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QJEEM21Mac538: JEE Main 2021 Mathematics - Class 11 JEEM21Mac538 Solution: The triangle of maximum area that can be inscribed in a given…

The triangle of maximum area that can be inscribed in a given circle of radius 'r' is :

Multiple Choice Options

  • A An equilateral triangle having each of its side of length 3\sqrt 3 r.
  • B An equilateral triangle of height 2r3{{2r} \over 3} .
  • C A right angle triangle having two of its sides of length 2r and r.
  • D An isosceles triangle with base equal to 2r.

Step-by-Step Verified Solution

Area of triangle ABC

JEE Main 2021 (Online) 26th February Evening Shift Mathematics - Properties of Triangle Question 19 English Explanation

A=12×BC×AMA = {1 \over 2} \times BC \times AM

=12×2r2−x2×(r+x)= {1 \over 2} \times 2\sqrt {{r^2} - {x^2}} \times (r + x)

A=(r+x)r2−x2A = (r + x)\sqrt {{r^2} - {x^2}}

dAdx=r2−x2−xr2−x2×(r+x){{dA} \over {dx}} = \sqrt {{r^2} - {x^2}} - {x \over {\sqrt {{r^2} - {x^2}} }} \times (r + x)

=r2−x2−rx−x2r2−x2=r2−rx−2x2r2−x2=−(x+r)(2x−r)r2−x2= {{{r^2} - {x^2} - rx - {x^2}} \over {\sqrt {{r^2} - {x^2}} }} = {{{r^2} - rx - 2{x^2}} \over {\sqrt {{r^2} - {x^2}} }} = {{ - (x + r)(2x - r)} \over {\sqrt {{r^2} - {x^2}} }}

dAdx=0⇒x=r2{{dA} \over {dx}} = 0 \Rightarrow x = {r \over 2}

Sign change of dAdx{{dA} \over {dx}} at x=r2x = {r \over 2}

⇒\Rightarrow A has maximum at x=r2x = {r \over 2}

BC=2r2−x2=3rBC = 2\sqrt {{r^2} - {x^2}} = \sqrt 3 r ,

AM=r+12rAM = r + {1 \over 2}r = 32r{3 \over 2}r

⇒AB=AC=3r\Rightarrow AB = AC = \sqrt 3 r

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