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JEE Main 201911 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
11 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Straight Lines

JEE Main 2019 · 11 January · Morning Shift · Mathematics · Straight Lines

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

In a triangle, the sum of lengths of two sides is x\mathrm{x} and the product of the lengths of the same two sides is yy. if x2−c2=y,x^{2}-c^{2}=y, where cc is the length of the third side of the triangle, then the circumradius of the triangle is

Options

  1. A32y\frac{3}{2} y
  2. Bc3\frac{c}{\sqrt{3}}
  3. Cc3\frac{c}{3}
  4. Dy3\frac{y}{\sqrt{3}}

Correct answer

B. c3\frac{c}{\sqrt{3}}

Step-by-step solution

Let two sides of triangle are aandba and b. a+b=xab=yx2−c2=y⇒(a+b)2−c2=ab⇒(a+b−c)(a+b+c)=ab⇒2(s−c)(2s)=ab⇒4s(s−c)=ab⇒s(s−c)ab=14⇒cos⁡2c2=14⇒cos⁡c=−12\Rightarrowc=120∘\begin{aligned}& a+b=x \\ & a b=y \\ & x^{2}-c^{2}=y \Rightarrow(a+b)^{2}-c^{2}=a b \\ \Rightarrow &(a+b-c)(a+b+c)=a b \\ \Rightarrow & 2(s-c)(2 s)=a b \\ \Rightarrow & 4 s(s-c)=a b \\ \Rightarrow & \frac{s(s-c)}{a b}=\frac{1}{4}\\ \Rightarrow & \cos ^{2} \frac{c}{2}=\frac{1}{4}\\ \Rightarrow & \cos c=-\frac{1}{2} \Rightarrowc=120^{{}^{\circ}} \end{aligned}\thereforeAreaoftriangleis,Area of triangle is,\Delta=\frac{1}{2} a b\left(\sin 120^{{}^{\circ}}\right)=\frac{\sqrt{3}}{4} a b\because \quad R=\frac{a b c}{4 \Delta}\because \quad \mathrm{R}=\frac{a b c}{\sqrt{3} a b}=\frac{c}{\sqrt{3}}
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