Question Number 171255 by mnjuly1970 last updated on 11/Jun/22
Commented by mr W last updated on 11/Jun/22
$${S}_{\Delta{APB}} =\frac{{AB}×{AP}\:\mathrm{cos}\:\theta}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{AD}×{AP}\:\mathrm{cos}\:\theta}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{AP}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{8}×\mathrm{8}}{\mathrm{4}}=\mathrm{16} \\ $$
Commented by mnjuly1970 last updated on 11/Jun/22
$${thx}\:{alot}\:\:\:{Sir}\:\:{W}\:.... \\ $$
Answered by infinityaction last updated on 11/Jun/22
Commented by mnjuly1970 last updated on 11/Jun/22
$${grateful}\:{sir} \\ $$