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pa help Guyss with solution #9&10


Thanks!!​


Pa Help Guyss With Solution 9amp10Thanks class=

Sagot :

Answer:

Step-by-step explanation:

1) QT = 12

2) m∠PQS = 90 - m∠RQS = 90 - 65 = 25°

3) m∠QRS = 90°

4) ST=15, PR= 2ST = 2(15) = 30

5) QR = 15, then RS = 15

6) m∠QRS = 120, m∠QRP = 60°

7) m∠TPQ + m∠TQP = 90

8) m∠RTS = 90

9) x =?

similar triangles:

 AP/(x+5) = RP / (3x-2)

since RP = 2AP

AP(3x-2) = 2(AP)(x+5)

3xAP - 2AP = 2xAP + 10AP

3xAP - 2xAP = 10AP + 2AP

 xAP= 12AP

 x = 12AP/AP= 12

OR  

 midpoint is 1/2 the size of the parallel side/base of the bigger triangle.

 3x-2 = 2(x+5)

 3x -2 = 2x+10

 3x-2x = 10+2

 x = 12

10) AB = (12+5) = (12+5) = 17