In a triangle PQR,S and T are the points on PQ and PR respectively if ST is perpendicular to PR, if area of triangle PQR is 320cm2
, PR=28cm, QS:PS=9:7, PT:TR=2:5 then find the length of ST?
Answers
Given : In a triangle PQR,S and T are the points on PQ and PR respectively if ST is perpendicular to PR, if area of triangle PQR is 320cm² , PR=28cm, QS:PS=9:7, PT:TR=2:5
To Find : the length of ST
Solution:
PR = 28 cm
PT : TR = 2 : 5
=> PT = (2/7) * 28 = 8 cm
& TR 20 cm
QS:PS = 9 : 7
let say PS = 7k & QS = 9k
Extend PQ and draw RM ⊥ PQ
ΔPTS ≈ ΔPRM ( as ∠P common , ∠PTS = ∠PRM = 90°)
=> PT / PR = PS/PM
=> 8 / 28 = 7k/PM
=> PM = 49k/2
PQ = 7K + 9k = 16K
Area of Δ PMR / Area of Δ PQR = PM/PQ
=> Area of Δ PMR / 320 = (49k/2)/16k
=> Area of Δ PMR = 490
Area of Δ PMR / Area of Δ PST = (PR/PT)²
=> 490 / Area of Δ PST = (7/2)²
=> Area of Δ PST = 40
(1/2) * PT * ST = 40
=> 8 * ST = 80
=> ST = 10 cm
length of ST = 10 cm
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