Question

The ΔPQR is right-angled at P, and PN is an altitude. If QN = 12 in and NR = 6 in, find PN, PQ, PR.

Answer 1

Answer:

PN = 6√2 in

PQ =  6√6 in

PR =  6√3 in

Step-by-step explanation:

Please see the attached image where the triangle and the known values are labeled.

From Right triangle altitude theorem, we have

$PN=\sqrt{QN\cdot NR}\\PN=\sqrt{12\cdot6}\\PN=6\sqrt2$

Now, in right angle triangle PNQ

$PQ=\sqrt{(6\sqrt2)^2+12^2}\\PQ=\sqrt{72+144}\\PQ=\sqrt{216}\\PQ=6\sqrt6$

Similarly, in triangle PNR,

$PR=\sqrt{(6\sqrt2)^2+6^2}\\PQ=\sqrt{72+36}\\PQ=\sqrt{108}\\PQ=6\sqrt3$

Therefore, we have

PN = 6√2 in

PQ =  6√6 in

PR =  6√3 in

Answer 2

In the right triangle PQR:
Hypotenuse: QR = QN + NR = 12 + 6 = 18 cm
PN² = QN · NR
PN² = 12 · 6 = 72
PN = √72 = √(36 · 2 ) = 6√2 cm
PQ² = (6√2)² + 12² = 72 + 144 = 216
PQ = √216 = √(36 · 6 ) = 6√6 cm
PR² = 18² - (6√6)² = 324 - 216 = 108
PR = √108 = √(36 · 3) = 6√3 cm
Answer:
PN = 6√2 cm,
PQ = 6√6 cm,
PR = 6√3 cm.