In a triangle PQR, S is a point on the side QR such that PS ⊥ QR, then which of the following options is true?

Correct — D. PS ⊥ QR cuts the triangle into two right triangles that share the leg PS.

In right ΔPSQ: PQ² = PS² + QS²
So PS² = PQ² − QS²

In right ΔPSR: PR² = PS² + SR²
So PS² = PR² − SR²

Equate the two: PQ² − QS² = PR² − SR²
Move QS² and SR² across: PR² + QS² = PQ² + SR² → option (d)

Why the others are wrong

• (a)
  PS² + QS² = PQ² + PR² — PS² + QS² is PQ² by Pythagoras in ΔPSQ. The option then says PQ² = PQ² + PR², which forces PR = 0.
• (b)
  PS² + PR² = PQ² + QR² — QR is not a leg of either right triangle. Expanding gives 2PS² + SR² on the left and PS² + QS² + (QS + SR)² on the right, which differ in general.
• (c)
  PQ² + PR² = QS² + SR² — Adding the two Pythagoras equations gives PQ² + PR² = QS² + SR² + 2PS². The option drops the 2PS², so it holds only when PS = 0.

A perpendicular from a vertex to the opposite side splits a triangle into two right triangles with a common leg — here PS.

Write Pythagoras in each and equate the shared leg:
PQ² − QS² = PS² = PR² − SR²

Rearranged, each slant side pairs with the base segment of the other right triangle: PR with QS, PQ with SR. That is PR² + QS² = PQ² + SR².

Key facts

• When PS ⊥ QR, PQ² − QS² = PS² = PR² − SR².
• Rearranged: PR² + QS² = PQ² + SR².
• Adding the two right-triangle equations instead gives PQ² + PR² = QS² + SR² + 2PS².

Common traps

• Adding the two Pythagoras equations instead of equating PS², which leaves a stray 2PS²
• Pairing each slant side with its own base segment (PQ with QS) instead of the far one
• Treating QR as a side of a right triangle, when the right angles sit at S

No directly related past PYQ was found.