Question

Answer 1

-△FDE vuông tại D có \(EF^2=DE^2+DF^2\Rightarrow EF=\sqrt{DE^2+DF^2}=\sqrt{6^2+8^2}=10\left(cm\right)\)

-△DHE∼△FDE∼△FHD \(\Rightarrow\dfrac{HE}{DE}=\dfrac{DE}{FE};\dfrac{FD}{FH}=\dfrac{FE}{FD}\)

\(\Rightarrow HE=\dfrac{DE^2}{FE}=\dfrac{6^2}{10}=3,6\left(cm\right);HF=\dfrac{DF^2}{FE}=\dfrac{8^2}{10}=6,4\left(cm\right)\).

-△DEF có: HN//DE (cùng vuông góc với DF)

\(\Rightarrow\dfrac{HN}{DE}=\dfrac{HF}{EF}\Rightarrow HN=\dfrac{DE.HF}{EF}=\dfrac{6.6,4}{10}=3,84\left(cm\right)\)

-△DEF có: MH//DF (cùng vuông góc với DE).

\(\Rightarrow\dfrac{MH}{DF}=\dfrac{EH}{EF}\Rightarrow MH=\dfrac{DF.EH}{EF}=\dfrac{8.3,6}{10}=2,88\left(cm\right)\).

-\(\widehat{DNH}=\widehat{MDN}=\widehat{DMH}=90^0\)\(\Rightarrow\)DNHM là hình chữ nhật.

-△DMN cân tại D có: \(MN^2=DM^2+DN^2\Rightarrow MN=\sqrt{DM^2+DN^2}=\sqrt{3,84^2+2,88^2}=4,8\left(cm\right)\)

\(\widehat{DPN}=180^0-\widehat{PDN}-\widehat{PND}=180^0-\widehat{HFD}-\widehat{HDF}=180^0-90^0=90^0\)

\(\Rightarrow\)△DPM∼△NDM (g-g) \(\Rightarrow\dfrac{DP}{ND}=\dfrac{DM}{NM}=\dfrac{PM}{DM}\)

\(\Rightarrow DP=\dfrac{ND.DM}{NM}=\dfrac{2,88.3,84}{4,8}=2,304\left(cm\right);PM=\dfrac{DM^2}{MN}=\dfrac{3,84^2}{4,8}=3,072\left(cm\right)\)

\(S_{DMP}=\dfrac{1}{2}PM.DP=\dfrac{1}{2}.2,304.3,072=3,538944\left(cm^2\right)\)