Answer:
Step-by-step explanation:
1). In ΔABC,
For angle A as a reference angle,
Opposite side of ∠A → BC
Adjacent side of ∠A → AC
Hypotenuse → AB
2). By using tangent ratio in ΔYWZ,
tan(72°) =
tan(72°) =
x =
x = 3.899
x ≈ 3.9
3). By using sine rule in ΔKNM,
tan(x°) =
tan(x) =
x =
x = 19.03
x ≈ 19.0°
4). By applying cosine ratio in ΔJKF,
tan(70°) =
tan(70) =
x =
x = 21.838
x ≈ 21.8
Answer:
Step-by-step explanation:
No direction or other information about the desired parallelogram is given here, so we drew one arbitrarily. Likewise for the segment cutting it in half. It is convenient to have the bases of the trapezoids be the sides of the parallelogram that are 5 units apart.
The area of one trapezoid is ...
A = (1/2)(b1 +b2)h = (1/2)(3+5)·5 = 20 . . . . square units
The sum of the trapezoid base lengths is necessarily the length of the base of the parallelogram, so the area of the trapezoid is necessarily 1/2 the area of the parallelogram. (The area is necessarily half the area of the parallelogram also because the problem has us divide the parallelogram into two identical parts.)
