Question-223317

Question Number 223317 by fantastic last updated on 21/Jul/25

Answered by A5T last updated on 21/Jul/25

Commented by A5T last updated on 21/Jul/25

((DA×20)/2)=150⇒DA=15⇒CD=(√(20^2 +15^2 ))=25 cos(90+∠CDA)=−sin∠CDA=−((20)/(25))=((−4)/5) AE^2 =DE^2 +AD^2 −2DE×ADcos(90+∠CDA) ⇒20^2 =DE^2 +15^2 −24DE⇒DE=(√(319))−12 AH⊥CD⇒HG=HD=((DG)/2)⇒CH=25−((DG)/2) AC^2 =CH×DC⇒20^2 =(25−((DG)/2))×25⇒DG=18 ⇒[DEFG]=18((√(319))−12)

$$\frac{\mathrm{DA}×\mathrm{20}}{\mathrm{2}}=\mathrm{150}\Rightarrow\mathrm{DA}=\mathrm{15}\Rightarrow\mathrm{CD}=\sqrt{\mathrm{20}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} }=\mathrm{25} \\ $$$$\mathrm{cos}\left(\mathrm{90}+\angle\mathrm{CDA}\right)=−\mathrm{sin}\angle\mathrm{CDA}=−\frac{\mathrm{20}}{\mathrm{25}}=\frac{−\mathrm{4}}{\mathrm{5}} \\ $$$$\mathrm{AE}^{\mathrm{2}} =\mathrm{DE}^{\mathrm{2}} +\mathrm{AD}^{\mathrm{2}} −\mathrm{2DE}×\mathrm{ADcos}\left(\mathrm{90}+\angle\mathrm{CDA}\right) \\ $$$$\Rightarrow\mathrm{20}^{\mathrm{2}} =\mathrm{DE}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} −\mathrm{24DE}\Rightarrow\mathrm{DE}=\sqrt{\mathrm{319}}−\mathrm{12} \\ $$$$\mathrm{AH}\bot\mathrm{CD}\Rightarrow\mathrm{HG}=\mathrm{HD}=\frac{\mathrm{DG}}{\mathrm{2}}\Rightarrow\mathrm{CH}=\mathrm{25}−\frac{\mathrm{DG}}{\mathrm{2}} \\ $$$$\mathrm{AC}^{\mathrm{2}} =\mathrm{CH}×\mathrm{DC}\Rightarrow\mathrm{20}^{\mathrm{2}} =\left(\mathrm{25}−\frac{\mathrm{DG}}{\mathrm{2}}\right)×\mathrm{25}\Rightarrow\mathrm{DG}=\mathrm{18} \\ $$$$\Rightarrow\left[\mathrm{DEFG}\right]=\mathrm{18}\left(\sqrt{\mathrm{319}}−\mathrm{12}\right) \\ $$

Commented by fantastic last updated on 21/Jul/25

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Commented by A5T last updated on 21/Jul/25

Commented by A5T last updated on 21/Jul/25

Commented by fantastic last updated on 21/Jul/25

can you please tell me how to higher up the decimals. mine only gives 1 decimal

$${can}\:{you}\:{please}\:{tell}\:{me}\:{how}\:{to}\: \\ $$$${higher}\:{up}\:{the}\:{decimals}. \\ $$$${mine}\:{only}\:{gives}\:\mathrm{1}\:{decimal} \\ $$

Commented by fantastic last updated on 21/Jul/25

$${Thanks} \\ $$

Commented by A5T last updated on 21/Jul/25

Click on the settings button in the top right corner, then click on the General menu, then click on the Rounding menu to change the number of decimal places or significant figures.

$$\mathrm{Click}\:\mathrm{on}\:\mathrm{the}\:\mathrm{settings}\:\mathrm{button}\:\mathrm{in}\:\mathrm{the}\:\mathrm{top}\:\mathrm{right}\:\mathrm{corner}, \\ $$$$\mathrm{then}\:\mathrm{click}\:\mathrm{on}\:\mathrm{the}\:\mathrm{General}\:\mathrm{menu},\:\mathrm{then}\:\mathrm{click} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{Rounding}\:\mathrm{menu}\:\mathrm{to}\:\mathrm{change}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of} \\ $$$$\mathrm{decimal}\:\mathrm{places}\:\mathrm{or}\:\mathrm{significant}\:\mathrm{figures}. \\ $$

Answered by fantastic last updated on 21/Jul/25

Commented by fantastic last updated on 21/Jul/25

(1/2)×AB×AD=150 AD=((150×2)/(20))=15 BD=(√(AD^2 +AB^2 ))=(√(15^2 +20^2 ))=(√(625))=25 Let DH=x So x×25=CD×DL[L is bottommost point] x=(((20−15)×(15+20))/(25))=7 BH=25+7=32 AL⊥BH⇒DL=(((32)/2)−7)=9 BD∥EJ So AK⊥EJ So EK=KJ DEKI rectangle EK=9=KJ Let KI=y AL×BD×(1/2)=150 ⇒AL=12 So AE^2 =EK^2 +(12+x)^2 ⇒x=(√(319))−12 Area of rectangle=(9+9)×((√(319))−12) =18((√(319))−12) sq.units

$$\frac{\mathrm{1}}{\mathrm{2}}×{AB}×{AD}=\mathrm{150} \\ $$$${AD}=\frac{\mathrm{150}×\mathrm{2}}{\mathrm{20}}=\mathrm{15} \\ $$$${BD}=\sqrt{{AD}^{\mathrm{2}} +{AB}^{\mathrm{2}} }=\sqrt{\mathrm{15}^{\mathrm{2}} +\mathrm{20}^{\mathrm{2}} }=\sqrt{\mathrm{625}}=\mathrm{25} \\ $$$${Let}\:{DH}={x} \\ $$$${So}\:{x}×\mathrm{25}={CD}×{DL}\left[{L}\:{is}\:{bottommost}\:{point}\right] \\ $$$${x}=\frac{\left(\mathrm{20}−\mathrm{15}\right)×\left(\mathrm{15}+\mathrm{20}\right)}{\mathrm{25}}=\mathrm{7} \\ $$$${BH}=\mathrm{25}+\mathrm{7}=\mathrm{32} \\ $$$${AL}\bot{BH}\Rightarrow{DL}=\left(\frac{\mathrm{32}}{\mathrm{2}}−\mathrm{7}\right)=\mathrm{9} \\ $$$${BD}\parallel{EJ}\: \\ $$$${So}\:{AK}\bot{EJ} \\ $$$${So}\:{EK}={KJ} \\ $$$${DEKI}\:{rectangle} \\ $$$${EK}=\mathrm{9}={KJ} \\ $$$${Let}\:{KI}={y} \\ $$$${AL}×{BD}×\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{150} \\ $$$$\Rightarrow{AL}=\mathrm{12} \\ $$$${So}\:{AE}^{\mathrm{2}} ={EK}^{\mathrm{2}} +\left(\mathrm{12}+{x}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{x}=\sqrt{\mathrm{319}}−\mathrm{12} \\ $$$${Area}\:{of}\:{rectangle}=\left(\mathrm{9}+\mathrm{9}\right)×\left(\sqrt{\mathrm{319}}−\mathrm{12}\right) \\ $$$$=\mathrm{18}\left(\sqrt{\mathrm{319}}−\mathrm{12}\right)\:{sq}.{units} \\ $$

Answered by mr W last updated on 21/Jul/25

Commented by fantastic last updated on 21/Jul/25

$${Thanks} \\ $$

Commented by mr W last updated on 21/Jul/25

R=20 h=((2×150)/(20))=15 c=((Rh)/( (√(R^2 +h^2 ))))=((20×15)/( (√(20^2 +15^2 ))))=12 (√(h^2 −((a/2))^2 ))=c=12 ⇒a=2(√(15^2 −12^2 ))=18 (b+c)^2 +((a/2))^2 =R^2 ⇒b=(√(R^2 −((a/2))^2 ))−12=(√(20^2 −9^2 ))−12=(√(319))−12 A_(red) =ab=18((√(319))−12)

$${R}=\mathrm{20} \\ $$$${h}=\frac{\mathrm{2}×\mathrm{150}}{\mathrm{20}}=\mathrm{15} \\ $$$${c}=\frac{{Rh}}{\:\sqrt{{R}^{\mathrm{2}} +{h}^{\mathrm{2}} }}=\frac{\mathrm{20}×\mathrm{15}}{\:\sqrt{\mathrm{20}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} }}=\mathrm{12} \\ $$$$\sqrt{{h}^{\mathrm{2}} −\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} }={c}=\mathrm{12} \\ $$$$\Rightarrow{a}=\mathrm{2}\sqrt{\mathrm{15}^{\mathrm{2}} −\mathrm{12}^{\mathrm{2}} }=\mathrm{18} \\ $$$$\left({b}+{c}\right)^{\mathrm{2}} +\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\Rightarrow{b}=\sqrt{{R}^{\mathrm{2}} −\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} }−\mathrm{12}=\sqrt{\mathrm{20}^{\mathrm{2}} −\mathrm{9}^{\mathrm{2}} }−\mathrm{12}=\sqrt{\mathrm{319}}−\mathrm{12} \\ $$$${A}_{{red}} ={ab}=\mathrm{18}\left(\sqrt{\mathrm{319}}−\mathrm{12}\right) \\ $$

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