Question

drawing triangles from side lengths practice pages
4 jacy has pieces of steel that are 8 in. and 12 in. long. he wants to weld the pieces together with a third piece of steel to make a triangular frame. what is the range of lengths he can use for the third piece? show your work.
5 juanita has three straws that are 5 in. long and three straws that are 10 in. long. name one combination of three straw lengths she can use to form a triangle and one combination that cannot form a triangle. explain your thinking.
6 noma has an eraser that is 1\frac{1}{2} in. long, a pencil that is 5\frac{7}{8} in. long, and a marker that is 7\frac{5}{8} in. long. can she arrange these items into a triangle? explain your thinking.
7 an artist uses scrap metal to build sculptures. they want to build a sculpture with triangles. the artist finds metal rods that are 20 in., 27 in., 47 in., and 67 in. long. which combinations of three of these lengths can they use to form a triangle? justify your answer.

Explanation:

Step1: Recall triangle - inequality theorem

The sum of the lengths of any two sides of a triangle must be greater than the length of the third side. Let the two given side - lengths be \(a\) and \(b\), and the third side - length be \(x\). Then \(|a - b|\lt x\lt a + b\).

Step2: Solve for Jacy's problem (Question 4)

Given \(a = 8\) in and \(b = 12\) in.
First, calculate \(|a - b|\): \(|8 - 12|=4\).
Then, calculate \(a + b\): \(8+12 = 20\).
So the range of the length of the third piece \(x\) is \(4\lt x\lt20\).

Step3: Solve for Juanita's problem (Question 5)

Combination that forms a triangle: Choose two 10 - in straws and one 5 - in straw.
Check the triangle - inequality: \(10+5>10\), \(10 + 10>5\), and \(5+10>10\).
Combination that does not form a triangle: Choose three 5 - in straws.
Because \(5 + 5=10\), which does not satisfy the condition that the sum of the lengths of any two sides must be greater than the length of the third side.

Step4: Solve for Noma's problem (Question 6)

Convert the mixed - numbers to improper fractions. The eraser length \(a = 1\frac{1}{2}=\frac{3}{2}\) in, the pencil length \(b = 5\frac{7}{8}=\frac{47}{8}\) in, and the marker length \(c=\frac{61}{8}\) in.
\(a + b=\frac{3}{2}+\frac{47}{8}=\frac{12 + 47}{8}=\frac{59}{8}\lt\frac{61}{8}=c\).
So she cannot arrange these items into a triangle.

Step5: Solve for the artist's problem (Question 7)

Check combinations:

• Consider the combination of 20 in, 27 in, and 47 in: \(20+27 = 47\), does not form a triangle.
• Consider the combination of 20 in, 27 in, and 67 in: \(20+27=47\lt67\), does not form a triangle.
• Consider the combination of 20 in, 47 in, and 67 in: \(20 + 47=67\), does not form a triangle.
• Consider the combination of 27 in, 47 in, and 67 in: \(27+47 = 74>67\), \(27+67>47\), \(47+67>27\), forms a triangle.

Answer:

1. The range of the length of the third piece of steel for Jacy is \(4\lt x\lt20\) inches.
2. A combination that forms a triangle: two 10 - in straws and one 5 - in straw; A combination that does not form a triangle: three 5 - in straws.
3. Noma cannot arrange the items into a triangle.
4. The combination of 27 in, 47 in, and 67 in can be used to form a triangle.