Answer:
2(x^2 + 4x - 28) = 0.
The length of the missing sides are:
4√2 units.
or 5.66 units ( to the nearest hundredth).
Step-by-step explanation:
Applying the Pythagoras Theorem:
8^2 = (x + 2)^2 + (x + 2)^2
2(x^2 + 4x + 4) = 64
2x^2 + 8x + 8 - 64 = 0
2x^2 + 8x - 56 = 0
2(x^2 + 4x - 28) = 0 models the situation.
Solving:
x = [- 4 +/- √(4^2-4*1*-28)] / 2
= (-4 +/- √128) / 2
= (-4 + 8√2) / 2 , (-4 - 8√2) / 2 (we ignore this negative root).
= -2 + 4√2.
This is 3.66 to the nearest hundredth.
So the length of the 2 equal sides is 2 + (- 2 + 4√2) = 4√2.
or 5.66 to the nearest hundredth.
