Question

3. In following figure, seg PS i seg RQ, seg QT 1
seg PR. If RQ = 6, PS = 6 and PR = 12, then find
QT​

Answer 1

Given:

PS ⊥ RQ

QT ⊥ PR

RQ = 6, PS = 6 and PR = 12

With base PR and height QT, A(Δ PQR) = `1/2 xx "PR" xx "QT"`

With base QR and height PS, A(Δ PQR) =`1/2 xx "QR" xx "PS"`

Let we know that "The ratio of areas of two triangles is equal to the ratio of the product of their base and corresponding heights."

⇒ In Δ PQR base RQ , heights = PS

`("A" (triangle "PQR"))/("A" (triangle "PQR")) = (1/2xx"PR"xx"QT")/(1/2xx"QR"xx"PS")`

`=> 1 = ("PR"xx"QT")/("QR"xx"PS")`

`=> "PR"xx"QT" = "QR"xx"PS"`

QT = `("QR"xx"PS")/"PR"`

`= (6 xx 6)/12`

`= 36/12`

QT = `3/1`

Hence, the measure of side QT is 3 units.

Answer 2

$\huge \tt \pink {answer}$

Given:

PS ⊥ RQ

QT ⊥ PR

RQ = 6, PS = 6 and PR = 12

With base PR and height QT, A(Δ PQR) =

12×PR×QT

With base QR and height PS, A(Δ PQR) =

12× QR×PS

Let we know that "The ratio of areas of two triangles is equal to the ratio of the product of their base and corresponding heights."

⇒ In Δ PQR base RQ , heights = PS

A(△PQR)A(△PQR)=12×PR×QT12×

⇒1=PR×QTQR×PS

⇒PR×QT=QR×PS

QT = QR×PSPR

=6×612

=3612

QT = 31

Hence, the measure of side QT is 3 units.