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 $a$ and $b$ be the lengths of two sides of a triangle.According to the given condition,$a+b=x$ and $ab=y$.Given $x^2-c^2=y$,we substitute $x=a

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $a$ and $b$ be the lengths of two sides of a triangle.According to the given condition,$a+b=x$ and $ab=y$.Given $x^2-c^2=y$,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,$\cos C = \frac{a^2+b^2-c^2}{2ab}$.Therefore,$\cos C = -\frac{1}{2}$.Since $C$ is an angle in a triangle,$C = 120^\circ$ or $\frac{2\pi}{3}$ radians.The circumradius $R$ is given by $R = \frac{c}{2\sin C}$.$R = \frac{c}{2\sin(120^\circ)} = \frac{c}{2(\frac{\sqrt{3}}{2})} = \frac{c}{\sqrt{3}}$.

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

Which of the following is true in a triangle $ABC$?

If $ABC$ is a right-angled triangle with $90^{\circ}$ at $C$ and $a > b$,then $\frac{a^2+b^2}{a^2-b^2} \sin (A-B) = $

The area of the triangle $ABC$ is $10\sqrt{3} \text{ cm}^2$,angle $B$ is $60^{\circ}$ and its perimeter is $20 \text{ cm}$. Then $\ell(AC) = $ (in $\text{ cm}$)

The sum of the radii of the inscribed and circumscribed circles for an $n$-sided regular polygon of side $a$ is:

In $\Delta ABC$,if $a = 2$,$b = 4$,and $\angle C = 60^\circ$,then $\angle A$ and $\angle B$ are equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module