sides of a triangle are 3x+4y,4x+3y,5x+5y units

[sumit saurav]

the sides of a triangle are 3x+4y,4x+3y,5x+5y units,where x>0 y>0 the triangle is:
a)right angled
b)acute angled
c)obtuse angled
d)isoceles

 

[pka]

the sides of a triangle are 3x+4y,4x+3y,5x+5y units,where x>0 y>0 the triangle is:
a)right angled
b)acute angled
c)obtuse angled
d)isoceles

Do you know the law of cosines? \(\displaystyle \cos \left( \theta \right) = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}\)

If the cosine is positive for all combinations then the triangle is acute.

If the cosine is negative for any combination then the triangle is obtuse.

If the cosine is zero for any combination then the triangle is a right triangle.

By combination I mean rearranging \(\displaystyle a,b,~\&c\).

You might try \(\displaystyle a=(3x+4y),~b=(4x+3y),~\&~c=(5x+5y)\) first.

 

[lookagain]

the sides of a triangle are 3x+4y,4x+3y,5x+5y units,where x>0 y>0 the triangle is:
a)right angled
b)acute angled
c)obtuse angled
d)isoceles

Suppose you want to cut down on algebra. There's symmetry in the coefficients
of the x and y terms. Pick two different values for x and y, say 1 and 2, respectively,
to see if you can eliminate the "isosceles" choice.

Here is my test example: x = 1, y = 2 yields sides of 11, 10, and 15 units

So, it's not necessarily isosceles. Eliminate choice d). (If you had chosen x = y = 1,
choice d) would be in the running until looking at further information.)


Here is the conclusion:

A) If the sum of the squares of the two shortest sides is greater than the square of the longest side, it is an acute triangle.

B) If the sum of the squares of the two shortest sides equals the square of the longest side, it is a right triangle.

C) If the sum of the squares of the two shortest sides is less than the square of the longest side, it is an obtuse triangle.



You can have your answer if you apply the above rules to my example of the triangle with sides of 11, 10, and 15 units.

 

[sumit saurav]

thnks you very much

 

[Bob Brown MSEE]

Do you know the law of cosines? \(\displaystyle \cos \left( \theta \right) = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}\)
You might try \(\displaystyle a=(3x+4y),~b=(4x+3y),~\&~c=(5x+5y)\) first.

This is the only case to test if you notice that c is the longest side. Along with lookagain's comment, testing any unequal values for x and y will be sufficient.

Only one test is required because there is no choice in the answer set that allows for more than one of the remaining two cases.