Question

In an adjoining fig seg PS perpendicular seg RQ seg QT perpendicular seg PR If RQ = 6, PS = 6 and PR = 12 then find QT ​

Answer 1

Answer:

Step-by-step explanation:

RQ=6, PS=6 & PR=12..... (GIVEN)

AREA OF TRIANGLE =1/2x BASE x HEIGHT

AREA OF TRIANGLE =1/2 x 6 x 6

THERFORE, AREA OF TRIANGLE PQR =18 SQ UNITS.....(1)

ALSO, AREA OF TRIANGLE PQR =1/2 x PR x QT

THERFORE, 18 =1/2 x 12 x QT.... (FROM 1)

THEREFORE QT =18 DIVIDE BY 6

THEREFORE QT = 3

DONE.....

hope it helps

:)

Answer 2

Answer:

Given:

PS ⊥ RQ

QT ⊥ PR

RQ = 6, PS = 6 and PR = 12

With base PR and height QT,

A(△PQR)=½×PR×QT

With base QR and height PS,

A(△PQR)=½×QR×PS

∴A(△PQR)/A(△PQR)=½×PR×QT/½×QR×PS

⇒1=PR×QT/QR×PS

⇒PR×QT=QR×PS

⇒QT=QR×PS/PR

=6×6/12

=3

Hence, the measure of side QT is 3 units.