Question Number 186337 by normans last updated on 03/Feb/23
Commented by normans last updated on 03/Feb/23
$$\boldsymbol{{the}}\:\boldsymbol{{volume}}\:\boldsymbol{{of}}\:\boldsymbol{{the}}\:\boldsymbol{{cube}}\:\boldsymbol{{is}}\:\mathrm{300} \\ $$$$\boldsymbol{{the}}\:\boldsymbol{{cube}}\:\boldsymbol{{is}}\:\boldsymbol{{cut}}\:\boldsymbol{{with}}\:\boldsymbol{{a}}\:\boldsymbol{{plane}}. \\ $$$$\boldsymbol{{the}}\:\boldsymbol{{volume}}\:\boldsymbol{{of}}\:\boldsymbol{{the}}\:\boldsymbol{{cut}}\:\boldsymbol{{part}}???? \\ $$
Answered by mr W last updated on 03/Feb/23
Commented by mr W last updated on 03/Feb/23
$${a}={edge}\:{length}\:{of}\:{cube} \\ $$$${volume}\:{of}\:{cube}\:{V}={a}^{\mathrm{3}} =\mathrm{300} \\ $$$$\frac{{b}}{{a}}=\frac{\mathrm{0}.\mathrm{4}{a}}{\mathrm{0}.\mathrm{5}{a}} \\ $$$$\Rightarrow{b}=\mathrm{0}.\mathrm{8}{a} \\ $$$$\frac{{h}}{{h}+{a}}=\frac{\mathrm{0}.\mathrm{4}{a}}{\mathrm{0}.\mathrm{5}{a}}=\mathrm{0}.\mathrm{8} \\ $$$$\Rightarrow{h}=\mathrm{4}{a} \\ $$$${V}_{{cut}} ={V}_{\mathrm{1}} +{V}_{\mathrm{2}} \\ $$$${V}_{\mathrm{1}} =\frac{\mathrm{0}.\mathrm{4}{a}×\mathrm{0}.\mathrm{8}{a}}{\mathrm{2}}×\frac{{a}}{\mathrm{3}}=\frac{\mathrm{4}{a}^{\mathrm{3}} }{\mathrm{75}} \\ $$$${V}_{\mathrm{2}} =\frac{\left(\mathrm{0}.\mathrm{4}{a}+\mathrm{0}.\mathrm{5}{a}\right)×{a}}{\mathrm{2}}×\frac{{a}}{\mathrm{3}}=\frac{\mathrm{3}{a}^{\mathrm{3}} }{\mathrm{20}} \\ $$$${V}_{{cut}} =\frac{\mathrm{4}{a}^{\mathrm{3}} }{\mathrm{75}}+\frac{\mathrm{3}{a}^{\mathrm{3}} }{\mathrm{20}}=\frac{\mathrm{61}{a}^{\mathrm{3}} }{\mathrm{300}}=\frac{\mathrm{61}{V}}{\mathrm{300}}=\mathrm{61}\:\checkmark \\ $$$${or} \\ $$$${V}_{{cut}} =\frac{\mathrm{0}.\mathrm{5}{a}×{a}}{\mathrm{2}}×\frac{\mathrm{4}{a}+{a}}{\mathrm{3}}−\frac{\mathrm{0}.\mathrm{4}{a}×\mathrm{0}.\mathrm{8}{a}}{\mathrm{2}}×\frac{\mathrm{4}{a}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{61}{a}^{\mathrm{3}} }{\mathrm{300}}=\frac{\mathrm{61}{V}}{\mathrm{300}}=\mathrm{61}\:\checkmark \\ $$
Commented by normans last updated on 03/Feb/23
$$\boldsymbol{{nice}}\:\boldsymbol{{Sir}} \\ $$