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Question Number 194624 by BaliramKumar last updated on 11/Jul/23

Commented by BaliramKumar last updated on 11/Jul/23

Please Help only answer no solution BF=? FH=? HJ=? JC=? & AE=? EG=? GI=? ID=?

$$\mathrm{Please}\:\mathrm{Help}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{only}\:\mathrm{answer}\:\mathrm{no}\:\mathrm{solution} \\ $$$$\mathrm{BF}=? \\ $$$$\mathrm{FH}=? \\ $$$$\mathrm{HJ}=? \\ $$$$\mathrm{JC}=? \\ $$$$\& \\ $$$$\mathrm{AE}=? \\ $$$$\mathrm{EG}=? \\ $$$$\mathrm{GI}=? \\ $$$$\mathrm{ID}=? \\ $$

Commented by BaliramKumar last updated on 11/Jul/23

yes sir approx answer need

$$\mathrm{yes}\:\mathrm{sir} \\ $$$$\mathrm{approx}\:\mathrm{answer}\:\mathrm{need} \\ $$

Answered by Frix last updated on 11/Jul/23

The area of ABCD is A=((AB^(−) +CD^(−) )/2)×AD^(−) A_1 =A_2 =A_3 =A_4 =(A/4) Put AD on the x−axis with A=(0/0) BC: f(x)=((154x)/(575))+122 (A/4)=∫_0 ^a f(x)dx=∫_a ^b f(x)dx=∫_b ^c f(x)dx=∫_c ^(575) f(xx)dx ...should be easy to solve

$$\mathrm{The}\:\mathrm{area}\:\mathrm{of}\:{ABCD}\:\mathrm{is}\:\mathbb{A}=\frac{\overline {{AB}}+\overline {{CD}}}{\mathrm{2}}×\overline {{AD}} \\ $$$${A}_{\mathrm{1}} ={A}_{\mathrm{2}} ={A}_{\mathrm{3}} ={A}_{\mathrm{4}} =\frac{\mathbb{A}}{\mathrm{4}} \\ $$$$\mathrm{Put}\:{AD}\:\mathrm{on}\:\mathrm{the}\:{x}−\mathrm{axis}\:\mathrm{with}\:{A}=\left(\mathrm{0}/\mathrm{0}\right) \\ $$$${BC}:\:{f}\left({x}\right)=\frac{\mathrm{154}{x}}{\mathrm{575}}+\mathrm{122} \\ $$$$\frac{\mathbb{A}}{\mathrm{4}}=\underset{\mathrm{0}} {\overset{{a}} {\int}}{f}\left({x}\right){dx}=\underset{{a}} {\overset{{b}} {\int}}{f}\left({x}\right){dx}=\underset{{b}} {\overset{{c}} {\int}}{f}\left({x}\right){dx}=\underset{{c}} {\overset{\mathrm{575}} {\int}}{f}\left({xx}\right){dx} \\ $$$$...\mathrm{should}\:\mathrm{be}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solve} \\ $$

Commented by BaliramKumar last updated on 12/Jul/23

Thanks

$$\mathrm{Thanks} \\ $$

Commented by mahdipoor last updated on 12/Jul/23

your solution is with approximation A=90 versus my solution is exact its reason for our difference answer like you,i got A=90 and my ans was AE=193,41... and etc .

$${your}\:{solution}\:{is}\:{with}\:{approximation}\:{A}=\mathrm{90} \\ $$$${versus}\:{my}\:{solution}\:{is}\:{exact}\: \\ $$$${its}\:{reason}\:{for}\:{our}\:{difference}\:{answer}\: \\ $$$${like}\:{you},{i}\:{got}\:{A}=\mathrm{90}\:{and}\:{my}\:{ans}\:{was}\: \\ $$$${AE}=\mathrm{193},\mathrm{41}...\:{and}\:{etc}\:. \\ $$

Commented by Frix last updated on 12/Jul/23

I get AE≈193 EG≈148 GI≈124 ID≈109 EF≈174 GH≈213 IJ≈247 BF≈200 FH≈153 HJ≈129 JC≈113

$$\mathrm{I}\:\mathrm{get} \\ $$$${AE}\approx\mathrm{193}\:\:{EG}\approx\mathrm{148}\:\:{GI}\approx\mathrm{124}\:\:{ID}\approx\mathrm{109} \\ $$$${EF}\approx\mathrm{174}\:\:{GH}\approx\mathrm{213}\:\:{IJ}\approx\mathrm{247} \\ $$$${BF}\approx\mathrm{200}\:\:{FH}\approx\mathrm{153}\:\:{HJ}\approx\mathrm{129}\:\:{JC}\approx\mathrm{113} \\ $$

Commented by Frix last updated on 12/Jul/23

Obviously we have 2 angles of 90° at A & D and the length 575 is an approximate value.

$$\mathrm{Obviously}\:\mathrm{we}\:\mathrm{have}\:\mathrm{2}\:\mathrm{angles}\:\mathrm{of}\:\mathrm{90}°\:\mathrm{at}\:{A}\:\&\:{D} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{length}\:\mathrm{575}\:\mathrm{is}\:\mathrm{an}\:\mathrm{approximate}\:\mathrm{value}. \\ $$

Commented by mahdipoor last updated on 12/Jul/23

we must take one of the A=90 or a length is approximate , mr BK and i consider length is exact ,for exact answer its important (which one is approximatio?) , but for approximation answer not important

$${we}\:{must}\:{take}\:{one}\:{of}\:{the}\:{A}=\mathrm{90}\:{or}\:{a}\:{length} \\ $$$${is}\:{approximate}\:, \\ $$$${mr}\:{BK}\:{and}\:{i}\:{consider}\:{length}\:{is}\:{exact}\:,{for}\: \\ $$$${exact}\:{answer}\:{its}\:{important}\:\left({which}\:{one}\:{is}\:{approximatio}?\right)\:, \\ $$$${but}\:{for}\:{approximation}\:{answer}\:{not}\:{important} \\ $$$$ \\ $$

Answered by mahdipoor last updated on 12/Jul/23

wrong: determine vertical line (BH) , H∈DC BHC is Δ and ∠H=90 (BH^2 +HC^2 )^(1/2) =BC⇒ BH=AD=575 , HC=DC−AB=154 , ⇒(575^2 +154^2 )=595⇒595.24...=595

$${wrong}: \\ $$$${determine}\:{vertical}\:{line}\:\left({BH}\right)\:,\:{H}\in{DC} \\ $$$${BHC}\:{is}\:\Delta\:{and}\:\angle{H}=\mathrm{90} \\ $$$$\left({BH}^{\mathrm{2}} +{HC}^{\mathrm{2}} \right)^{\mathrm{1}/\mathrm{2}} ={BC}\Rightarrow \\ $$$${BH}={AD}=\mathrm{575}\:,\:{HC}={DC}−{AB}=\mathrm{154}\:, \\ $$$$\Rightarrow\left(\mathrm{575}^{\mathrm{2}} +\mathrm{154}^{\mathrm{2}} \right)=\mathrm{595}\Rightarrow\mathrm{595}.\mathrm{24}...=\mathrm{595}\: \\ $$

Commented by BaliramKumar last updated on 12/Jul/23

∠A=∠D ≈ 90°

$$\angle\mathrm{A}=\angle\mathrm{D}\:\approx\:\mathrm{90}° \\ $$

Commented by mahdipoor last updated on 12/Jul/23

∠A=90+α , tanα=a ; ∠B=90+β , tanβ=b AB=x , h_1 =AH_1 ,H_1 ∈EF ; h_2 =AH_2 ,H_2 ∈GH;h_3 ,... A≡total Area A_n =(n/4)A=((x+(x+h_n (a+b)))/2)×h_n (note: A_(n ) in eq ≡A_1 +A_2 +...+A_n in shape ) ⇒(a+b)h_n ^2 +(2x)h_n −((nA)/2)⇒h_n >0 h_n =((−2x+(√(4x^2 +2nA(a+b))))/(2(a+b))) now you can determine other ... in this case: , h_n ⋍449(√(.957+n))−429.82 AE=h_1 cosα⋍198.3 EG=AF−AE=h_2 cosα−198.3=144 ...

$$\angle{A}=\mathrm{90}+\alpha\:,\:{tan}\alpha={a}\:;\:\:\angle{B}=\mathrm{90}+\beta\:,\:{tan}\beta={b} \\ $$$${AB}={x}\:,\:{h}_{\mathrm{1}} ={AH}_{\mathrm{1}} ,{H}_{\mathrm{1}} \in{EF}\:;\:{h}_{\mathrm{2}} ={AH}_{\mathrm{2}} ,{H}_{\mathrm{2}} \in{GH};{h}_{\mathrm{3}} ,... \\ $$$${A}\equiv{total}\:\:{Area} \\ $$$${A}_{{n}} =\frac{{n}}{\mathrm{4}}{A}=\frac{{x}+\left({x}+{h}_{{n}} \left({a}+{b}\right)\right)}{\mathrm{2}}×{h}_{{n}} \\ $$$$\left({note}:\:{A}_{{n}\:} {in}\:{eq}\:\equiv{A}_{\mathrm{1}} +{A}_{\mathrm{2}} +...+{A}_{{n}} \:{in}\:{shape}\:\right) \\ $$$$\Rightarrow\left({a}+{b}\right){h}_{{n}} ^{\mathrm{2}} +\left(\mathrm{2}{x}\right){h}_{{n}} −\frac{{nA}}{\mathrm{2}}\Rightarrow{h}_{{n}} >\mathrm{0} \\ $$$${h}_{{n}} =\frac{−\mathrm{2}{x}+\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{2}{nA}\left({a}+{b}\right)}}{\mathrm{2}\left({a}+{b}\right)} \\ $$$${now}\:{you}\:{can}\:{determine}\:{other}\:... \\ $$$${in}\:{this}\:{case}:\:, \\ $$$${h}_{{n}} \backsimeq\mathrm{449}\sqrt{.\mathrm{957}+{n}}−\mathrm{429}.\mathrm{82} \\ $$$${AE}={h}_{\mathrm{1}} {cos}\alpha\backsimeq\mathrm{198}.\mathrm{3} \\ $$$${EG}={AF}−{AE}={h}_{\mathrm{2}} {cos}\alpha−\mathrm{198}.\mathrm{3}=\mathrm{144} \\ $$$$... \\ $$

Commented by BaliramKumar last updated on 12/Jul/23

I calculate AE = 193.41 Link by integration

$$\mathrm{I}\:\mathrm{calculate}\:\mathrm{AE}\:=\:\mathrm{193}.\mathrm{41}\:\mathrm{Link}\:\:\:\:\:\:\: \\ $$$$\mathrm{by}\:\mathrm{integration} \\ $$

Commented by Frix last updated on 12/Jul/23

∠A, D ≈ 90°? makes no sense. In this case the horizontal lines cannot be parallel.

$$\angle{A},\:{D}\:\approx\:\mathrm{90}°?\:\mathrm{makes}\:\mathrm{no}\:\mathrm{sense}.\:\mathrm{In}\:\mathrm{this}\:\mathrm{case} \\ $$$$\mathrm{the}\:\mathrm{horizontal}\:\mathrm{lines}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{parallel}. \\ $$

Commented by BaliramKumar last updated on 12/Jul/23

Commented by BaliramKumar last updated on 12/Jul/23

To be divided equally among four people.

$$ \\ $$To be divided equally among four people.

Commented by Frix last updated on 12/Jul/23

If the 2 angles are not 90° there are infinite possibilities to divide

$$\mathrm{If}\:\mathrm{the}\:\mathrm{2}\:\mathrm{angles}\:\mathrm{are}\:\mathrm{not}\:\mathrm{90}°\:\mathrm{there}\:\mathrm{are}\:\mathrm{infinite} \\ $$$$\mathrm{possibilities}\:\mathrm{to}\:\mathrm{divide} \\ $$

Commented by BaliramKumar last updated on 12/Jul/23

yes but 3 red line & 2 side line should be approx parallel

$$\mathrm{yes} \\ $$$$\mathrm{but}\:\mathrm{3}\:\mathrm{red}\:\mathrm{line}\:\&\:\mathrm{2}\:\mathrm{side}\:\mathrm{line}\:\:\mathrm{should}\:\mathrm{be}\:\:\mathrm{approx}\:\mathrm{parallel} \\ $$

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