Published by:
CGP EDU Academic Team
Published on: August 14, 2026
Let P is any point inside the triangle ABC of side lengths 6, 5, 5 units and p 1 , p 2 , p 3 be the lengths of perpendiculars drawn from P to the sides of triangle. Find the maximum value of p 1 .p 2 .p 3 .
Text Solution
Verified by ExpertsThe correct answer is:
CHECK THE SOLUTION.
Sol. Area Δ Δ ABC = Area Δ Δ BCP + Area Δ Δ CPA + Area Δ Δ APB
12 = 3p 1 + 
p 2 + 
p 3
Applying AM ≥ ≥ G.M
.
──────────────────────────────────────────────────────────────────────────────────────────
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems
The vertices of a triangle OBC are O(0,0) B(–3,–1) and C(–1,–3). Find the equation of line parallel…
A variable line, drawn through the point of intersection of the straight lines = 1 and = 1, meet…
From the vertices A, B, C of a triangle ABC, perpendiculars AD, BE, CF are drawn to any straight li…
A triangle is formed by the lines whose equations are AB : x + y – 5 = 0, BC : x + 7y – 7 = 0 and …
The coordinates of the feet of from the vertices of a Δ on the opposite sides are (20, 25), (8, 16…
Let in Δ PAB, A is (0, 0), B is (a, 0) and P is variable such that ∠ ∠ PBA is equal to three times …