301
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Actually, all three are.
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Perimeter = 2L + 2W 350 = 2(93) + 2W 350 = 186 + 2W 350 - 186 = 186 - 186 + 2W 164 = 2W 164/2 = 2W/2 82 = W So the width is 82 cm.
60.0625 square feet.
The diameter is 93.
The diameter is 29.6
In any triangle whose sides have measures a,b, and c, the triangular inequality tells us: a + b > c a + c > b b + c > a Since the triangles here have two sides with the same measure, we will suppose that a = b and the above becomes a + a > c a + c > a a + c > a or 2a > c c > 0 c > 0 So we have 1) 2a > c > 0 Since the perimeter is 93, we have a + a + c = 93 2) 2a + c = 93 Solve 2) for c, c = 93 - 2a Substitute in 1) 1) 2a > 93 - 2a 4a > 93 a > 23.25 and since a is an integer a > 23 Solve 2) for 2a 2a = 93 - c Substitute 93 - c for 2a in 1) 93 - c > c 93 > 2c 46.5 > c and since c is an integer, 47 > c or c < 47 So we have a > 23 c < 47 Let p be the positive integer amount that a is greater than 23 Let q be the positive integer amount that a is less than 47 So a = 23 + p and c = 47 - q Substituting those into 2) 2a + c = 93 2(23 + p) + (47 - q) = 93 46 + 2p + 47 - q = 93 93 + 2p - q = 93 2p - q = 0 2p = q So now we have a = 23+p and c = 47-2p Since c > 0 47 - 2p > 0 -2p > -47 p < 23.5 and since p is an integer p < 24 And p > 0 so 0 < p < 24 So there are 23 values p can take on, any of the integers which are greater than 0 and less than 24. That's 1 through 23, so there are 23 possible isoceles triangles with integer sides that have perimeter 93. Here they all are: p a = 23+p b = a c = 47-2p ---------------------------------- 1. 24 24 45 2. 25 25 43 3. 26 26 41 4. 27 27 39 5. 28 28 37 6. 29 29 35 7. 30 30 33 8. 31 31 31 9. 32 32 29 10. 33 33 27 11. 34 34 25 12. 35 35 23 13. 36 36 21 14. 37 37 19 15. 38 38 17 16. 39 39 15 17. 40 40 13 18. 41 41 11 19. 42 42 9 20. 43 43 7 21. 44 44 5 22. 45 45 3 23. 46 46 1 Edwin