in ∆PQR, angle PQR =90° .QS perpendicular to PR and ST perpendicular to QR. if PQ =6 cm and PS =3 cm, find the area∆QST/area∆RST​

2 thoughts on “in ∆PQR, angle PQR =90° .QS perpendicular to PR and ST perpendicular to QR. if PQ =6 cm and PS =3 cm, find the area∆QST/area∆RST​”

1. Answer:

   Final Answer : 1/3

   Steps and Understanding :

   1) In PQS,

   cos (P) = PS/PQ

   => cos(P) = 3/6 = 1/2

   => P = 60°

   2) Now, finding other angles as shown in pic by using properties of triangle.

   3) In PQS,

   tan(60°) = QS/PS

   => QS = 3* tan(60°) = 3√3cm

   In QST,

   sin(60°) = ST/QS

   => ST =( 3√3) * (√3/2 )

   => ST = 9/2 cm.

   cos(60°) = QT/QS

   => 1/2 = QT/QS

   => QT = 3√3 * (1/2)

   => QT = 3√3/2 cm

   4) In RST,

   tan(30°) = ST/TR

   => 1/√3 = (9/2)/ TR

   => TR = 9√3/2 cm.

   5) Now, Finally

   area ( QST) / area (RST) =

   ( 1/2 * QT* ST )/( 1/2 * RT* ST)

   =QT/RT

   (3√3/2)/(3√3/2)

   =1/3

   Hope this will help you

   Plz mark this answer as brainliest

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