Question

topic 2: parallelograms

1. if ebcd is a parallelogram, eb = 16, ed = 25, bf = 11, ec = 34, m∠bed = 55°, m∠cdb = 67°, and m∠bce = 24°, find each missing measure.

bc = ____ m∠edc = ____
bd = ____ m∠ebd = ____
fc = ____ m∠bec = ____
cd = ____ m∠dbc = ____

1. find m∠n.
2. find m∠r.

1. in parallelogram abcd, if ed = 7x - 13 and bd = 16x - 38, find bd.

topic 3: rectangles

1. if abcd is a rectangle, ad = 9, ac = 22, and m∠bca = 66°, find each missing measure.

bc = ____ m∠adc = ____
ab = ____ m∠bac = ____
bd = ____ m∠cdb = ____
ec = ____ m∠aeb = ____

1. if pqrs is a rectangle, pr = 9x + 1, and qs = 13x - 11, find tr.
2. if defg is a rectangle, m∠deg = (4x - 5)°, and m∠fge = (6x - 21)°, find m∠dge.

Explanation:

Problem 14

Step1: Use parallelogram side property

In parallelogram $EBCD$, $BC=ED$, $CD=EB$.
$BC=25$, $CD=16$

Step2: Use diagonal bisector property

Diagonals bisect each other: $BD=2\times EB=2\times16=32$; $FC=EC-BF=34-11=23$

Step3: Find $\angle EDC$

$\angle EDC=\angle BED=55^\circ$ (alternate interior angles)

Step4: Find $\angle EBD$

$\angle EBD=\angle CDB=67^\circ$ (alternate interior angles)

Step5: Calculate $\angle BEC$

In $\triangle BCE$, $\angle BEC=180^\circ-\angle EBC-\angle BCE$. $\angle EBC=\angle EBD+\angle DBC$, first find $\angle DBC=\angle BED=55^\circ$ (alternate interior angles), so $\angle EBC=67^\circ+55^\circ=122^\circ$. $\angle BEC=180^\circ-122^\circ-24^\circ=34^\circ$

Step6: Calculate $\angle DBC$

$\angle DBC=\angle BED=55^\circ$ (alternate interior angles)

Step1: Set parallel angles equal

In parallelogram $KLMN$, consecutive angles are supplementary: $(8x+17)+(12x-39)=180$

Step2: Solve for $x$

$20x-22=180 \implies 20x=202 \implies x=10.1$

Step3: Find $\angle N$

$\angle N=\angle L=8x+17=8\times10.1+17=80.8+17=97.8^\circ$

Step1: Set parallel angles equal

In parallelogram $RSTU$, alternate interior angles: $5x-8=14x-4$

Step2: Solve for $x$

$-9x=4 \implies x=-\frac{4}{9}$ (Note: This indicates a typo, assuming $14x-41$ instead: $5x-8=14x-41 \implies 9x=33 \implies x=\frac{11}{3}$)

Step3: Find $\angle R$

$\angle R=180^\circ-(5x-8)=180^\circ-(5\times\frac{11}{3}-8)=180^\circ-(\frac{55}{3}-\frac{24}{3})=180^\circ-\frac{31}{3}=169.67^\circ$
(Using corrected problem: if $14x-41$, $\angle R=169.67^\circ$; original problem has invalid $x$, so assuming typo correction)

Answer:

$BC=25$, $BD=32$, $FC=23$, $CD=16$
$\angle EDC=55^\circ$, $\angle EBD=67^\circ$, $\angle BEC=34^\circ$, $\angle DBC=55^\circ$

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Problem 15