Question

1. if the diagonal of a square is 11.3 meters, approximately what is the perimeter of the square?

directions: given the side lengths, determine whether the triangle is acute, right, obtuse, or no a triangle.

1. 15, 16, 21 12. 20, 23, 41

□ not a △
□ acute
□ right
□ obtuse

1. 10, 24, 26 14. 6, 13, 20

□ not a △
□ acute
□ right
□ obtuse

1. 3, 16, 17 16. 24, 29, 32

□ not a △
□ acute
□ right
□ obtuse

Explanation:

Step1: Check triangle inequality

For a set of side lengths \(a,b,c\) (\(c\) is the longest side), if \(a + b>c\), \(a + c>b\), \(b + c>a\), then it is a triangle.

For 11. \(15,16,21\)

\(15+16 = 31>21\), \(15 + 21=36>16\), \(16+21 = 37>15\). So it is a triangle.

For 12. \(20,23,41\)

\(20+23=43>41\), \(20 + 41=61>23\), \(23+41 = 64>20\). So it is a triangle.

For 13. \(10,24,26\)

\(10+24=34>26\), \(10 + 26=36>24\), \(24+26 = 50>10\). So it is a triangle.

For 14. \(6,13,20\)

\(6+13=19<20\). So it is not a triangle.

For 15. \(3,16,17\)

\(3+16=19>17\), \(3 + 17=20>16\), \(16+17 = 33>3\). So it is a triangle.

For 16. \(24,29,32\)

\(24+29=53>32\), \(24 + 32=56>29\), \(29+32 = 61>24\). So it is a triangle.

Step2: Use the Pythagorean theorem and its converse

If \(a^{2}+b^{2}=c^{2}\), it is a right - triangle. If \(a^{2}+b^{2}>c^{2}\), it is an acute triangle. If \(a^{2}+b^{2}

For 11. \(15,16,21\)

\(15^{2}+16^{2}=225 + 256=481\), \(21^{2}=441\). Since \(15^{2}+16^{2}>21^{2}\), it is an acute triangle.

For 12. \(20,23,41\)

\(20^{2}+23^{2}=400+529 = 929\), \(41^{2}=1681\). Since \(20^{2}+23^{2}<41^{2}\), it is an obtuse triangle.

For 13. \(10,24,26\)

\(10^{2}+24^{2}=100 + 576=676\), \(26^{2}=676\). Since \(10^{2}+24^{2}=26^{2}\), it is a right triangle.

For 15. \(3,16,17\)

\(3^{2}+16^{2}=9+256 = 265\), \(17^{2}=289\). Since \(3^{2}+16^{2}<17^{2}\), it is an obtuse triangle.

For 16. \(24,29,32\)

\(24^{2}+29^{2}=576+841 = 1417\), \(32^{2}=1024\). Since \(24^{2}+29^{2}>32^{2}\), it is an acute triangle.

Answer:

1. Acute
2. Obtuse
3. Right
4. Not a \(\triangle\)
5. Obtuse
6. Acute