Two diagonals of a parallelogram have lengths 6 and 10. What is the largest possible length of the shortest side of the parallelogram?

Guest Jul 3, 2021

#1**+2 **

here is a representative image i made:

since both diagonals cot eachother in the middle, they form 4 triangles; by looking at the image, we can say that:

$ \overline{AO}=\frac{6}{2} \ \therefore \ \overline{CO}=\frac{6}{2} $

$ \overline{DO}=\frac{10}{2} \ \therefore \ \overline{BO}=\frac{10}{2} $

now there are two little rules for triangles

1) sum of any two sides $>$ third side.

2) difference of any two sides $<$ third side. ( commonly used only to prove whether the triangle exists or not )

lets consider the $\triangle BOC $ where side $\overline{BC}=$third side (the third and largest side of the triangle, but the smallest side of the parallelogram)

1) $ \frac{10}{2}+\frac{6}{2} > \text{ third side} $

$ 5+3> \text{ third side} $

$ 8> \text{ third side} $

changing the sides, would also change the 'direction' of the less-than-sign, so logically, it would become larger-than-sign:

$ \text{ third side} <8 $

UsernameTooShort Jul 3, 2021