Question

From the figure PT is an altitude of the triangle PQR in which PQ = 25 cm, PR = 17 cm and PT = 15 cm. If QR = $x$ cm.
Calculate $x$.

Answer 1

From the figure, we have $QR=QT+TR$.
To find : QT and TR.
In the right angled triangle PTQ,
$\angle PTQ={90}^o$ [PT is attitude]
By Pythagoras Theorem, $P{Q}^2=P{T}^2+Q{T}^2$
$\therefore P{Q}^2-P{T}^2=Q{T}^2$
$\therefore Q{T}^2={25}^2-{15}^2=625-225=400$ ...(1)
Similarly, in the right angled triangle PTR,
by Pythagoras Theorem, $P{R}^2=P{T}^2+T{R}^2$
$\therefore T{R}^2=P{R}^2-P{T}^2$
$={17}^2-{15}^2$
$=289-225=64$
$TR=\sqrt{64}=8cm$ ...(2)
From (1) and (2)
QR = QT + TR = 20 + 8 = 28 cm.