In a ∆PQR, PQ = PR and PT is perpendicular to QR. If PQ = 17 cm, PT = 15 cm, then what is the measure (in cm) of QR?

Correct — D. PQ = PR makes ∆PQR isosceles, and the perpendicular from the apex to the base also bisects that base.

Rule: PT ⊥ QR with PQ = PR ⇒ QT = TR

In right triangle PTQ, PQ is the hypotenuse:
QT² = PQ² − PT² = 17² − 15²
= 289 − 225 = 64
QT = 8 cm

QR = 2 × QT = 2 × 8 = 16 cm → option (d).

Why the others are wrong

• (a)
  8 — 8 is QT, half the base. The perpendicular bisects QR, so what the question asks for is 2 × 8.
• (b)
  32 — 32 doubles the half-base twice over. QR = 2 × QT = 16, and nothing here calls for a factor of 4.
• (c)
  24 — 24 needs each half of the base to be 12. That would make PQ = √(12² + 15²) = √369 ≈ 19.2 cm, not the 17 cm the stem fixes.

In an isosceles triangle the perpendicular dropped from the apex to the base is at once the median, the altitude and the angle bisector. That one fact turns half the figure into the whole of it.

Here it splits ∆PQR into two congruent right triangles with hypotenuse 17 and one leg 15, and 8, 15, 17 is a Pythagorean triple, so the remaining side comes out whole.

The base is then twice the leg you found, never the leg itself.

Key facts

• In an isosceles triangle the perpendicular from the apex to the base bisects that base.
• 8² + 15² = 17², so 8, 15, 17 is a Pythagorean triple.
• QT = TR = 8 cm here, which makes QR = 16 cm.

Common traps

• Stopping at QT = 8 and answering the half-base
• Treating 17 as a leg and computing √(289 + 225) instead of the difference
• Reading PT as one of the equal sides rather than as the perpendicular

No directly related past PYQ was found.