$
b) $MN = 12$
c) $JL = 26$
d) Perimeter of $\triangle JKL = 84$

1. $x = 12$
2. $DH = 76$
3. $x = 4$
4. $x = 17$

7.
$m\angle DEC = 72^\circ$
$m\angle BCE = 17.75^\circ$
$m\angle ADE = 72^\circ$
$m\angle EDB = 108^\circ$
$m\angle DBC = 17.75^\circ$