Question Number 183886 by Acem last updated on 31/Dec/22
Answered by HeferH last updated on 31/Dec/22
$${A}\:=\:\left(\frac{{BC}\:+\:{AD}}{\mathrm{2}}\right)\mathrm{15} \\ $$$$\:{AE}\:=\:\sqrt{\mathrm{17}^{\mathrm{2}} −\mathrm{15}^{\mathrm{2}} }\:=\:\sqrt{\mathrm{2}\centerdot\mathrm{32}}\:=\:\mathrm{8} \\ $$$$\:{B}'{D}\:=\:\sqrt{\mathrm{39}^{\mathrm{2}} −\mathrm{15}^{\mathrm{2}} }\:=\:\mathrm{36} \\ $$$$\:{AD}\:=\:{AE}\:+\:{ED}\:=\:\mathrm{8}\:+\:\mathrm{36}\:−\:{BC}\Rightarrow \\ $$$$\:{A}\:\:=\left(\frac{\mathrm{8}+\mathrm{36}}{\mathrm{2}}\right)\mathrm{15}\:=\:\mathrm{66}\centerdot\mathrm{5}=\:\mathrm{330}{u} \\ $$
Commented by Acem last updated on 31/Dec/22
$${Very}\:{well}\:{Sir}!\: \\ $$
Answered by mr W last updated on 31/Dec/22
Commented by mr W last updated on 31/Dec/22
![[ABCD]=[ΔABF]=((AF×BG)/2) =((((√(17^2 −15^2 ))+(√(39^2 −15^2 )))×15)/2) =((44×15)/2)=330 ✓](Q183896.png)
$$\left[{ABCD}\right]=\left[\Delta{ABF}\right]=\frac{{AF}×{BG}}{\mathrm{2}} \\ $$$$=\frac{\left(\sqrt{\mathrm{17}^{\mathrm{2}} −\mathrm{15}^{\mathrm{2}} }+\sqrt{\mathrm{39}^{\mathrm{2}} −\mathrm{15}^{\mathrm{2}} }\right)×\mathrm{15}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{44}×\mathrm{15}}{\mathrm{2}}=\mathrm{330}\:\checkmark \\ $$
Commented by manxsol last updated on 31/Dec/22
![the formula would then be Areatrap(d1,d2,h)= [(√(d1^2 −h^2 )) +(√(d2^2 −h^2 ]))×(h/2)](Q183905.png)
$${the}\:{formula}\:{would}\:{then}\:{be} \\ $$$${Areatrap}\left({d}\mathrm{1},{d}\mathrm{2},{h}\right)= \\ $$$$\left[\sqrt{{d}\mathrm{1}^{\mathrm{2}} −{h}^{\mathrm{2}} \:}\:+\sqrt{\left.{d}\mathrm{2}^{\mathrm{2}} −{h}^{\mathrm{2}} \right]}×\frac{{h}}{\mathrm{2}}\right. \\ $$
Commented by manxsol last updated on 31/Dec/22
Commented by Acem last updated on 31/Dec/22
$${Very}\:{well}\:{Sir}!\: \\ $$
Answered by mr W last updated on 31/Dec/22
![alternative: cos α=((15)/(17)) ⇒sin α=((√(17^2 −15^2 ))/(17))=(8/(17)) cos β=((15)/(39)) ⇒sin α=((√(39^2 −15^2 ))/(39))=((36)/(39)) [ABCD]=(1/2)×17×39×sin (α+β) =(1/2)×17×39×((8/(17))×((15)/(39))+((15)/(17))×((36)/(39))) =((15×44)/2)=330 ✓](Q183897.png)
$${alternative}: \\ $$$$\mathrm{cos}\:\alpha=\frac{\mathrm{15}}{\mathrm{17}}\:\Rightarrow\mathrm{sin}\:\alpha=\frac{\sqrt{\mathrm{17}^{\mathrm{2}} −\mathrm{15}^{\mathrm{2}} }}{\mathrm{17}}=\frac{\mathrm{8}}{\mathrm{17}} \\ $$$$\mathrm{cos}\:\beta=\frac{\mathrm{15}}{\mathrm{39}}\:\Rightarrow\mathrm{sin}\:\alpha=\frac{\sqrt{\mathrm{39}^{\mathrm{2}} −\mathrm{15}^{\mathrm{2}} }}{\mathrm{39}}=\frac{\mathrm{36}}{\mathrm{39}} \\ $$$$\left[{ABCD}\right]=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{17}×\mathrm{39}×\mathrm{sin}\:\left(\alpha+\beta\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{17}×\mathrm{39}×\left(\frac{\mathrm{8}}{\mathrm{17}}×\frac{\mathrm{15}}{\mathrm{39}}+\frac{\mathrm{15}}{\mathrm{17}}×\frac{\mathrm{36}}{\mathrm{39}}\right) \\ $$$$=\frac{\mathrm{15}×\mathrm{44}}{\mathrm{2}}=\mathrm{330}\:\checkmark \\ $$
Commented by Acem last updated on 31/Dec/22
$${Very}\:{well}\:{Sir}!\: \\ $$$$\:{and}\:{please}\:{explanation}\:{about}\:\mathrm{sin}\:\left(\alpha+\beta\right)\:{thanks} \\ $$
Commented by mr W last updated on 31/Dec/22
Commented by mr W last updated on 31/Dec/22
$${A}=\frac{\mathrm{1}}{\mathrm{2}}{d}_{\mathrm{1}} ×{d}_{\mathrm{2}} ×\mathrm{sin}\:\theta \\ $$$$\theta=\alpha+\beta \\ $$
Commented by Acem last updated on 31/Dec/22
$${Excellent}!\:{Thank}\:{you}\:{dear}\:{friend} \\ $$
Answered by Acem last updated on 31/Dec/22
