Question

which of the three sets could not be the lengths of the sides of a triangle {6,10,12} {5,7,10} {4,4,9} {2,3,3}

Answer 1

The triangle theorem which represents the relationship between sides is a-b<c<a+b

First you have four sets.

1) 10-6<12<10+6 => 4< 12 < 16 true

2) 7-5<10<7+5 = 2<10<12 true

3) 4-4<9<4+4 => 0<9<8 not true

4) 3-2<3<3+2 => 1<3<5 true

The set No 3 can not form triangle.

Good luck!!!

Answer 2

So for this, we will be applying the Triangle Inequality Theorem, which states that the sum of 2 sides must be greater than the third side for it to be a triangle. If any inequality turns out to be false, then the set cannot be a triangle.

`A+B>C\\A+C>B\\B+C>A`

First Option: {6, 10, 12}

Let A = 6, B = 10, and C = 12:

`6+10>12\\16>12\ \textsf{true}\\\\6+12>10\\18>10\ \textsf{true}\\\\12+10>6\\22>6\ \textsf{true}`

Second Option: {5, 7, 10}

Let A = 5, B = 7, and C = 10

`5+7>10\\12>10\ \textsf{true}\\\\5+10>7\\15>7\ \textsf{true}\\\\10+7>5\\17>5\ \textsf{true}`

Third Option: {4, 4, 9}

Let A = 4, B = 4, and C = 9

`4+4>9\\8>9\ \textsf{false}\\\\4+9>4\\13>4\ \textsf{true}\\\\4+9>4\\13>4\ \textsf{true}`

Fourth Option: {2, 3, 3}

Let A = 2, B = 3, and C = 3

`2+3>3\\5>3\ \textsf{true}\\\\2+3>3\\5>3\ \textsf{true}\\\\3+3>2\\6>2\ \textsf{true}`

Conclusion:

Since the third option had an inequality that was false, the third option cannot be a triangle.