Question Number 156335 by cortano last updated on 10/Oct/21
Commented by cortano last updated on 10/Oct/21
$$\mathrm{find}\:\mathrm{the}\:\mathrm{perimeter}\:\mathrm{ABCD} \\ $$
Answered by mr W last updated on 10/Oct/21
Commented by mr W last updated on 10/Oct/21
$${AE}^{\mathrm{2}} =\left({a}+\mathrm{7}\right)^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} −\mathrm{2}×\mathrm{3}×\left({a}+\mathrm{7}\right)\:\mathrm{cos}\:\mathrm{60}° \\ $$$$\:\:\:\:\:\:\:\:\:={a}^{\mathrm{2}} +\mathrm{11}{a}+\mathrm{37} \\ $$$${DE}^{\mathrm{2}} ={a}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} −\mathrm{2}×\mathrm{4}×{a}\:\mathrm{cos}\:\mathrm{120}° \\ $$$$\:\:\:\:\:\:\:\:\:={a}^{\mathrm{2}} +\mathrm{4}{a}+\mathrm{16} \\ $$$${DA}^{\mathrm{2}} =\mathrm{7}^{\mathrm{2}} ={a}^{\mathrm{2}} +\mathrm{11}{a}+\mathrm{37}+{a}^{\mathrm{2}} +\mathrm{4}{a}+\mathrm{16}−\sqrt{\left({a}^{\mathrm{2}} +\mathrm{11}{a}+\mathrm{37}\right)\left({a}^{\mathrm{2}} +\mathrm{4}{a}+\mathrm{16}\right)} \\ $$$$\mathrm{2}{a}^{\mathrm{2}} +\mathrm{15}{a}+\mathrm{4}=\sqrt{\left({a}^{\mathrm{2}} +\mathrm{11}{a}+\mathrm{37}\right)\left({a}^{\mathrm{2}} +\mathrm{4}{a}+\mathrm{16}\right)} \\ $$$$… \\ $$$$\Rightarrow{a}=\mathrm{2} \\ $$$${perimeter}\:{ABCD}=\mathrm{2}+\mathrm{2}×\mathrm{7}+\mathrm{9}=\mathrm{25} \\ $$
Commented by Tawa11 last updated on 10/Oct/21
$$\mathrm{great}\:\mathrm{sir} \\ $$
Commented by Tawa11 last updated on 10/Oct/21
$$\mathrm{Sir},\:\mathrm{please}\:\mathrm{help}\:\mathrm{on}\:\:\mathrm{Q156270} \\ $$
Commented by mr W last updated on 10/Oct/21
$${what}\:{are}\:{you}\:{talking}\:{about}? \\ $$
Commented by Tawa11 last updated on 10/Oct/21
$$\mathrm{I}\:\mathrm{mean}\:\mathrm{help}\:\mathrm{me}\:\mathrm{with}\:\mathrm{the}\:\mathrm{question}.\:\:\:\mathrm{Q156270} \\ $$
Commented by mr W last updated on 10/Oct/21
$${i}\:{don}'{t}\:{know}\:{what}\:{that}\:{question}\:{is} \\ $$$${about}.\:{i}\:{can}'{t}\:{help}\:{you}\:{by}\:{things}\:{which} \\ $$$${i}\:{don}'{t}\:{understand}. \\ $$
Commented by Tawa11 last updated on 10/Oct/21
$$\mathrm{Alright}.\:\mathrm{Thanks}\:\mathrm{sir}. \\ $$