a = 6
b = 2√2
c = ?
y = 135°
z tw. cos.
c^2 = a^2 + b^2 - 2ab * cosy
c^2 = 6^2 + (2√2)^2 - 2* 6 * 2√2 * cos135°
c^2 = 36 + 8 - 24√2 * (-cos45°)
c^2 = 46 - 24√2 * (-√2/2)
c^2 = 46 + 48/2
c^2 = 46 + 24
c^2 = 70
c = √70
√70 ≈ 8,367, wiec najkrostszy bok to bok b.
P = 1/2 * 6 * 2√2 * sin135* = 6√2 * sin135° = 6√2 * sin45° = 6√2 * √2/2 = 12/2 = 6
P = 1/2 * b * hb
6 = 1/2 * 2√2 * hb
hb = 12/2√2 = 6/√2 = 6√2/2 = 3√2
odp: wysokosc padajaca na najkrotszy z bokow trojkata jest rowna 3√2, a pole tego trojkaya wynosi 6.
