Question
The missing side length on
this triangle is
units long.
6.4
2.3
?
this triangle is
units long.
6.4
2.3
?
Answers
Answered by
GPT-4o mini
To find the missing side length in a triangle with side lengths of 6.4 units and 2.3 units, we need more information about the type of triangle and which side the missing length refers to (e.g., if it is a right triangle).
If we are assuming that this is a right triangle and you're seeking the third side using the Pythagorean theorem, you could calculate it if you know which lengths are the legs and which is the hypotenuse.
Assuming \( a = 2.3 \) and \( b = 6.4 \), and that these are the legs of a right triangle, the hypotenuse \( c \) can be found with:
\[
c = \sqrt{a^2 + b^2}
\]
Calculating that:
\[
c = \sqrt{(2.3)^2 + (6.4)^2} = \sqrt{5.29 + 40.96} = \sqrt{46.25} \approx 6.8
\]
If the missing side is one of the legs instead and 6.4 is the hypotenuse, the calculation would be:
\[
? = \sqrt{c^2 - a^2} = \sqrt{(6.4)^2 - (2.3)^2} = \sqrt{40.96 - 5.29} = \sqrt{35.67} \approx 5.96
\]
Please specify if this is a right triangle and which side length you are looking to find, or provide additional details if this is to be solved using a different method!
If we are assuming that this is a right triangle and you're seeking the third side using the Pythagorean theorem, you could calculate it if you know which lengths are the legs and which is the hypotenuse.
Assuming \( a = 2.3 \) and \( b = 6.4 \), and that these are the legs of a right triangle, the hypotenuse \( c \) can be found with:
\[
c = \sqrt{a^2 + b^2}
\]
Calculating that:
\[
c = \sqrt{(2.3)^2 + (6.4)^2} = \sqrt{5.29 + 40.96} = \sqrt{46.25} \approx 6.8
\]
If the missing side is one of the legs instead and 6.4 is the hypotenuse, the calculation would be:
\[
? = \sqrt{c^2 - a^2} = \sqrt{(6.4)^2 - (2.3)^2} = \sqrt{40.96 - 5.29} = \sqrt{35.67} \approx 5.96
\]
Please specify if this is a right triangle and which side length you are looking to find, or provide additional details if this is to be solved using a different method!