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Question Number 67692 by mr W last updated on 30/Aug/19

Answered by mr W last updated on 30/Aug/19

Commented by mr W last updated on 30/Aug/19

AD=(√(7^2 +24^2 ))=25  FC=(√(25^2 −15^2 ))=20  ((HC)/(FC))=((FC)/(BC))  ⇒HC=((20^2 )/(25))=16  ((HF)/(HC))=((BF)/(FC))  ⇒HF=((16×15)/(20))=12  ((FG)/(GD))=((AE)/(ED))  ⇒FG=((16×7)/(24))=((14)/3)  AB=HF+FG=12+((14)/3)=((50)/3)

$${AD}=\sqrt{\mathrm{7}^{\mathrm{2}} +\mathrm{24}^{\mathrm{2}} }=\mathrm{25} \\ $$$${FC}=\sqrt{\mathrm{25}^{\mathrm{2}} −\mathrm{15}^{\mathrm{2}} }=\mathrm{20} \\ $$$$\frac{{HC}}{{FC}}=\frac{{FC}}{{BC}} \\ $$$$\Rightarrow{HC}=\frac{\mathrm{20}^{\mathrm{2}} }{\mathrm{25}}=\mathrm{16} \\ $$$$\frac{{HF}}{{HC}}=\frac{{BF}}{{FC}} \\ $$$$\Rightarrow{HF}=\frac{\mathrm{16}×\mathrm{15}}{\mathrm{20}}=\mathrm{12} \\ $$$$\frac{{FG}}{{GD}}=\frac{{AE}}{{ED}} \\ $$$$\Rightarrow{FG}=\frac{\mathrm{16}×\mathrm{7}}{\mathrm{24}}=\frac{\mathrm{14}}{\mathrm{3}} \\ $$$${AB}={HF}+{FG}=\mathrm{12}+\frac{\mathrm{14}}{\mathrm{3}}=\frac{\mathrm{50}}{\mathrm{3}} \\ $$

Commented by Prithwish sen last updated on 30/Aug/19

(1/2)×HF×BC=(1/2)×BF×FC  ⇒HF= ((15×20)/(25)) = 12

$$\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{HF}×\mathrm{BC}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{BF}×\mathrm{FC} \\ $$$$\Rightarrow\mathrm{HF}=\:\frac{\mathrm{15}×\mathrm{20}}{\mathrm{25}}\:=\:\mathrm{12} \\ $$

Commented by Prithwish sen last updated on 30/Aug/19

Beautiful

$$\mathrm{Beautiful} \\ $$

Commented by TawaTawa last updated on 30/Aug/19

Wow. God bless you sir

$$\mathrm{Wow}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by behi83417@gmail.com last updated on 30/Aug/19

FD=24  or  ED=24  ?  can it be solved if:  FD=24  ?

$$\mathrm{FD}=\mathrm{24}\:\:\mathrm{or}\:\:\mathrm{ED}=\mathrm{24}\:\:? \\ $$$$\mathrm{can}\:\mathrm{it}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{if}:\:\:\mathrm{FD}=\mathrm{24}\:\:? \\ $$

Commented by mr W last updated on 30/Aug/19

ED=24.

$${ED}=\mathrm{24}. \\ $$

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