Question

6) Se considera trapezul isoscel cu AB || CD, masura unghiului A = 45. Aflati aria trapezului stiind ca AD = 20cm.

Answer 1

Raspuns:

${\boxed{\boxed{\mathbf{A=400\,cm^2}}}}$

Explicație pas cu pas:

$\mathbf{Fie: \:\; d(D;AB)=M \:\; si \:\; d(C;AB)=N}$

$\mathbf{In \:\; \triangle{AMD}-dreptunghic\:\; avem:}$

$\mathbf{\angle{M}=90^{\circ}}$

$\mathbf{\angle{A}=45^{\circ}}$

$\mathbf{\implies \triangle{AMD}-isoscel\:\; dreptunghic}$

$\mathbf{\implies AM=DM=x}$

$\mathbf{AD^2=AM^2+DM^2 }$

$\mathbf{\implies x^2+x^2=20^2 \implies 2x^2=400\:|:2}$

$\mathbf{x^2=200 \implies x=\sqrt{200} =10\sqrt{2} \, cm}$

$\mathbf{\implies T\,\: diag. \: ca\: \angle{DCM}=\angle{NCM}=\dfrac{90}{2} =45^{\circ}}$

$\mathbf{DN \cap CM=\{O}\}$

$\mathbf{In \:\; \triangle{DOC}\:\; avem\:\; doua \:\;unghiuri\:\;de\:\;45^{\circ}\implies \angle{O}=90^{\circ}}$

$\mathbf{\implies d_1 \bot d_2 \implies MNCD-patrat}$

$\mathbf{\implies AM=MD=DC=CN=MN=NB=10\sqrt{2} \, cm}$

$\mathbf{AB=AM+MN+NB=3 \cdot 10\sqrt{2} =30\sqrt{2} }$

$\mathbf{\implies A=\dfrac{(B+b)\cdot h}{2}=\dfrac{(AB+CD)\cdot DM}{2} }$

$\mathbf{\implies A=\dfrac{(30\sqrt{2}+10\sqrt{2})\cdot 10\sqrt{2} }{2} }$

$\mathbf{\implies A=\dfrac{40\sqrt{2}\cdot 10\sqrt{2} }{2}^{(2}=20\sqrt{2}\cdot 10\sqrt{2} }$

$\mathbf{\implies A=(20\cdot 10)\sqrt{2} ^2=200\cdot 2= 400\,cm^2}$

Bafta! :)