No. But they can be the lengths (not leangths!) of the three sides.
With sides of 5 and 12, you can make a triangle with any perimeter you want between 24 and 34. If you call them "legs" because they are the sides of a right triangle, then the hypotenuse is 13, and the perimeter is 30.
If its a right angle it will be 34 units in length because it complies with Pythagoras' theorem.
Doesn't it depend on what type of triangle it is? And which sides you are measuring? And which side it's laying on?
1 Let the sides be x+14 and x 2 So: 0.5*(x+14)*x = 240 which transposes to x2+14x-480 3 Solving the above quadratic equation gives x a positive value of 16 4 Therefore the sides are 30 and 16 5 Using Pythagoras: 302+162 = 1156 and its square root is 34 6 Hypotenuse = 34 cm 7 Perimeter = 34+30+16 = 80 cm
No. But they can be the lengths (not leangths!) of the three sides.
To test use PYTHAGORAS ; h^2 = a^2 + b^2 34' is the longest side (h) , hence 34^(2) = 1156 a & b are 16 & 30 Hence 16^() + 30^(2) = 256 + 900 = 1156 , which equatres with 34^(2) . Thereby satisfies the Pythagorean Equation. Hence it is a Right angled triangle. So the sides 16,30, & 34 form a right triangle.
A triangle whose sides are 16, 30, and 35 in length is not a right triangle, becausethe square of the length of the longest side is not equal to the sum of the squaresof the lengths of the other two sides.But if the 35 were a 34 instead, then it wouldbe.
With sides of 5 and 12, you can make a triangle with any perimeter you want between 24 and 34. If you call them "legs" because they are the sides of a right triangle, then the hypotenuse is 13, and the perimeter is 30.
4x2 - 30 = 34 Add 30 to both sides 4x2 = 64 Divide both sides by 4 x2 = 16 Take the square root of both sides: x = 4 or -4
c2 = a2 + b2, where c is the hypotenuse, and a and b are the other two sides. c2 = 162 + 302 c2 = 1156 √c2 = √1156 c = 34
Oh, what a happy little question! It looks like you have a special triangle there with side lengths 30, 32, and 34. Since the sum of the two shorter sides must be greater than the longest side for a triangle to exist, let's check if that's true here. If we add 30 and 32, we get 62, which is indeed greater than 34, so you can paint a beautiful triangle with those side lengths!
The length is 32. Since an isosceles triangle is symmetrical, the height divides it into two right triangles. The hypotenuse of one of these triangles is 34, one of the legs is 30 (the height), and the other leg is half of the base, which we'll call x. Using the Pythagorean Theorem, we know x^2+30^2=34^2, or x^2+900=1156. This gives us x^2=256, or x=16 (it could also be -16, but this is absurd because it's a length). We said x was half the base, so the base itself is 2x=2*16=32.
pythagoreon thereom 34sq - 30sq = 256 sqrt = 16 base 16+16 = 32 half length of base = SQR(34^2 - 30^2) =SQR(1156-900) =SQR(256) =16 base length = 2x 16=32
If its a right angle it will be 34 units in length because it complies with Pythagoras' theorem.
Doesn't it depend on what type of triangle it is? And which sides you are measuring? And which side it's laying on?
1 Let the sides be x+14 and x 2 So: 0.5*(x+14)*x = 240 which transposes to x2+14x-480 3 Solving the above quadratic equation gives x a positive value of 16 4 Therefore the sides are 30 and 16 5 Using Pythagoras: 302+162 = 1156 and its square root is 34 6 Hypotenuse = 34 cm 7 Perimeter = 34+30+16 = 80 cm