In triangle ABC,if AB: BC: CA = 6: 4: 5,then | vedclass

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

Question

(B) Let the sides be $c = 6k$,$a = 4k$,and $b = 5k$. The semi-perimeter $s = \frac{6k + 4k + 5k}{2} = \frac{15k}{2}$. Using Heron's formula,the area $

Vedclass · Accepted Answer

$16: 7$

Vedclass · Answer

(B) Let the sides be $c = 6k$,$a = 4k$,and $b = 5k$. The semi-perimeter $s = \frac{6k + 4k + 5k}{2} = \frac{15k}{2}$. Using Heron's formula,the area $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{\frac{15k}{2} 	imes \frac{7k}{2} 	imes \frac{5k}{2} 	imes \frac{3k}{2}} = \sqrt{\frac{1575k^4}{16}} = \frac{15\sqrt{7}k^2}{4}$. The circumradius $R = \frac{abc}{4\Delta} = \frac{(6k)(4k)(5k)}{4 	imes \frac{15\sqrt{7}k^2}{4}} = \frac{120k^3}{15\sqrt{7}k^2} = \frac{8k}{\sqrt{7}}$. The inradius $r = \frac{\Delta}{s} = \frac{15\sqrt{7}k^2}{4} 	imes \frac{2}{15k} = \frac{\sqrt{7}k}{2}$. Therefore,$\frac{R}{r} = \frac{8k}{\sqrt{7}} 	imes \frac{2}{\sqrt{7}k} = \frac{16}{7}$.

In $\triangle ABC$,if $AB: BC: CA = 6: 4: 5$,then $R: r =$

Similar Questions

In a $\Delta ABC,$ if $b = 20, c = 21$ and $\sin A = 3/5$,then $a = $

If $A, B, C$ are the angles of a triangle with $\tan \frac{A}{2}=\frac{1}{3}$ and $\tan \frac{B}{2}=\frac{2}{3}$,then the value of $\tan \frac{C}{2}$ is:

In a triangle $ABC$, if $r_1=2 r_2=3 r_3$, then $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=$

Sides of a triangle are $2 \ cm$,$\sqrt{6} \ cm$,and $(\sqrt{3} + 1) \ cm$. The smallest angle of the triangle is .....$^o$

If the sides of a triangle are $3, 5, 7$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module