Question

PQR = 90°, seg Qs | side PR, PS = 4, PQ = 6. Find x,y, and z.​

Answer 1

Given :  in triangle PQR angle PQR=90,seg QS is perpendicular segPR

To Find :  x,y,z

Solution:

in ΔQSR  &  ΔPSQ

∠QSR = ∠PSQ = 90°

∠SRQ  = ∠SQP   ( as ∠SRQ = 90°  -∠P  , ∠SQP = 90°  -∠P)

=> ΔQSR ≈  ΔPSQ    ( AA  Criteria )

QS/PS = SR/SQ  = QR/PQ

=> x/10 = 8/x  =  y/z

=> x² = 80

=> x = 4√5

=> 4√5/10 =  y/z

=> 2/√5 = y/z

Area of Triangle

= (1/2) * y * z  = (1/2) * PR * SQ

PR = PS + SR = 10 + 8 = 18

=> yz  = 18x

=> yz = 18(4√5)

=> yz = 72√5

yz = 72√5

y/z  = 2/√5

=> y² = 144 = 12

Z = 6√5

x =  4√5

y = 12

z = 6√5

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