Question

Cerinta poate fi gasita in poza

Answer 1

Răspuns:

B) 18 cm²

Explicație pas cu pas:

T.Pitagora:

AC² = BC²-AB² = 10²-8² = 100-64 = 6²

=> AC = 6 cm

$\dfrac{A'C}{A'B} = \dfrac{3}{7} \iff \dfrac{A'C}{A'B + A'C} = \dfrac{3}{7 + 3}$

$\dfrac{A'C}{BC} = \dfrac{3}{10} \iff \dfrac{A'C}{10} = \dfrac{3}{10} \implies \bf A'C = 3 \ cm$

$\dfrac{B'A}{B'C} = \dfrac{1}{5} \iff \dfrac{B'A + B'C}{B'C} = \dfrac{1 + 5}{5}$

$\dfrac{AC}{B'C} = \dfrac{6}{5} \iff \dfrac{6}{B'C} = \dfrac{6}{5} \implies \bf B'C = 5 \ cm$

$\sin C = \dfrac{AB}{BC} = \dfrac{8}{10} = \dfrac{4}{5}$

$\mathcal{A}_{\triangle A'B'C} = \dfrac{A'C \cdot B'C \cdot \sin C}{2} = \dfrac{3 \cdot 5 \cdot \dfrac{4}{5}}{2} = 6 \ cm^{2}$

$\mathcal{A}_{\triangle ABC} = \dfrac{AB \cdot AC}{2} = \dfrac{8 \cdot 6}{2} = 24 \ cm^{2}$

$\mathcal{A}_{AB'A'B} = \mathcal{A}_{\triangle ABC} - \mathcal{A}_{\triangle A'B'C} = 24 - 6 = \bf 18 \ cm^{2}$