Question Number 70516 by TawaTawa last updated on 05/Oct/19
Commented by TawaTawa last updated on 05/Oct/19
ABCD and DEFK are squares DK=6cm A(BEK)=?
Answered by mr W last updated on 05/Oct/19
![let AD=AB=a [ABCD]=a^2 [DKE]=(6^2 /2) [BAK]=((a(a+6))/2) [BCE]=((a(a−6))/2) A_(shade) =[ABCD]+[DKE]−[BAK]−[BCE] ⇒A_(shade) =a^2 +(6^2 /2)−((a(a+6))/2)−((a(a−6))/2) =(6^2 /2) =18](Q70550.png)
$${let}\:{AD}={AB}={a} \\ $$$$\left[{ABCD}\right]={a}^{\mathrm{2}} \\ $$$$\left[{DKE}\right]=\frac{\mathrm{6}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\left[{BAK}\right]=\frac{{a}\left({a}+\mathrm{6}\right)}{\mathrm{2}} \\ $$$$\left[{BCE}\right]=\frac{{a}\left({a}−\mathrm{6}\right)}{\mathrm{2}} \\ $$$${A}_{{shade}} =\left[{ABCD}\right]+\left[{DKE}\right]−\left[{BAK}\right]−\left[{BCE}\right] \\ $$$$\Rightarrow{A}_{{shade}} ={a}^{\mathrm{2}} +\frac{\mathrm{6}^{\mathrm{2}} }{\mathrm{2}}−\frac{{a}\left({a}+\mathrm{6}\right)}{\mathrm{2}}−\frac{{a}\left({a}−\mathrm{6}\right)}{\mathrm{2}} \\ $$$$=\frac{\mathrm{6}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$=\mathrm{18} \\ $$
Commented by TawaTawa last updated on 05/Oct/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by TawaTawa last updated on 05/Oct/19
$$\mathrm{Sir}\:\mathrm{am}\:\mathrm{trying}\:\mathrm{to}\:\mathrm{see}\:\mathrm{where}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{come}\:\mathrm{from} \\ $$$$\mathrm{Please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{label}\:\mathrm{so}\:\mathrm{that}\:\mathrm{i}\:\mathrm{can}\:\mathrm{study}\:\mathrm{it}.\:\mathrm{Sorry}\:\mathrm{for}\:\mathrm{disturbing}. \\ $$
Commented by mr W last updated on 05/Oct/19
$${comments}\:{added}. \\ $$
Commented by TawaTawa last updated on 05/Oct/19
$$\mathrm{Wow},\:\mathrm{now}\:\mathrm{i}\:\mathrm{understand}\:\mathrm{sir}.\:\mathrm{Thanks}\:\mathrm{for}\:\mathrm{every}\:\mathrm{time}.\:\mathrm{God}\:\mathrm{bless} \\ $$$$\mathrm{you}\:\mathrm{sir} \\ $$