IIT JEE , AIEEE Forum - Mathematics , Trignometry

sboosy

r=4RsinA/2sinB/2sinC/2

but maximum value of sinA/2sinB/2sinC/2 is 1/8

so

2r<=R

r =

/s

so

2 (

/s)<=R

<=sR/2

s=4RcosA/2cosB/2cosC/2

replacing R from above

< = s²/8cosA/2cosB/2cosC/2

s = 3+5+c/2

s² = (c+8)²/4

so

< = (c+8)²/(4*8cosA/2cosB/2cosC/2)

but max of cosA/2cosB/2cosC/2 is 3

3/8

substituing this

we get

<= (c²+16c+64)/12

3

which is option A

nivedh_89

the above method is the perfect way 2 solve this prob..........but for JEE, use the substitution method.......
now when 3 n 5 r 2 sides of the triangle, c maybe 4 (this is just 1 posssibility).....

now area is 6......!!!! (right angled triangle)

now substitute value of c in the options.....u will get option A....!!!!!!

sriram.a

For a triangle of given perimeter the area is maximum for a equilateral triangle.

In the triangle perimeter is (c+5+3)=c+8

For the equilateral triangle side is (c+8)/3

so, area of given triangle is $<\frac{\sqrt{3}}{4}a^2$ $\Rightarrow$ $\frac{\sqrt{3}}{4}(\frac{c+8}{3})^2=\frac{c^2+16c+64}{12\sqrt{3}}$

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