IDNLearn.com offers a reliable platform for finding accurate and timely answers. Get the information you need from our experts, who provide reliable and detailed answers to all your questions.
Sagot :
To determine which sets of side lengths form a right triangle, we use the Pythagorean Theorem. The Pythagorean Theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. That is, if [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are the side lengths of a right triangle with [tex]\(c\)[/tex] being the hypotenuse, then:
[tex]\[a^2 + b^2 = c^2\][/tex]
Let’s analyze each set of side lengths one by one.
### Set 1: [tex]\(3, 14, \sqrt{205}\)[/tex]
- [tex]\(a = 3\)[/tex]
- [tex]\(b = 14\)[/tex]
- [tex]\(c = \sqrt{205}\)[/tex]
Check:
[tex]\[3^2 + 14^2 = 9 + 196 = 205\][/tex]
[tex]\((\sqrt{205})^2 = 205\] Since \(205 = 205\)[/tex], this set of side lengths forms a right triangle.
### Set 2: [tex]\(6, 11, \sqrt{158}\)[/tex]
- [tex]\(a = 6\)[/tex]
- [tex]\(b = 11\)[/tex]
- [tex]\(c = \sqrt{158}\)[/tex]
Check:
[tex]\[6^2 + 11^2 = 36 + 121 = 157\][/tex]
[tex]\((\sqrt{158})^2 = 158\] Since \(157 \neq 158\)[/tex], this set of side lengths does not form a right triangle.
### Set 3: [tex]\(19, 180, 181\)[/tex]
- [tex]\(a = 19\)[/tex]
- [tex]\(b = 180\)[/tex]
- [tex]\(c = 181\)[/tex]
Check:
[tex]\[19^2 + 180^2 = 361 + 32400 = 32761\][/tex]
[tex]\[181^2 = 32761\][/tex]
Since [tex]\(32761 = 32761\)[/tex], this set of side lengths forms a right triangle.
### Set 4: [tex]\(3, 19, \sqrt{380}\)[/tex]
- [tex]\(a = 3\)[/tex]
- [tex]\(b = 19\)[/tex]
- [tex]\(c = \sqrt{380}\)[/tex]
Check:
[tex]\[3^2 + 19^2 = 9 + 361 = 370\][/tex]
[tex]\((\sqrt{380})^2 = 380\] Since \(370 \neq 380\)[/tex], this set of side lengths does not form a right triangle.
### Set 5: [tex]\(2, 9, \sqrt{85}\)[/tex]
- [tex]\(a = 2\)[/tex]
- [tex]\(b = 9\)[/tex]
- [tex]\(c = \sqrt{85}\)[/tex]
Check:
[tex]\[2^2 + 9^2 = 4 + 81 = 85\][/tex]
[tex]\((\sqrt{85})^2 = 85\] Since \(85 = 85\)[/tex], this set of side lengths forms a right triangle.
### Summary
The sets of side lengths that form right triangles are:
- [tex]\(3, 14, \sqrt{205}\)[/tex]
- [tex]\(19, 180, 181\)[/tex]
- [tex]\(2, 9, \sqrt{85}\)[/tex]
[tex]\[a^2 + b^2 = c^2\][/tex]
Let’s analyze each set of side lengths one by one.
### Set 1: [tex]\(3, 14, \sqrt{205}\)[/tex]
- [tex]\(a = 3\)[/tex]
- [tex]\(b = 14\)[/tex]
- [tex]\(c = \sqrt{205}\)[/tex]
Check:
[tex]\[3^2 + 14^2 = 9 + 196 = 205\][/tex]
[tex]\((\sqrt{205})^2 = 205\] Since \(205 = 205\)[/tex], this set of side lengths forms a right triangle.
### Set 2: [tex]\(6, 11, \sqrt{158}\)[/tex]
- [tex]\(a = 6\)[/tex]
- [tex]\(b = 11\)[/tex]
- [tex]\(c = \sqrt{158}\)[/tex]
Check:
[tex]\[6^2 + 11^2 = 36 + 121 = 157\][/tex]
[tex]\((\sqrt{158})^2 = 158\] Since \(157 \neq 158\)[/tex], this set of side lengths does not form a right triangle.
### Set 3: [tex]\(19, 180, 181\)[/tex]
- [tex]\(a = 19\)[/tex]
- [tex]\(b = 180\)[/tex]
- [tex]\(c = 181\)[/tex]
Check:
[tex]\[19^2 + 180^2 = 361 + 32400 = 32761\][/tex]
[tex]\[181^2 = 32761\][/tex]
Since [tex]\(32761 = 32761\)[/tex], this set of side lengths forms a right triangle.
### Set 4: [tex]\(3, 19, \sqrt{380}\)[/tex]
- [tex]\(a = 3\)[/tex]
- [tex]\(b = 19\)[/tex]
- [tex]\(c = \sqrt{380}\)[/tex]
Check:
[tex]\[3^2 + 19^2 = 9 + 361 = 370\][/tex]
[tex]\((\sqrt{380})^2 = 380\] Since \(370 \neq 380\)[/tex], this set of side lengths does not form a right triangle.
### Set 5: [tex]\(2, 9, \sqrt{85}\)[/tex]
- [tex]\(a = 2\)[/tex]
- [tex]\(b = 9\)[/tex]
- [tex]\(c = \sqrt{85}\)[/tex]
Check:
[tex]\[2^2 + 9^2 = 4 + 81 = 85\][/tex]
[tex]\((\sqrt{85})^2 = 85\] Since \(85 = 85\)[/tex], this set of side lengths forms a right triangle.
### Summary
The sets of side lengths that form right triangles are:
- [tex]\(3, 14, \sqrt{205}\)[/tex]
- [tex]\(19, 180, 181\)[/tex]
- [tex]\(2, 9, \sqrt{85}\)[/tex]
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Find clear answers at IDNLearn.com. Thanks for stopping by, and come back for more reliable solutions.