Question

Urgent va rog!dau coroana si punctee

Answer 1

a)Facem cu Pitagora

$CB=\sqrt{AB^{2} +AC^{2}} = \sqrt{625} =25$

b)Tot Pitagora

MN^2 = NP^2-PM^2 = 192 - 144 = 48 MN=V48

c)Imparti CB la 2, afli CD si aplici pitagora

AD^2=AC^2 - DC^2

d)Aplici teorema lui pitagora, la fel ca la primul, nu mai fac calculele

Answer 2

$\it a)\ ABC-dreptunghic,\ m(\hat A)=90^o,\ \stackrel{T.Pitagora}{\Longrightarrow}\ BC^2=AB^2+AC^2 \Rightarrow\\ \\ \Rightarrow\ BC^2=15^2+20^2=225+400=625=25^2 \Rightarrow BC=25\ cm\\ \\ b)\ MNP-dreptunghic,\ m(\hat M)=90^o,\ \stackrel{T.Pitagora}{\Longrightarrow}\ MN^2=NP^2-MP^2 \Rightarrow\\ \\ \Rightarrow\ MN^2=(8\sqrt3)^2-12^2=192-144=48\Rightarrow MN=\sqrt{48}=\sqrt{16\cdot3}=4\sqrt3$

$\it c)\ ABC-isoscel,\ \ AB=AC,\ \ AD- \^{i}n\breve{a}l\c{\it t}ime\ \Rightarrow AD- median\breve{a}\Rightarrow AD=\dfrac{24}{2}=12cm$

$\it \it ABD-dreptunghic,\ m(\hat D)=90^o,\ \stackrel{T.Pitagora}{\Longrightarrow}\ AD^2=AB^2-BD^2 \Rightarrow\\ \\ \Rightarrow\ AD^2=20^2-12^2=400-144=256=16^2 \Rightarrow AD=16\ cm$