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Show that the triangle of maximum are...

Show that the triangle of maximum area that can be inscribed in a given circle is an equilateral triangle.

Text Solution

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Let 2r be the base and h be the heightof triangle, which is inscribed in a circle of radius R
Area of triangle=`1/2`(base)(height)
`A=1/2(2r)(h)=rh…(1)`

Area being positive quantity, A will be maximum or minimum if `A^(2)` is
maximum or minimum.
`Z = A^(2) = r^(2)h^(2) "…………"(2)`
Now In triangle OLB `BL^(2) = OB^(2) - OL^(2)`
In `DeltaOBD`
`Z = A^(2) = r^(2)h^(2) , r^(2) = R^(2) - underset("put in (2)")((h - R)^(2) rArr r^(2) = 2hR - h^(2))`
`Z = h^(2)(2hR - h^(2))`
`rArr Z = (2h^(3)R - h^(4))`
`rArr (dZ)/(dh) = 6h^(2) R - 4h^(3)"......."(3)`
For maxima/minima`(dZ)/(dh) = 0`
`rArr 6h^(2) R - 4h^(3) = 0`
`rArr 4 R = 4h (h ne 0)`
`rArr h = (3R)/(2)`
differentiating (3) w.r.t . h
`rArr (d^(2)Z)/(dh^(2)) = 12 R - 12 h^(2)`
`rArr (d^(2)Z)/(dh^(2))]_(h = (3R)/(2)) = 12((3R)/(2))R - 12 ((3R)/(2))^(2)`
`= 18^(2) - 27R^(2) = - ve`
so, `Z = A^(2)` is maxima when `h = (3R)/(2)`
`rArr A` is maximum when `h = (3R)/(2)`
when `h = (3R)/(2), r^(2) = 2hR - h^(2) = 2R. (3R)/(2) - ((3R)/(2))^(2)`
`r^(2) = (3R^(2))/(4)`
`r = (sqrt(3)R)/(2)`
`tan theta = h/r = ((3R)/(2))/((sqrt(3)R)/(2)) = sqrt(3) theta = (pi)/(3)`
Triangle ABC is equilateral triangle
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Knowledge Check

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

    A
    An isoscles triangle with base equal to 2r
    B
    An equilateral of height `(2r)/3`
    C
    an equilateral trianlge having each of its side of length `sqrt3r`
    D
    `A right angle triangle having two of its sides of length 2r and r.
  • The triangle of maximum area inscribed in a circle is :

    A
    equilateral
    B
    isosceles
    C
    right-angled
    D
    none of these
  • Rectangle of maximum area that can be inscribed in an equilateral triangle of side awill have area=

    A
    `(a^2sqrt3)/2`
    B
    `(a^2sqrt3)/4`
    C
    `(a^2 sqrt3)/8`
    D
    None
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