Question

Ajutor la problema 3.Ofer 100 puncte​

Answer 1

Răspuns: c) 48 m

Explicație pas cu pas :

CE - drumul cel mai scurt

În ΔABC :

m(∡A) = 90° => (Teorema lui Pitagora) BC²=60²+80²

BC²=3600+6400 => BC²=10000 => BC=√10000 => BC=100 m

D-mijloc BC => BD=DC = 50 m

A ΔABC = (60*80)/2 = 2400 m²

D - mijloc BC => AD-mediană

A ΔABD = A ΔADC

A Δ ADC=A ΔABC/2 =2400/2=1200 m²

A ΔADC=(EC*AD)/2

AD-mediana => AD=BD=DC=50 m

1200 m² =(EC*50)/2=> 50*EC=1200*2 => 50*EC=2400 => EC=2400/50 => CE=48 m

Distanța parcursă de Ștefan este egală cu :

c) 48 m

Answer 2

$\it Deoarece \ (60,\ 80,\ 100)\ -\ triplet\ pitagoreic\ \Rightarrow\ BC=100\ m\\ \\ AD-median\breve a\ \Rightarrow\ AD=BD=DC=100:2=50\ m.\\ \\ Fie\ CE\perp AD\ drumul\ cel\ mai\ scurt .\\ \\ Din\ \Delta ECD\ \Rightarrow\ sin\widehat{CDE} =\dfrac{CE}{CD}\ \Rightarrow\ sin\widehat{CDE}=\dfrac{CE}{50}\ \ \ \ \ \ (1)$

$\it Din\ \Delta ABC \Rightarrow sinB=\dfrac{AC}{BC}=\dfrac{80}{100}=0,8\ \ \ \ \ \ (2)\\ \\ \\ Teorema\ sinusurilor\ pentru\ \Delta ABD \Rightarrow \dfrac{AB}{sin\widehat{BDA}}=\dfrac{AD}{sinB} \Rightarrow \\ \\ \\ sin\widehat{BDA}=\dfrac{AB\cdot sinB}{AD}\ \stackrel{(2)}{=}\dfrac{60\cdot0,8}{50}=\dfrac{48}{50}\ \ \ \ \ (3)$

$\it \widehat{CDE}=\widehat{BDA}\ (opuse\ la\ v\hat arf) \Rightarrow sin\widehat{CDE}=sin\widehat{BDA}\ \ \ \ \ (4)\\ \\ (1),\ (3),\ (4) \Rightarrow \dfrac{CE}{50}=\dfrac{48}{50} \Rightarrow CE=48\ m$