Answer:
5 length
Step-by-step explanation:
The diagram attached shows two equilateral triangles ABC & CDE. Since both squares share one side of the square BDFH of length 10, then their lengths will be 5 each. To obtain the largest square inscribed inside the original square BDFH, it makes sense to draw two other equilateral triangles AGH & EFG at the upper part of BDFH with length equal to 5.
So, the largest square that can be inscribe in the space outside the two equilateral triangles ABC & CDE and within BDFH is the square ACEG.
Remove the radical by raising each side to the index of the radical.
x > 29/4
Hello from MrBillDoesMath!
Answer:
20
Discussion:
Let (x1,y1) = (-8, -6) and (x2,y2) = (4,10). Per the distance formula,
Distance = sqrt ( (x1-x2)^2 + (y1-y2)^2 ) so
Distance = sqrt ( (-8-4)^2 + (-6-10)^2 )
= sqrt( (12)^2 + (-16)^2 )
= sqrt( 144 + 256 )
= sqrt(400)
= 20
Thank you,
MrB
Answer:
63°
Step-by-step explanation:
TSU = 180 - 117 = 63