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Yes because the sum of its 2 smaller sides is greater than its longest side.

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βˆ™ 9y ago
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βˆ™ 9y ago

Yes

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Q: Could the segments 9 4 11 form a triangle?
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Related questions

Could the segments 6 5 and 12 form a triangle?

No. To form a triangle the sum of the shorter two sides MUST be greater than the longer side. 6 + 5 = 11 < 12 → cannot be a triangle.


Could the segments 9 4 and 11 form a triangle?

Yes, the sum of any two sides is always greater than the third side: 9 + 4 > 11 4 + 11 > 9 11 + 9 > 4


Could 9 and 4 And 11 form a triangle?

Yes because the sum of its two shortest sides is greater than its longest side.


Is the angles 11 60 and 61 a right triangle?

If you mean dimensions of 11, 60 and 61 then yes they will form a right angle triangle.


Can 11 centimeters 15 centimeters and 17 centimeters form a triangle?

11 centimeters 15 centimeters and 17 centimeters can form a triangle . It is because some of any two sides of triangle is greater than the third side . a + b >c always.


Which set of side lengths can form a triangle?

11, 4, 8


Does 4 inches 8 inches and 11 inches form a triangle?

Yes, it does. Your triangle has an area of 12.285 square inches and a perimeter of 23 inches.


Can 9 4 and 11 be triangle?

9, 4, and 11 are three dimensionless numbers. Yes, they can represent the lengths of the sides of a triangle. You can take three straight sticks, cut them to lengths of 9, 4, and 11 inches, then lay them down on a table so that the ends hook up and they form a triangle.


What is 11 4ths in simplest form?

11/4 is in its simplest form. You could also represent it as 2¾


Do 8 and 11 and 13 form a right triangle?

No, it fails the Pythagoras test: 64 + 121 is not equal to 169.


Do the side lengths 15 11 and 2 x square root of 26 form a right triangle?

Yes.


Can 5 11 4 be the sides of a triangle?

No it is not possible because the sum of the lengths of the two sides has to be greater than the length of the third side. 5 + 4 = 9 which is less than 11, so we can't form a triangle with these sides.