Лице на квадрат

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[ammornil]

А)
[tex]\triangle MBC \cong \triangle NCD \begin{cases} BC=DC \\ BM=CN \\ \angle MBC = \angle BCD \end{cases} \Rightarrow \begin{cases} \angle CMB = \angle CND = \angle 1 \\ \angle MCB = \angle NDC = \angle 2 \\ \angle 1 + \angle 2 = 90 ^\circ \end{cases}[/tex]

[tex]\angle DCM = \angle BMC = \angle 1[/tex] -кръстни

[tex]\triangle DPC: \ \angle PCD + \angle PDC = \angle 1 + \angle 2= 90 ^\circ \Rightarrow \angle DPC = 90 ^\circ[/tex]

Б) [tex]DF=10\ cm, OQ=6\ cm[/tex]
Диагоналите на квадрата са и ъглополовящи, следователно [tex]\angle ODC = \angle OCB = 45 ^\circ \Rightarrow \angle ODF = \angle OCQ = 45 ^\circ - \angle 2[/tex]

[tex]\triangle FOD \cong \triangle QOC \begin{cases} \angle FOD = \angle QOC \\ \angle ODF = \angle OCQ \\ DO=CO \end{cases} \Rightarrow OF = OQ = 6cm[/tex]

[tex]\triangle FOD - правоъгълен \ \Rightarrow OD^2+OF^2=DF^2 \Rightarrow OD=\sqrt{DF^2-OF^2} =\sqrt{100-36}=\sqrt{64}=8cm[/tex]

[tex]S_{_{ABCD}}=\frac{d^2}{2}, d=2.OD \Rightarrow S_{_{ABCD}}=\frac{16^2}{2}=128\ cm^2[/tex]


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