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JEE Advanced 2014Paper 2
Exam
JEE Advanced
Year
2014
Paper
Paper 2
Subject
Mathematics
Chapter
Properties of Triangles

JEE Advanced 2014 · Paper 2 · Mathematics · Properties of Triangles

In a triangle the sum of two sides is x and the product of the same...

In a triangle the sum of two sides is x and the product of the same two sides is y. If x2−c2=y,{x}^{2} - {c}^{2} = y, (where c is the third side of the triangle) then the ratio of the inradius to the circumradius of the triangle is

Options

  1. A\dfrac{3y}{2x\left(x + c\right)}
  2. B\dfrac{3y}{2c\left(x + c\right)}
  3. C\dfrac{3y}{4x\left(x + c\right)}
  4. D\dfrac{3y}{4c\left(x + c\right)}

Correct answer

B. \dfrac{3y}{2c\left(x + c\right)}

Step-by-step solution

Let x=a+bx = a + b
y=aby = ab
x2−c2=y{x}^{2} - {c}^{2} = y
⇒a2+b2−c22ab=−12=cos⁡(120o)⇒ \dfrac{{a}^{2} + {b}^{2} - {c}^{2}}{2ab} = - \dfrac{1}{2} = \cos \left({120}^{o}\right)
⇒∠C=2π3⇒ ∠C = \dfrac{2\pi}{3}
⇒R=abc4∆,r=∆s⇒ R = \dfrac{abc}{4 ∆}, r = \dfrac{∆}{s}
⇒rR=4∆2s( abc )⇒ \dfrac{r}{R} = \dfrac{4{∆}^{2}}{s\left(\text{ abc }\right)} = \dfrac{4{\left[\dfrac{1}{2}absin \left(\dfrac{2\pi}{3}\right)\right]}^{2}}{\dfrac{x + c}{2}y . c}
rR=3y2c(x+c)\dfrac{r}{R} = \dfrac{3y}{2c\left(x + c\right)}
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