Question

The lengths of the sides of a right triangle are the integers a, b and c, and these integers have no
common factor. If a < b < c and (C-a): b = 5:7, then find the value of (a+b+c)​/3

Answer 1

The value of (a+b+c)/3 is 28

Step-by-step explanation:

Given a, b, c integers are lengths of sides of right angled triangle

And, a < b < c

We know that in a right angled triangle, hypotenuse is the greatest side

Thus, by pythagoras theorem

$c^2=a2+b^2$

$\implies c^2-a^2=b^2$

$\implies (c-a)(c+a)=b^2$   ............. (1)

But given that

$\frac{c-a}{b}=\frac{5}{7}$

or, $c-a=\frac{5}{7}b$    .................... (2)

Therefore,  from (1)

$\frac{5}{7}b(c+a)=b^2$

$\implies c+a=\frac{7}{5}b$   .....................(3)

Adding equations (2) and (3)

$2c=\frac{5b}{7}+\frac{7b}{5}$

$\implies 2c=\frac{74b}{35}$

$\implies c=\frac{37b}{35}$

$\implies \frac{c}{b}=\frac{37}{35}$

Since there are no common factors between c and b we can safely say that

$c=37, b=35$

Again from (3)

$\frac{37b}{35}+a=\frac{7b}{5}$

$\implies a=\frac{7b}{5}-\frac{37b}{35}$

$\implies a=\frac{12b}{35}$

$\implies \frac{a}{b}=\frac{12}{35}$

Since there are no common factors between a and b

Therefore,

$a=12, b=35$

Thus,

$\frac{a+b+c}{3}=\frac{12+35+37}{3}=\frac{84}{3}=28$

Hope this answer is helpful.

Know More:

Q: The lengths of the sides of a right triangle are the integers a, b and c, and these integers have no  common factor. If a < b < c and (C-a): b = 4:5, then find the value of (b + c-a).

Click Here: https://brainly.in/question/10665256

Answer 2

HELLO DEAR,

 

Given that length of side of a right triangle are the integers a,b,c and these have no common  factor.

Given that length of side of a right triangle are the integers a,b,c and these have no common  factor.And a < b < c and (c-a) : b = 5 : 7.

To find the value of (a + b +  c )/3.

 SOLUTION : it is a right triangle So,

$\sf{C^{2} = a^{2} + b^{2}}$

$\sf{C=\sqrt{a^{2} +b^{2} } ............( 1 )}$                    

   and

$\sf{(c-a) : b= 5: 7}$

$\sf{(c-a)/b = 5/7..........( 1 )}$                      

 by putting the value equation i) in equation ii)                            

$\sf{7c - 7a = 5b}$

$\sf{7\sqrt{a^{2} + b^{2}}  - 7a = 5b}$

$\sf{7\sqrt{a^{2} + b^{2}} = 5b + 7b}$                        

 On squaring both side,                            

$\sf{49( a^{2} + b^{2} )= 25b^{2} + 49a^{2}  + 70 ab}$

$\sf{49b^{2} = 25b^{2} + 70 ab}$

$\sf{24b^{2} = 70 ab}$

$\sf{\frac{a}{b} = \frac{24}{70}}$

$\sf{\frac{a}{b} = \frac{12}{35}}$

        From above a= 12 , b= 35.    By putting the value of a= 12 and b= 35 in above relation ( c- a ) / b= 5/7    

 => (C- 12) /35 = 5/ 7      

=> 7c- 84 = 175      

=> 7c = 259        

=> c= 259/ 35        

=> c= 37.

So, a= 12 ,b= 35 ,c= 37.

Therefore,

( a+ b+ c) / 3 = (12+35+37)/3                                        = 84/3                                      

=> (a + b + c)  = 28. Answer

I HOPE IT'S HELP YOU DEAR,

THANKS