Question

In adjoining figure, seg PS ⊥ seg RQ, seg QT ⊥ seg PR. If RQ = 6, PS = 6 and PR = 12, then Find QT. 

Answer 1

`{:("seg PS ⊥ seg RQ"), ("seg QT ⊥ seg PR"), ("RQ = 6, PS = 6 and PR = 12"):}   }"Given"`

With base PR and height QT,

A(Δ PQR) = `1/2` × PR × QT   ...(1)

With base QR and height PS,

A(Δ PQR) = `1/2` × QR × PS   ...(2)   

The ratio of areas of two triangles is equal to the ratio of the product of their base and corresponding heights.

From (1) and (2),

`"A(Δ PQR)"/"A(Δ PQR)" = (1/2 × "PR × QT")/(1/2 × "QR ×PS")`

`1/1 = ("PR"×"QT")/("QR"×"PS")`

PR × QT = QR × PS

QT = `("QR × PS")/"PR"`

QT = `(6 × 6)/12`

QT = `3/1`

Hence, the measure of side QT is 3 units.