NEED HELP SOS Find an integer, x, such that 5, 10, and x represent the lengths of the sides of an obtuse triangle. A. 4 B. 5 C. 6 D. 11

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NEED HELP SOS Find an integer, x, such that 5, 10, and x represent the lengths of the sides of an obtuse triangle. A. 4 B. 5 C. 6 D. 11

Mathematics
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In an obtuse triangle c^2 > a^2 + b^2
so if we take c = 10 what inequality can we write?
10^2 > 5^2 + b^2 can you solve this for b^2? then for b?

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Other answers:

100 + 25
not +
100 > 25 + 36 for 6
i thought the asnwer would be 6
b^2 < 100 - 25 b^2 < 75
b < sqrt75 b < 8.66
im really confused
b cannot be 4 or 5 because that would not make a triangle and 11 > 8.66 therefore the answer is 6
8 isnt an option
oh i get it
i have one more question
ok
how do i find an acute triangle
Using Pythagorean inequalities, determine which set of the given three sides produces an acute triangle.
for an acute angle , if c is the longest side, c^2 < a^2 + b^2
3, 4, 5 6,12,20 4,6,10, 5,12,18 ALL NUMBERS TO SQUARE ROOT
- and dont forget for all triangles the sum of any 2 sides must exceed the length of the other side
what
i dont know how to do it cause the numbers are squared
what exactly is the second question?
Using Pythagorean inequalities, determine which set of the given three sides produces an acute triangle.
6 minutes till time runs out on this question:/
well the first one is a right angled triangle because 5^2 = 3^2 + 4^2
what about C
rembember all the numbers are squared
the 2nd is not a triangle? can you see why? neither is the third why?
4, 6,10 4x4 = 16 6x6=36
and 10^2 = 100 but 4+6 = 10 so its not a triangle..
oh
you cant draw a triangle with sides this length
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- they give a straight line
i see
also 18 > 5 + 12 so that's not a triangle also
i got the answer wrong
the only one which is a triangle is A but that's a right-angled triangle because it fits the Pythagoras theorem.
what is the correct answer?

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