Question

12. Aflaţi aria triunghiului ABC, dacă KA = 120°, AB=AC = 20 cm şi BC = 20√3 cm.

Answer 1

$\boxed{\tiny A_{ABC} = \sqrt{p(p - a)(p - b)(p - c)} }\\ \\ \tiny p = \tfrac{a + b + c}{2} = \tfrac{20 \sqrt{3} + 20 + 20}{2} = 10 \sqrt{3} + 20 \: \\ \\ \:\tiny A_{ABC} = \sqrt{(10 \sqrt{3} + 20)(10 \sqrt{3} + 20 - 20 \sqrt{3})(10 \sqrt{3} + 20 - 20)(10 \sqrt{3} + 20 - 20) } \\ \tiny A_{ABC} = \sqrt{(10 \sqrt{3} + 20)( - 10 \sqrt{3} + 20) \times 10 \sqrt{3} \times 10 \sqrt{3} } \\ \tiny A_{ABC} = \sqrt{(10 \sqrt{3} + 20)( - 10 \sqrt{3} + 20) \times 300 } \\ \tiny A_{ABC} = 10 \sqrt{(20 + 10 \sqrt{3})(20 - 10 \sqrt{3}) \times 3 } \\ \tiny A_{ABC} = 10 \sqrt{(400 - 100 \times 3) \times 3} \\\tiny A_{ABC} = 10 \sqrt{100 \times 3} \\\boxed{ \tiny A_{ABC} = 100 \sqrt{3} \: \: cm {}^{2}}$

Answer 2

buna, sper că te am ajutat !!