Question

Calculați aria trapezului.

Bună seară!
Numele meu este Ana-Maria Grigorescu.
Mă puteți ajuta și pe mine la acest ex. ?
Vă rog frumos din tot sufletul!
Mulțumesc anticipat.​

Answer 1

Răspuns: $\bf A_{ABCD}=400~cm^{2}$

Explicație pas cu pas:

Salutare!

$\bf Ducem~ d(D;AB)=M$

$\bf d(C;AB)=N$

$\bf In~\triangle{AMD}-dreptunghic~ avem:$

$\bf \angle{AMD}=90^{\circ}$

$\bf \angle{DAM}=45^{\circ}$

$\bf \angle{MDA}=45^{\circ}$  

$\bf \implies \triangle{AMD}~este ~dreptunghic ~isoscel$

$\bf \implies AM=DM=x$

$\bf In~ \triangle{AMD}~aplicam~teorema~lui ~Pitagora$

$\bf{AD^{2}=AM^{2}+DM^{2}$    

$\bf x^{2}+x^{2}=20^{2}$

$\bf 2\cdot x^2=400~~~\bigg|:2$

$\bf x^2=200\implies x = 10\sqrt{2}~cm$

$\bf AM = DM = 10\sqrt{2}~cm$

$\bf Analog~ si~ in~ cazul ~\triangle{BNC}-dreptunghic~$

$\bf \implies T~diag.~ca~\angle{DCM}=\angle{NCM}=\dfrac{90}{2} =45^{\circ}$  

$\bf DN \cap CM=\{O\}$  

$\bf In~\triangle{DOC}~ avem~doua~unghiuri~de~45^{\circ}\implies \angle{O}=90^{\circ}$

$\bf d_1 \bot d_2 \implies MDCN-patrat$  

$\bf AM=MD=DC=MN=CN=NB=10\sqrt{2}~cm$

$\bf AB=AM+MN+NB$

$\bf AB=3 \cdot 10\sqrt{2}$

$\bf AB=30\sqrt{2}$

$\bf A_{ABCD}=\dfrac{(Baza~mare+baza~mica)\cdot Inaltimea}{2}$

$\bf A_{ABCD} =\dfrac{(AB+CD)\cdot DM}{2}$

$\bf A_{ABCD}=\dfrac{(30\sqrt{2}+10\sqrt{2})\cdot 10\sqrt{2} }{2}$

$\bf A_{ABCD}=\dfrac{(30\sqrt{2}+10\sqrt{2})\cdot \not10\sqrt{2} }{\not2}$

$\bf A_{ABCD}=(30\sqrt{2}+10\sqrt{2})\cdot 5\sqrt{2}$

$\bf A_{ABCD}=40\sqrt{2}\cdot 5\sqrt{2}$

$\bf A_{ABCD}=200\cdot 52$

$\boxed{\bf A_{ABCD}=400~cm^{2}}$

#copaceibrainly