Question

x, y, z are sides of a triangle and x+y+z=16. How many such triangles are possible.

Answer 1

ANSWER:-

18

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Answer 2

Given:

x, y, z are sides of triangle and $x+y+z=16$.

To find how many triangles.

Triangle inequality

$x+y>z Eq1\\y+z>x Eq 2\\x+z>y Eq 3$

$x+y+z=16$

$x+y=16-z$

From Eq 1

$16-z>z$

$2z<16\\z<8$

Similarly,

$x<8,y<8$

Sum is 16 but $z<8$, Choose the values of x, y and z

Such that $x<8,y<8,z<8$

For $x=1$no values of y and z

For $x=2$, y and z are 7 and 7

For $x=3$, y and z are 6 and 7

For $x=4$, y and z are 5 and 7

For $x=5$, y and z are 4 and 7

For $x=6$, y and z are 3 and 7

For $x=7$, y and z are 2 and 7

So, total 6 triangles for x.

So as 6 for y and 6 for z.

Total triangles are 18.