Question

Trapezul isoscel ABCD are AB=24cm,CD=18cm iar masura unghiului A=60°.Determina:
a)Aria trapezului.
b)Lungimile diagonalelor trapezului
c)Distanta de la punctul C la dreapta AD.

Answer 1

$fie\,AB=24cm\,si\,CD=18cm\,,\,AD\equiv{BC}\,,\,m(<A)=60^0$
[tex]a)...\\
din\,triunghiul \,dr.AD'D\,AD'=3\,si\,AD=2\cdot3=6cma,\\
\rightarrow\,DD'=\sqrt{36-9}=3\sqrt3cm\,\\
A=\frac{24+18}{2\cdot3\sqrt3}=63\sqrt3cm^2,\\
b)...\\
diagonala\,BD=\sqrt{D'B^2+D'D^2}=\sqrt{21^2+27}=6\sqrt13cm,\\
c)...\\
AD\cap{CE}=\{E\}\,CE=d(C,AD)=...\\ m(\hat{A})=m(\hat{EDC})=60,\rightarrow\,DE=18:2=9cm,\\ CE=\sqrt{18^2-9^2}=9\sqrt3cm, [/tex]