What kind of triangle has the relations between a, b, c, sides: (3^1/2)*a - (3^1/2)*b = c a + b = (3^1/2)*c

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william1941's profile pic

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

For the triangle we assume that the sides are a, b and c. We are given :

(3^1/2)*a - (3^1/2)*b = c and a + b = (3^1/2)*c

a + b = (3^1/2)*c

=> c = a / (3^1/2) + b/ (3^1/2)

So (3^1/2)*a - (3^1/2)*b = c

=> (3^1/2)*a - (3^1/2)*b = a / (3^1/2) + b/ (3^1/2)

=> 3 ( a - b) = a + b

=> 3a - 3b = a + b

=> 2a - 4b = 0

=> a = 2b

Now substituting a= 2b in (3^1/2)*a - (3^1/2)*b = c

=> (3^1/2)*2b - (3^1/2)*b = c

=> (3^1/2)*b = c

=> c = (3^1/2)*b

So we have a= 2b and c = (3^1/2)*b

Now b^2+ c^2 = b^2 + 3b^2 = 4b^2, which we have already derived.

Therefore the triangle is a right angled triangle.

Top Answer

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll try to determine the length of b and the length of c, considering the length of a side.

We'll multiply the second relation with the value (3^1/2) and after that we'll add the equivalent obtained relation to the first one.

(3^1/2)*a + (3^1/2)*b + (3^1/2)*a - (3^1/2)*b = 3*c + c

We'll group the same terms:

2*(3^1/2)*a = 4*c

(3^1/2)*a = 2 *c

c= [(3^1/2)*a]/2

With the c value written in function of "a" value, we'll go in the second relation and substitute it:

a + b = [(3^1/2)*(3^1/2)*a]/2

a + b = 3*a/2

We'll have the same denominator on the left side of the equality:

2*a + 2*b = 3*a

2*b = 3*a - 2*a

2*b = a

b = a/2

If the triangle is a right one, then, using the Pythagorean theorem, we'll have the following relation between the sides of triangle:

a^2 = b^2 + c^2

Now, we have to plug in the values of "b" and "c", in the relation above:

a^2 = a^2/4 + 3*a^2/4

a^2 = 4*a^2/4

a^2 = a^2

We've shown that the equality is true, so the triangle is right, where "a" is hypotenuse and "b","c" are cathetus.

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

3^(1/2)a - 3^(1/2)b = c........(1)

a+b = 3^(1/2)............(2)

(1) could be rewritten as:

a-b  = c/3^(1/2)..............(3).

Adding (2) and (3) we get:

2a = {3^(1/2) +1/3(1/2) }c

a =  {(3^(1/2) + 1/3^(1/2)}c/2 =  4c/sqrt3

Similarly b = {3^(1/2) - 1/3^(1/2)}c/2 = 2c/sqrt3.

 

 

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