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# help geometry

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

Jul 3, 2021

#1
+139
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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:

$\text{ third side} <8$

Jul 3, 2021
#2
+11864
+1

Hello Guest!

The greatest possible length of the shortest side of the parallelogram is greater than two by the smallest possible length.

!

Jul 3, 2021