I can't solve this problem... help please!

Draw a perpendicular line from the upper left corner to the base. 

That makes a right triangle with one side = 3 and the hypotenuse = 7.

Use Pythagoras' Theorem to determine the third side.

x² + 3²  = 7²

x² = 7² – 3²

x² = 49 – 9

x = square root of 40  =  2 sqrt(10)     A square root is plus and minus of course, but we disregard the negative answer.

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Drop a perpendicular height from the right of the side length of 5 to the base of 8. This would create a right-triangle with a hypotenuse of 7 units and a measure of one of the legs, 3 units. Using the Pythagorean theorem, we have \(7^2-3^2=49-9=\sqrt{40}=2\sqrt{10}.\)

Draw a perpedicular to the base from the top left vertex

Call the intersection point of the base and this perpendicular, E

Call the top left vertex D  and the bottom left vertex, F

So....we have right triangle DEF

EF [ 8 - 5]  = 3   = a leg

And the hypotenuse  FD  = 7

So...DE  is the missing leg = missing side

So...using the Pythagorean Theorem

DE  = √ (7^2-  3^2]  = √ [ 49 - 9 ]  = √40  =  √4 *√10  =  2√10  = missing side

Thanks, everyone!