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Question Number 144888 by imjagoll last updated on 30/Jun/21

Commented by imjagoll last updated on 30/Jun/21

find the shaded area.

$$\mathrm{find}\:\mathrm{the}\:\mathrm{shaded}\:\mathrm{area}. \\ $$

Answered by nimnim last updated on 30/Jun/21

Commented by mr W last updated on 30/Jun/21

how can you say BC//AD?

$${how}\:{can}\:{you}\:{say}\:{BC}//{AD}? \\ $$

Commented by nimnim last updated on 30/Jun/21

Ok. I′m done. Sir, Please can you show me the correct way

$$\mathrm{Ok}.\:\mathrm{I}'\mathrm{m}\:\mathrm{done}.\:\mathrm{Sir},\:\mathrm{Please}\:\mathrm{can}\:\mathrm{you}\:\mathrm{show}\:\mathrm{me}\:\mathrm{the}\:\mathrm{correct}\:\mathrm{way} \\ $$

Answered by mr W last updated on 30/Jun/21

Commented by mr W last updated on 30/Jun/21

BC^2 =(4+5)^2 +3^2 =90  AC^2 =BC^2 −3^2 =81  AC=9  DE=(4/(4+5))×3=(4/3)  EF=(5/(4+5))×3=(5/3)  A_(shade) =(1/2)×9×3−(1/2)×4×(4/3)+(1/2)×5×(5/3)      =15

$${BC}^{\mathrm{2}} =\left(\mathrm{4}+\mathrm{5}\right)^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} =\mathrm{90} \\ $$$${AC}^{\mathrm{2}} ={BC}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} =\mathrm{81} \\ $$$${AC}=\mathrm{9} \\ $$$${DE}=\frac{\mathrm{4}}{\mathrm{4}+\mathrm{5}}×\mathrm{3}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${EF}=\frac{\mathrm{5}}{\mathrm{4}+\mathrm{5}}×\mathrm{3}=\frac{\mathrm{5}}{\mathrm{3}} \\ $$$${A}_{{shade}} =\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{9}×\mathrm{3}−\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{4}×\frac{\mathrm{4}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{5}×\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\:\:\:\:=\mathrm{15} \\ $$

Commented by nimnim last updated on 01/Jul/21

Thank you Sir. May the good Lord bless you.

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir}.\:\mathrm{May}\:\mathrm{the}\:\mathrm{good}\:\mathrm{Lord}\:\mathrm{bless}\:\mathrm{you}. \\ $$

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