Question

more practice with circumcenter & incenter directions: if w is the circumcenter of δqrs find each measure. 1. rs 2. rq 3. ws 4. qw 5. tw 6. iw directions: if p is the circumcenter of δabc, find each measure. 7. find bc. 8. find ap. 9. find dp. 10. find fc.

Explanation:

Step1: Solve for x (QES circumcenter)

Circumcenter implies $QW=EW=SW$. Use $QW=EW$:
$$32x - 24 = 19$$
$$32x = 43$$
$$x = \frac{43}{32}$$

Step2: Calculate RS

RS is twice the segment $rS$:
$$RS = 2 \times 21 = 42$$

Step3: Calculate TQ

TQ is half of $QE$:
$$TQ = \frac{32x}{2} = 16x = 16 \times \frac{43}{32} = \frac{43}{2} = 21.5$$

Step4: Calculate WS

WS equals QW:
$$WS = 32x - 24 = 19$$

Step5: Calculate QV

QV equals QE:
$$QV = 32x = 32 \times \frac{43}{32} = 43$$

Step6: Calculate TW

TW is $QW - WT$. First find WT: $WT=EW - 19=0$? No, $WT = QW - TQ = (32x-24) - 16x = 16x -24 = 21.5 -24 = -2.5$ (absolute value 2.5, but correct: $TW = TQ - WT$ correction: $TQ=16x$, $QW=32x-24$, so $TW = QW - TQ = 16x -24 = 21.5 -24 = -2.5$, so $TW=2.5$)

Step7: Calculate WV

WV equals 19 (given on diagram)

Step8: Solve for x (Problem7, ABC circumcenter)

Circumcenter: $PE=PF$, so $13x-51=10x-33$
$$13x-10x = 51-33$$
$$3x=18$$
$$x=6$$
BC is twice $EC$: $EC=10x-33=60-33=27$, so $BC=2\times27=54$

Step9: Solve for x (Problem8, ABC circumcenter)

Circumcenter: $BE=FC$, so $11x-14=6x+1$
$$11x-6x=14+1$$
$$5x=15$$
$$x=3$$
$AP=BP=CP$, $BP=11x-14=33-14=19$, so $AP=19$

Step10: Solve for x (Problem9, ABC circumcenter)

Circumcenter: $AD=DP$, so $2x+14=8x-52$
$$14+52=8x-2x$$
$$66=6x$$
$$x=11$$
$DP=2x+14=22+14=36$

Step11: Solve for x (Problem10, ABC circumcenter)

Circumcenter: $AD=BD$, so $3x+2=5x-16$
$$2+16=5x-3x$$
$$18=2x$$
$$x=9$$
$FC=14$ (given, and $FC=AF$, so $FC=14$)

Answer:

1. $RS=42$
2. $TQ=21.5$
3. $WS=19$
4. $QV=43$
5. $TW=2.5$
6. $WV=19$
7. $BC=54$
8. $AP=19$
9. $DP=36$
10. $FC=14$