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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

$$letAD=AB=a$$$$\left[ABCD\right]=a^2$$$$\left[DKE\right]=\frac{6^2}{2}$$$$\left[BAK\right]=\frac{a(a+6)}{2}$$$$\left[BCE\right]=\frac{a(a-6)}{2}$$$$A_{shade}=\left[ABCD\right]+\left[DKE\right]-\left[BAK\right]-\left[BCE\right]$$$$\Rightarrow A_{shade}=a^2+\frac{6^2}{2}-\frac{a(a+6)}{2}-\frac{a(a-6)}{2}$$$$=\frac{6^2}{2}$$$$=18$$

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

$$commentsadded.$$

Commented by TawaTawa last updated on 05/Oct/19

$$\mathrm{Wow},\mathrm{now}\mathrm{i}\mathrm{understand}\mathrm{sir}.\mathrm{Thanks}\mathrm{for}\mathrm{every}\mathrm{time}.\mathrm{God}\mathrm{bless}$$$$\mathrm{you}\mathrm{sir}$$

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