Atriangle has 2 sides that equal 5 inches. the other side equals 4 inches

Isosceles triangle

Step-by-step explanation:

It is assumed that the triangle is a right angle triangle, implying that Pythagoras theorem becomes applicable.

If the first side is represented by x and the second side by y,

x + y = 5

The hypotenuse is always the longest side hence it can be presumed to be 4 since both x and y cannot be more than 5.

Therefore using pythagoras equation:

hyp^2 = sqrt(opp^2 + adj^2)

this implies that

therefore

16 = sqrt (x^2 + y^2) ----------------- (i)

since x + y = 5, y = 5 - x ------------------- (ii)

substituting (ii) into eqn (i)

16 = sqrt ( x^2 + (5-X)^2)

16 = sqrt( x^2 + 25 -10x + x^2)

16 = sqrt( x^2 + x^2 - 10x + 25)

16 = sqrt( 2x^2 - 10x + 25)

squaring both sides

256 = 2x^2 - 10x + 25

this can be refined as

2x^2 - 10x - 231 = 0

Solving using quadratic equation:

where a = 2, b = -10 and c = -231

x = 13.53  or -8.53.

Since x is a side of a triangle that cannot be negative,

x = 13.53 but x can never be greater than 5 since x + y = 5

This implies that the triangle is not a right angle triangle; it is most likely an  isosceles triangle with two sides of equal length such that x = y = 2.5 inches while the longest side; the hypotenuse is 4 inches.

substituting for x in

0 is its x and y coordinate