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Question
AB is a diameter of a circle and C is any point on the circle. Show that the area of ∆ABC is maximum, when it is isosceles.
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Solution
Let AB be the diameter and C be any point on the circle with radius r.
∠ACB = 90° ......[angle in the semi-circle is 90°]
Let AC = x
∴ BC = `sqrt("AB"^2 - "AC"^2)`
⇒ BC = `sqrt((2"r")^2 - x^2)`
⇒ BC = `sqrt(4"r"^2 - x^2)` ....(i)
Now area of ∆ABC
A = `1/2 xx "AC" xx "BC"`
⇒ A = `1/2 x * sqrt(4"r"^2 - x^2)`
Squaring both sides, we get
A2 = `1/4 x^2 (4"r"^2 - x^2)`
Let A2 = Z
∴ Z = `1/4 x^2(4"r"^2 - x^2)`
⇒ Z = `1/4(4x^2"r"^2 - x^4)`
Differentiating both sides w.r.t. x, we get
`"dZ"/"dx" = 1/4 [8x"r"^2 - 4x^3]` ....(ii)
For local maxima and local minima `"dZ"/"dx"` = 0
∴ `1/4 [8x"r"^2 - 4x^3]` = 0
⇒ `x[2"r"^2 - x^2]` = 0
x ≠ 0
∴ 2r2 – x2 = 0
⇒ x2 = 2r2
⇒ x = `sqrt(2)"r"`
= AC
Now from equation (i) we have
BC = `sqrt(4"r"^2 - 2"r"^2)`
⇒ BC = `sqrt(2"r"^2)`
⇒ BC = `sqrt(2)"r"`
So AC = BC
Hence, ∆ABC is an isosceles triangle.
Differentiating equation (ii) w.r.t. x, we get
`("d"^2"Z")/("dx"^2) = 1/4 [8"r"^2 - 12x^2]`
Put x = `sqrt(2)"r"`
∴ `("d"^2"Z")/("dx"^2) = 1/4 [8"r"^2 - 12 xx 2"r"^2]`
= `1/4[8"r"^2 - 24"r"^2]`
= `1/4 xx (-16"r"^2)`
= `-4"r"^2 < 0` maxima
Hence, the area of ∆ABC is maximum when it is an isosceles triangle.
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