A área de um trapézio é dada por:
[tex]\boxed{\sf Area~=~\dfrac{(B+b)\cdot h}{2}}\\\\\\\sf Onde:~~\left\{\begin{array}{ccl}\sf B&\sf :&\sf Base~maior\\\sf b&\sf :&\sf Base~menor\\\sf h&\sf :&\sf Altura\end{array}\right.[/tex]
Do desenho, podemos extrair os dados:
[tex]\boxed{\begin{array}{ccl}\sf B&\sf =&\sf 17\\\sf b&\sf =&\sf x\\\sf h&\sf =&\sf 4\sqrt{3}\end{array}}[/tex]
Substituindo estes dados na fórmula da área do trapézio, temos:
[tex]\sf 60\sqrt{ 3}~=~\dfrac{(17+x)\cdot 4\sqrt{3}}{2}\\\\\\60\sqrt{3}\cdot 2~=~(17+x)\cdot 4\sqrt{3}\\\\\\\dfrac{120\sqrt{3}}{4\sqrt{3}}~=~17+x\\\\\\17+x~=~\dfrac{120~~\backslash\!\!\!\!\!\!\sqrt{3}}{4~~ \backslash\!\!\!\!\!\!\sqrt{3}}\\\\\\17+x~=~\dfrac{120}{4}\\\\\\17+x~=~30\\\\\\x~=~30-17\\\\\\\boxed{\sf x~=~13~m}[/tex]
[tex]\Huge{\begin{array}{c}\Delta \tt{\!\!\!\!\!\!\,\,o}\!\!\!\!\!\!\!\!\:\,\perp\end{array}}Qualquer~d\acute{u}vida,~deixe~ um~coment\acute{a}rio[/tex]
