I'll answer problem 19
=======================================================
Part (a)
The tickmarks on segments DE and DC indicate they are the same length. We're also told that |DC| = |DE| = 5.
So DE is also 5 cm long. Triangle DAE is a right triangle with the 90 degree angle at angle A. Let x be the length of segment AE.
We'll use the pythagorean theorem to find x
a^2 + b^2 = c^2
(AE)^2 + (AD)^2 = (DE)^2
x^2 + 4^2 = 5^2
x^2 + 16 = 25
x^2 = 25-16
x^2 = 9
x = sqrt(9)
x = 3
Segment AE is 3 cm long
=======================================================
Part (b)
For any rectangle, the opposite sides are parallel and the same length. This means AB = DC = 5 cm.
We found earlier that AE = 3 cm, so,
AE+EB = AB
3+EB = 5
EB = 5-3
EB = 2
Segment EB is 2 cm long
=======================================================
Part (c)
We know that AD = BC = 4 cm, because the opposite sides are the same length.
Earlier in part (b), we found that segment EB was 2 cm long.
Triangle EBC is a right triangle with legs EB = 2 and BC = 4. Let's apply the pythagorean theorem to find EC. Let y = EC.
a^2 + b^2 = c^2
(EB)^2 + (BC)^2 = (EC)^2
2^2 + 4^2 = y^2
4 + 16 = y^2
20 = y^2
y^2 = 20
y = sqrt(20)
y = sqrt(4*5)
y = sqrt(4)*sqrt(5)
y = 2*sqrt(5)
y = 4.4721359
As an exact value, EC is sqrt(20) or 2*sqrt(5) cm long.
As an approximate value, EC is roughly 4.4721359 cm long.
I would ask your teacher if you should use the exact or approximate value.
