Assuming you mean side AB is 5:
If angle B is the right angle, side AC is the hypotenuse and is of length 6.
If angle A is the right angle, side BC is the hypotenuse and is of length sqrt(52 + 62) ~= 7.81
Angle C cannot be the right angle as then side AB would be the hypotenuse but the hypotenuse is the longest side and side AB is shorter than AC.
triangle ABC with rigth at C
Triangle ABC would be an isosceles. An isosceles triangle is defined as having two sides of equal length. This would also mean, then, that two angles in the triangle are also the same.
Let consider the right triangle ABC with hypotenuse AB and heigth AC then base is BC Pythagorean theorem states that AB2=AC2+BC2 so BC2=AB2-AC2 then BC=sqrt(AB2-AC2)
If CB is the hypotenuse, then AB measures, √ (62 - 52) = √ 11 = 3.3166 (4dp) If AB is the hypotenuse then it measures, √ (62 + 52) = √ 61 = 7.8102 (4dp)
Use Pythagoras' theorem to find the length of the 3rd side
triangle ABC with rigth at C
7.2
If the sides of the triangle are 20 and 15 then by using Pythagoras' theorem the length of the hypotenuse works out as 25 units of measurement.
Each leg is the square root of 180.5 which is about 13.435 rounded to 3 decimal places
61
In order to find length BC the length of AC or length of the hypotenuse must be given
A triangle has 3 sides and so the length of bc will depend on its perimeter.
15
You can make it whatever length you like by selecting the other sides appropriately.
Triangle ABC would be an isosceles. An isosceles triangle is defined as having two sides of equal length. This would also mean, then, that two angles in the triangle are also the same.
In a right angled triangle its hypotenuse when squared is equal to the sum of its squared sides which is Pythagoras' theorem for a right angle triangle.
The theorem is only true if the base is the side of different length.To see this consider the right angled isosceles triangle with sides 1, 1 and √2. If one of the sides of length 1 is the base, the height is obviously the other side of length 1, but it clearly does not meet the base at its mid-point to make it a median.So with an isosceles triangle ABC with sides AB & AC equal, angles ABC & ACB equal and side BC the base, we need to prove that the point X where the height (AX) meets BC is such that BX = CX.Consider triangles AXB and AXC.Angle AXB is a right angle, as is AXC (since AX is a height of triangle ABC).Side AB is the hypotenuse of triangle AXB; AC is the hypotenuse of triangle AXC; they are known to be equal (from isosceles triangle ABC)Side AX is common to both trianglesThus triangles AXB and AXC are congruent since we have a Right-angle, Hypotenuse, Side match.Thus XB must be the same length as XC, that is X is the mid-point of BC.As X is the mid-point of BC, AX is the median.