Question

practice
example 1
use the figure at the right.

1. if ( ab = 6 ), ( bc = 4 ), and ( ae = 9 ), find ( ed ).
2. if ( ab = 12 ), ( ac = 16 ), and ( ed = 5 ), find ( ae ).

example 2
determine whether ( overline{nr} parallel overline{pq} ). justify your answer.

1. ( pm = 18 ), ( pn = 6 ), ( qm = 24 ), and ( rm = 16 )
2. ( qm = 31 ), ( rm = 21 ), and ( pm = 4pn )

example 3
( overline{vr} ), ( overline{vz} ), and ( overline{zr} ) are midsegments of ( \triangle uwy ). find the value of ( x ).

1. figure of triangle with midsegments
2. figure of triangle with midsegments and angle ( 57^circ )
3. figure of triangle with midsegments and length ( 2.7 )
4. figure of triangle with angles ( 44^circ ), ( 76^circ ) and angle ( x^circ )

example 4

1. maps in mika’s town, cay street and bay street are parallel. find the value of ( x ), the distance from cay street to bay street along earl street.

figure of parallel streets with distances 1 km, 1.4 km, 0.8 km

Explanation:

Step1: Solve for ED (Q1)

By Triangle Proportionality Theorem:
$\frac{AE}{ED}=\frac{AB}{BC}$
Substitute values: $\frac{9}{ED}=\frac{6}{4}$
Cross-multiply: $6 \cdot ED = 9 \cdot 4$
$ED = \frac{36}{6} = 6$

Step2: Solve for AE (Q2)

First find $BC = AC - AB = 16 - 12 = 4$
By Triangle Proportionality Theorem:
$\frac{AE}{ED}=\frac{AB}{BC}$
Let $AE = x$, substitute values: $\frac{x}{5}=\frac{12}{4}$
$x = 5 \cdot 3 = 15$

Step3: Check $\overline{NR} \parallel \overline{PQ}$ (Q3)

Use converse of Triangle Proportionality Theorem:
Check $\frac{PN}{PM}=\frac{RM}{QM}$
$\frac{6}{18}=\frac{1}{3}$, $\frac{16}{24}=\frac{2}{3}$
$\frac{1}{3}
eq \frac{2}{3}$, so not parallel.

Step4: Check $\overline{NR} \parallel \overline{PQ}$ (Q4)

Let $PN = x$, so $PM = 4x$
$\frac{PN}{PM}=\frac{x}{4x}=\frac{1}{4}$
$\frac{RM}{QM}=\frac{21}{31}$
$\frac{1}{4}
eq \frac{21}{31}$, so not parallel.

Step5: Find x (Q5, midsegment)

Midsegment is half the parallel side:
$x = 2 \times 25 = 50$

Step6: Find x (Q6, midsegment)

Midsegment creates congruent angles:
$x = 57$

Step7: Find x (Q7, midsegment)

Midsegment is half the parallel side:
$x = \frac{27}{2} = 13.5$

Step8: Find x (Q8, midsegment)

Use triangle angle sum and midsegment:
First, $\angle VRW = 76^\circ$ (alternate interior), $\angle VWR = 44^\circ$
$x = 180 - 76 - 44 = 60$

Step9: Find x (Q9, parallel lines)

Set up proportion for similar triangles:
$\frac{x}{1.4}=\frac{0.8}{1}$
$x = 1.4 \times 0.8 = 1.12$

Answer:

1. $ED = 6$
2. $AE = 15$
3. $\overline{NR}$ is not parallel to $\overline{PQ}$ (ratios are not equal)
4. $\overline{NR}$ is not parallel to $\overline{PQ}$ (ratios are not equal)
5. $x = 50$
6. $x = 57$
7. $x = 13.5$
8. $x = 60$
9. $x = 1.12$ km