Question

in triangle pqr angle pqr is 90° QS is perpendicular to PR and ST is perpendicular to QR if PQ is equal to 6 cm and PS is equal to 3 cm find area of the triangle QST by area of the triangle RST

Answer 1

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
=
$\frac{ \frac{3 \sqrt{3} }{2} }{ \frac{9 \sqrt{3} }{2} } = \frac{1}{3}$

Hence, Required Ratio is 1/3 .

Answer 2

Answer:

Step-by-step explanation:

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

=

Hence, Required Ratio is 1/3 .