Question

Cine mă poate ajuta? Dau coroană! ​

Answer 1

Răspuns:

Explicație pas cu pas:

1. m(∡ABC) = 2*m(∡ACB) =  2*m(∡ECB) =

BE = bisectoare

⇒ m(∡ABC) = 2*m(∡EBC) =  2*m(∡ABE)

⇒ m(∡ABE) = m(∡ECB) = m(∡EBC) = 1/2*[180° - m(∡BEC)] = 1/2*(180° - 110°) = 1/2*70° = 35°

⇒ m(∡ABD) = m(∡ABC) = 2*m(∡EBC) = 2*35° = 70°

AD ⊥ BC ⇒ m(∡ADB) = 90°

⇒ m(∡BAF)  = m(∡BAD) =180° - [m(∡ABD) + m(∡ADB)] = 180° - (70° + 90°) = 180° + 160° = 20°

m(∡ABF) = m(∡ABE) = 35°

in Δ ABF:

m(∡AFB) = 180° - [m(∡BAF) + m(∡ABF)] = 180° - (20° + 35°) =180° - 55° = 125°

Raspuns :

d) 35°

2.

m(∡APD) = 180° - m(∡DPB) = 180° - {180° - [ m(∡PDB) + m(∡PBD)]} = 180° - 180° + m(∡PDB) + m (∡PBD) =  m(∡PDB) + m (∡PBD) = 40° + 35° = 75°

Raspuns :

c) 75°

3. foloim teorema medianei:

$m_a^2 = \frac{2b^2+ 2c^2-a^2}{4}$

si adunam pentru celelalte doua mediane:

$m_b^2 + m_c^2= \frac{2a^2+ 2c^2-b^2}{4} + \frac{2a^2+ 2b^2-c^2}{4} = \frac{4a^2+ b^2 + c^2}{4} = \frac{8a^2+ 2b^2 + 2c^2}{8} = \frac{9a^2+ 2b^2 + 2c^2-a^2}{8} = \frac{9a^2}{8} + \frac{2b^2 + 2c^2-a^2}{8} = \frac{9a^2}{8} + \frac{1}{2}* \frac{2b^2 + 2c^2-a^2}{4} = \frac{9a^2}{8} + \frac{1}{2}*m_a^2 = \frac{9a^2}{8} + \frac{m_a^2}{2} \\\\Deci\\\\\frac{m_a^2}{2} = m_b^2 + m_c^2 - \frac{9a^2}{8}\\\\m_a^2} = 2m_b^2 + 2m_c^2 - \frac{9a^2}{4}$

$m_a^2} = 2*81^2 + 2*108^2 - \frac{9*90^2}{4} = \\\\ 2*6561 + 2*11664- \frac{9*8100}{4} = 13122 + 23328 - 18225 = 18225 = 135^2\\\\m_a=135m$

Raspuns :

d) 135m