In a triangle,the sum of lengths of two sides | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the sides of the triangle be $a, b,$ and $c$. Given $a + b = x$ and $ab = y$.Given the condition $x^2 - c^2 = y$,we substitute $x = a + b$:$(a

Vedclass · Accepted Answer

$\frac{c}{\sqrt{3}}$

Vedclass · Answer

(B) Let the sides of the triangle be $a, b,$ and $c$. Given $a + b = x$ and $ab = y$.Given the condition $x^2 - c^2 = y$,we substitute $x = a + b$:$(a + b)^2 - c^2 = ab$$a^2 + b^2 + 2ab - c^2 = ab$$a^2 + b^2 - c^2 = -ab$Dividing by $2ab$,we get:$\frac{a^2 + b^2 - c^2}{2ab} = -\frac{1}{2}$By the Law of Cosines,$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$,so $\cos C = -\frac{1}{2}$.Thus,$C = 120^\circ$ or $\frac{2\pi}{3}$ radians.Then $\sin C = \sin(120^\circ) = \frac{\sqrt{3}}{2}$.Using the Sine Rule,$\frac{c}{\sin C} = 2R$,where $R$ is the circumradius:$R = \frac{c}{2 \sin C} = \frac{c}{2(\frac{\sqrt{3}}{2})} = \frac{c}{\sqrt{3}}$.

In a triangle,the sum of lengths of two sides is $x$ and the product of the lengths of the same two sides is $y$. If $x^2 - c^2 = y$,where $c$ is the length of the third side of the triangle,then the circumradius of the triangle is

Similar Questions

Let $ABC$ be a triangle in which $AB=BC$. Let $X$ be a point on $AB$ such that $AX:XB=AB:AX$. If $AC=AX$,then the measure of $\angle ABC$ equals (in $^{\circ}$)

In a $\Delta ABC$,$\frac{\cos^2 \left( \frac{B - C}{2} \right)}{(b + c)^2} + \frac{\sin^2 \left( \frac{B - C}{2} \right)}{(b - c)^2} = $ (in $/ a^2$)

In a triangle $ABC$,with usual notations,if $a=5, b=4$,and $\cos(A-B)=\frac{31}{32}$,then $c=$

In a $\Delta ABC$,if $\cos 3A + \cos 3B + \cos 3C = 1$,then:

In a right-angled $\triangle ABC$, the measures of the angles are in an Arithmetic Progression ($A$.$P$.). If its smallest side is $4 \text{ units}$, then the area of $\triangle ABC$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module