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A triangle with perimeter 7has integer side lengths. What is the maximum possible area of such a triangle?
Asked by vikasg13.hardware | 16 Jun, 2018, 06:25: AM
answered-by-expert Expert Answer

a +b + c = 7

Sum of any 2 arms must be greater than the third one , i.e. a+b >c, b+c >a , c+a >b

Area of triangle which is maximum. 7 can be summed up of 3 integers as:

i. 7= 1+1+5, as 1+1<5 , ∆ formation not possible.

ii.7=2+1+4, as 2+1<4,∆ formation not possible.

iii. 7=3+1+3, it is possible

iv.7 = 2+2+3, here it is also possible.

For iii)

area of ∆ =

begin mathsize 16px style square root of 3.5 open parentheses 3.5 minus 3 close parentheses open parentheses 3.5 minus 3 close parentheses open parentheses 3.5 minus 1 close parentheses end root..... h e r o n apostrophe s space f o r m u l a equals fraction numerator square root of 35 over denominator 4 end fraction equals 1.4790 end style

For iv)

Area of ∆ =

begin mathsize 16px style square root of 3.5 open parentheses 3.5 minus 2 close parentheses open parentheses 3.5 minus 2 close parentheses open parentheses 3.5 minus 3 close parentheses end root equals fraction numerator 3 square root of 7 over denominator 4 end fraction equals 1.9843 end style

Maximum possible area of ∆ is 3√7/4 sq.unit

 
 
Answered by Arun | 18 Jun, 2018, 10:33: AM

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