The length of the hypotenuse of a right triangle is 26 cm the length of one leg is 24 cm. find the length of the other leg?
is the answer 10?
Well, if the length of one leg is 24 cm and the length of the hypotenuse is 26 cm, then it seems like the triangle is suffering from a classic case of "leg envy." The poor other leg is feeling left out and wants to know its length.
To find the length of the other leg, we can use the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs.
So, in this case, we can set up the equation as follows: (length of the other leg)^2 + 24^2 = 26^2.
Simplifying that equation, we get: (length of the other leg)^2 + 576 = 676.
Subtracting 576 from both sides of the equation, we find: (length of the other leg)^2 = 100.
Taking the square root of both sides of the equation, we get: length of the other leg = +/- 10.
So, it turns out there are two possible lengths for the other leg: 10 cm and -10 cm. However, since we can't have negative lengths for physical objects, we can conclude that the length of the other leg is indeed 10 cm.
Congratulations, you are correct! Good job!
To find the length of the other leg in a right triangle, you can use the Pythagorean theorem. The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the two legs.
In this case, you are given that the length of the hypotenuse is 26 cm, and the length of one leg is 24 cm. Let's assume the length of the other leg is x cm.
Using the Pythagorean theorem, you can set up the following equation:
x^2 + 24^2 = 26^2
x^2 + 576 = 676
Subtracting 576 from both sides of the equation, you get:
x^2 = 676 - 576
x^2 = 100
Taking the square root of both sides, you get:
x = sqrt(100)
x = 10
Therefore, the length of the other leg is indeed 10 cm. So, your answer is correct.
Its 10 !!
Did you test if 10^2 + 24^2 = 26^2 ??
yes, but idk if that's the answer
also could you check another of my questions?