In right-angled triangle △PQR with ∠QPR = 90°, the hypotenuse QR = 26 cm. A point M lies on PR such that ∠PMR = 90° with PM = 6 cm and MR = 8 cm (so △PMR is right-angled at M). Find the area of △PQR (in cm^2).

Introduction / Context:This problem combines two right triangles. △PQR is right-angled at P, and a smaller right triangle △PMR sits along PR with legs PM = 6 and MR = 8. We use these to determine the legs of △PQR and hence its area.

Given Data / Assumptions:

• In △PQR, ∠QPR = 90° (right at P).
• QR = 26 cm (hypotenuse of △PQR).
• △PMR is right at M with PM = 6 cm and MR = 8 cm.
• M lies on PR, so PR is the hypotenuse of △PMR.

Concept / Approach:

• For △PMR: hypotenuse PR = sqrt(6^2 + 8^2) = 10 cm.
• In △PQR, legs are PQ and PR (right at P). With PR = 10 and hypotenuse QR = 26, use Pythagoras to get PQ.
• Area(△PQR) = (1/2) * PQ * PR.

Step-by-Step Solution:

Verification / Alternative check:Triple (10,24,26) is a scaled 5-12-13 triple (×2), confirming consistency. Area from legs is therefore 120 cm^2.

Why Other Options Are Wrong:

• 180, 150, 240 cm^2: Each implies incorrect leg lengths inconsistent with 10-24-26.
• None of these: Incorrect since we get an exact 120 cm^2.

Common Pitfalls:

• Misreading PR as 14 by adding 6 and 8; PR is hypotenuse √(6^2+8^2)=10.
• Using QR with wrong right-angle assumption; right angle is at P.

Final Answer:120 cm2