Explanation:
ABCD is a rectangle.
so, it has 2 pairs of equal and parallel sides.
all 4 angles are 90°.
both diagonals are equally long and intersect each other at their midpoint. so, all 4 half-diagonal segments are equally long.
(1)
2x + 4y = 4x - y = 36/2 = 18 cm
we can split this into 2 equations
2x + 4y = 4x - y
5y = 2x
x = 5y/2
and e.g.
4x - y = 18
now we use the first equation in the second
4(5y/2) - y = 18
20y/2 - y = 18
10y - y = 18
9y = 18
y = 2
x = 5y/2 = 5×2/2 = 5
(2)
I cannot draw here, but the shortest distance between a point and a line is always a height (a line segment from the point meeting it intersecting the line at a right angle, 90°).
(3)
e.g.
AE, AD/2 and the drawn height from E to AD create a right-angled triangle.
we know AD/2 = 15 cm
from (1) we know AE = 4x - y = 4×5 - 2 = 20 - 2 = 18 cm.
the height E to AD we get via Pythagoras
AE² = (AD/2)² + height²
height² = AE² - (AD/2)² = 18² - 15² = 324 - 225 = 99
height = sqrt(99) = sqrt(9×11) = 3×sqrt(11)
remember trigonometry ?
this height = sin(theta) × AE
sin(theta) = height / AE = 3×sqrt(11) / 18 = sqrt(11)/6 =
= 0.552770798...
theta = 33.55730976...° ≈ 34°
(4)
the diagonals of a square bisect the angles of the square, so that theta would become 90/2 = 45°.