TRUE OR FALSE: A triangle with side lengths 6, 4, and the square root of 52 is a right triangle.?

True





Check with the Pythagorean theorem:


6虏 + 4虏 = (鈭?2)虏TRUE OR FALSE: A triangle with side lengths 6, 4, and the square root of 52 is a right triangle.?
You're blocked, ingrate. Mine was right and mine was first.

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TRUE OR FALSE: A triangle with side lengths 6, 4, and the square root of 52 is a right triangle.?
One way to find out is to use the Cosine law.





Since in a right triangle, the 90 degree angle is the largest of the 3, we expect the side with length sqrt(52) to be the one opposite to the 90 degree angle.





Cosine Law:


a虏 = b虏 + c虏 - 2bc Cos(A)


A = angle opposite side a


Let a = 鈭?2; b = 6, c = 4


If Cos(A) = 0, then A = 90





Cos(A) = (a虏 - b虏 - c虏)/(-2bc)


Cos(A) = (52 - 36 - 16)/(-2(6)(4))


Cos(A) = (52 - 52)/(-42) = 0





Therefore A = 90 and thus making this a right triangle! %26lt;%26lt;%26lt;





The law of cosines generalizes the Pythagorean theorem, which holds only in right triangles: if the angle A is a right angle (of measure 90掳 or PI/2 radians), then cos(PI/2) = 0, and thus the law of cosines reduces to the Pythagorean theorem.
true





It satisfies the pythagorean theorem, which only right triangles can do: a^2 + b^2 = c^2 (36 + 16 = 52)
true


apply





base^2 +perpendicular^2=hypotenuse^2





here 6x6+4x4=36+16=52


sqrt of 52 is the length of hyp
6^2 + 4^2 = 36 + 16 = 52 = square of the 3rd side


so the triangle is a right-angled triangle
It's DEFINTLY TRUE
fasle: 6 squared plus 4 squared is not equal to 52(square root of 52 squared)
36+ 16 = 52





52 = 52





True!
True.





a^2 + b^2 = c^2





6^2 + 4^2 = (sqr root 52)^2
true
true
um... true?