In a triangle XYZ, let x,y,z be the lengths of sides opposite to the angles X,Y,Z, respectively and 2s = x + y + z. If s-x/4=s-y/3=s-z/2 and area of in circle of the triangle XYZ is 8π/2, then- [JEE Adv 2016 P1]

area of the triangle XYZ is 6{6}

the radius of circumcircle of the triangle XYZ is {35}{6{6}}

Test Index

Properties of Triangle

Section: Mathematics

Question : 7 of 15

Marks: +1, -0

2s = x + y + z

\frac{s-x}{4}=\frac{s-y}{3}=\frac{s-z}{2}

\frac{8\pi}{2}

Solution:

Let

\frac{s-x}{4}=\frac{s-y}{3}=\frac{s-z}{2}=k

S-x=4k\Rightarrow y+z-x=8k

S-y=3k\Rightarrow x+z-y=6k

S-z=2k\Rightarrow x+y-z=4k

\Rightarrow x=5k,\ y=6k,\ z=7k

\Rightarrow 3s-(x+y+z)=9k

s=9k

Let r be in radius

\Rightarrow \pi r^2=\frac{8\pi}{3}

\pi\left(\frac{\Delta}{s}\right)^2=\frac{8\pi}{3}

\Delta=\sqrt{\frac{8}{3}}\cdot S

\sqrt{S(S-x)(S-y)(S-y)}=\sqrt{\frac{8}{3}}\cdot S

\sqrt{9k\cdot 4k\cdot 3k\cdot 2k}=\sqrt{\frac{8}{3}}\cdot 9k

\sqrt{24\cdot 9k^2}=\sqrt{\frac{8}{3}}9k

k=1

\Rightarrow x=5,\ y=6,\ z=7

\Delta=\sqrt{\frac{8}{3}}\cdot 9k=\sqrt{\frac{8}{3}}\cdot 9=6\sqrt{6}

R = circumradius

=\frac{xyz}{4\Delta}=\frac{5\cdot 6\cdot 7}{4\cdot 6\sqrt{6}}=\frac{35}{4\sqrt{6}}

\frac{\sin X}{2}\frac{\sin Y}{w}\frac{\sin Z}{2} = \frac{r}{4R} = \frac{\frac{\Delta}{s}}{\frac{xyz}{\Delta}} = \frac{\Delta^2}{s\cdot xyz} = \frac{36\cdot 6}{9\cdot 5\cdot 6\cdot 7} = \frac{4}{35}

\cos Z = \frac{x^2+y^2-z^2}{2xy} = \frac{25+36-49}{2\cdot 5\cdot 6} = \frac{1}{5}

\sin^2\frac{X+Y}{2} = \cos^2\frac{Z}{2} = \frac{1+\cos Z}{2} = \frac{1+\frac{1}{5}}{2} = \frac{3}{5}

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