Question

In adjoining figure 1.14 segPS⊥segRQsegQT⊥segPR. If RQ = 6, PS = 6 and PR = 12, then find QT.

Answer 1

by using formula of area of triangle

Answer 2

Answer:

In adjoining figure 1.14 segPS⊥segRQsegQT⊥segPR. If RQ = 6, PS = 6 and PR = 12, then QT =  3

Step-by-step explanation:

RQ =6

PS = 6

PR = 12

Area of Δ PQR = (1/2) PR * QT

= (1/2) * 12 * QT

= 6 QT

Area of Δ PQR  = Area of Δ PRS - Area of Δ PQS

=> Area of Δ PQR  = (1/2)RS * PS - (1/2) QS * PS

=> Area of Δ PQR  = (1/2)(RQ + QS) * PS - (1/2) QS * PS

=> Area of Δ PQR  = (1/2)(RQ) * PS

=> Area of Δ PQR  = (1/2)(6) * 6

=> Area of Δ PQR  = 18

6QT = 18

=> QT = 3