Question

parallelogram pqrs has diagonals PR in SQ that intersect at T given s p equals 2 a + 5 + r q equals 5 a - 1 St equals 3 b - 3 + SQ equals 7 b - 9 what are the values of RQ and TQ

Answer 1

SP = 2a+5

RQ= 5a-1

ST = 3b-3

SQ = 7b-9

RQ=?

TQ=?

SP = RQ

2a+5 = 5a-1

Solve for a

5+1 = 5a-2a

6 = 3a

6/3 = a

2=a

RQ= 5a-1 = 5(2)-1 = 10-1 = 9

RQ= 9

ST + TQ = SQ

ST= TQ

TQ= 3b-3

3b-3+3b-3= 7b-9

Solve for b

6b-6 = 7b-9

-6+9 = 7b-6b

3=b

TQ = 3b-3= 3(3)-3= 9-3 =6

TQ= 6