Answer:
[tex]a = \frac{4\sqrt3}3[/tex]; [tex]d= 2\sqrt 2[/tex]
Step-by-step explanation:
First one is half an equilateral triangle. It means the hypotenuse is twice the shorter side, so 2a. Then let's apply the Pythagorean theorem:
[tex](2a)^2 = a^2 + 4^2\\4a^2 = a^2 + 16\\3a^2 = 16[/tex]
At this point is simply take the square root, use the positive value since they're lengths, and rearrange the fraction to have things look nice:
[tex]a^2 = \frac{16}3\\a= {4\over \sqrt3}\\a = {4\sqrt3\over \sqrt3 \sqrt3}}\\ a = \frac{4\sqrt3}3[/tex]
Second one is half a square - or if you prefer, it's an isosceles right triangle, both side lengths are 2. Then its Pythagoras again:
[tex]d^2 = 2^2 + 2^2\\d^2= 4+4 = 8\\[/tex]
Again, we just have to pick the square roots, and rearrange in a better looking way.
[tex]d= \sqrt 8 = \sqrt{4\cdot2} = 2\sqrt 2[/tex]
