Question

determine a area total e o volume de uma piramede regular de base quadrada sabendo que as medidas das crestas laterais e da base sao 15 cm e 18 cm respectivamente​

Answer 1

$\large\boxed{\begin{array}{l}\underline{\rm C\acute alculo\,da\,\acute area\,da\,base\,B}\\\sf B=\ell^2\\\sf B=18^2\\\sf B=324\,cm^2\\\underline{\rm C\acute alculo\,do\,ap\acute otema\,da\,pir\hat amide\!:}\\\sf a_p^2+\bigg(\dfrac{\ell}{2}\bigg)^2=g^2\\\sf a_p^2+\bigg(\dfrac{18}{2}\bigg)^2=15^2\\\sf a_p^2+81=225\\\\\sf a_p^2=225-81\\\\\sf a_p^2=144\\\\\sf a_p=\sqrt{144}\\\sf a_p=12\,cm\end{array}}$

$\boxed{\begin{array}{l}\underline{\rm C\acute alculo\,do\,ap\acute otema\,da\,base:}\\\sf m=\dfrac{\ell}{2}=\dfrac{18}{2}=9\,cm\\\underline{\rm C\acute alculo\,da\,altura\,da\,pir\hat amide\!:}\\\sf h^2+m^2=a_p^2\\\sf h^2+9^2=144\\\sf h^2+81=144\\\sf h^2=144-81\\\sf h^2=63\\\sf h=\sqrt{63}=\sqrt{9\cdot7}\\\sf h=3\sqrt{7}\,cm\\\underline{\rm C\acute alculo\,da\,\acute area\,lateral\!:}\\\sf A_l=\diagup\!\!\!4\cdot\dfrac{1}{\diagup\!\!\!2}\cdot18\cdot12\\\sf A_l=2\cdot18\cdot12\\\sf A_l=432\,cm^2\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm A\,partir\,daqui\,comec_{\!\!,}a\,as\,respostas}\\\underline{\rm do\,exerc\acute icio.}\\\dagger\bf \acute Area\,total:\\\sf A_t=A_l+B\\\sf A_t=432+324\\\sf A_t=756\,cm^2\\\dagger\dagger \bf volume:\\\sf V=\dfrac{1}{3}\cdot B\cdot h\\\\\sf V=\dfrac{1}{\backslash\!\!\!3}\cdot324\cdot\backslash\!\!\!3\sqrt{7}\\\\\sf V=324\sqrt{7}\,cm^3\end{array}}$