Question Number 38079 by ajfour last updated on 21/Jun/18
Commented by ajfour last updated on 21/Jun/18
$${Find}\:{coloured}\:{area}\:{ABCD}\:{in} \\ $$$${terms}\:{of}\:{a},{b},{c},{d}\:. \\ $$$${It}\:{is}\:{the}\:{area}\:{common}\:{to}\:{two} \\ $$$${squares}\:{of}\:{side}\:{length}\:{a}. \\ $$
Commented by Rasheed.Sindhi last updated on 22/Jun/18
$$\mathrm{What}'\mathrm{s}\:\mathrm{the}\:\mathrm{source}\:\mathrm{of}\:\mathrm{your}\:\mathrm{questions},\:\boldsymbol{\mathrm{sir}}\:\boldsymbol{\mathrm{ajfour}}? \\ $$
Commented by ajfour last updated on 22/Jun/18
$${I}\:{create}\:{the}\:{physics}\:{and}\:{geometry} \\ $$$${ones}\:{myself}.\:{Integration}\:{and} \\ $$$${algebra}\:{from}\:{books},\:{sometimes}. \\ $$
Commented by Rasheed.Sindhi last updated on 22/Jun/18
You're creative-mind Sir!
Answered by ajfour last updated on 21/Jun/18
![let sin θ = ((d−b)/a) x_A =c−(y_A −y_B )tan θ x_A = (a−b)tan θ area△ABD = (1/2)(a−x_A )(a−b) =(((a−b))/2)[a−c+(a−b)tan θ] y_C =b+(a−c)tan θ y_D −y_C =a−b−(a−c)tan θ area△BCD =(1/2)(y_D −y_C )(a−c) =(((a−c))/2)[a−b−(a−c)tan θ] Area ABCD = area△ABD +area△BCD = (1/2){(a−b)[a−c+(a−b)tan θ] +(a−c)[a−b−(a−c)tan θ]} =(1/2){a^2 −ac−ab−ac+bc+a^2 −ac−ab+bc +(a^2 −2ab+b^2 −a^2 +2ac−c^2 )tan θ} =(a^2 −ab−ac+bc) +(1/2)[(a−b)^2 −(a−c)^2 ](((d−b))/( (√(a^2 −(d−b)^2 )))) } . =(a−c)(a−b) +(c−b)(a−((b+c)/2))(((d−b))/( (√(a^2 −(d−b)^2 )))) .](https://www.tinkutara.com/question/Q38090.png)
$${let}\:\:\mathrm{sin}\:\theta\:=\:\frac{{d}−{b}}{{a}} \\ $$$${x}_{{A}} ={c}−\left({y}_{{A}} −{y}_{{B}} \right)\mathrm{tan}\:\theta \\ $$$$\:\:{x}_{{A}} =\:\left({a}−{b}\right)\mathrm{tan}\:\theta \\ $$$${area}\bigtriangleup{ABD}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({a}−{x}_{{A}} \right)\left({a}−{b}\right)\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\left({a}−{b}\right)}{\mathrm{2}}\left[{a}−{c}+\left({a}−{b}\right)\mathrm{tan}\:\theta\right] \\ $$$${y}_{{C}} ={b}+\left({a}−{c}\right)\mathrm{tan}\:\theta \\ $$$${y}_{{D}} −{y}_{{C}} ={a}−{b}−\left({a}−{c}\right)\mathrm{tan}\:\theta \\ $$$${area}\bigtriangleup{BCD}\:=\frac{\mathrm{1}}{\mathrm{2}}\left({y}_{{D}} −{y}_{{C}} \right)\left({a}−{c}\right) \\ $$$$\:\:\:\:\:=\frac{\left({a}−{c}\right)}{\mathrm{2}}\left[{a}−{b}−\left({a}−{c}\right)\mathrm{tan}\:\theta\right] \\ $$$${Area}\:{ABCD}\:=\:{area}\bigtriangleup{ABD} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{area}\bigtriangleup{BCD} \\ $$$$\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left\{\left({a}−{b}\right)\left[{a}−{c}+\left({a}−{b}\right)\mathrm{tan}\:\theta\right]\right. \\ $$$$\left.\:\:\:\:\:+\left({a}−{c}\right)\left[{a}−{b}−\left({a}−{c}\right)\mathrm{tan}\:\theta\right]\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{{a}^{\mathrm{2}} −{ac}−{ab}−{ac}+{bc}+{a}^{\mathrm{2}} −{ac}−{ab}+{bc}\right. \\ $$$$\left.\:\:+\left({a}^{\mathrm{2}} −\mathrm{2}{ab}+{b}^{\mathrm{2}} −{a}^{\mathrm{2}} +\mathrm{2}{ac}−{c}^{\mathrm{2}} \right)\mathrm{tan}\:\theta\right\} \\ $$$$=\left({a}^{\mathrm{2}} −{ab}−{ac}+{bc}\right) \\ $$$$\left.\:+\frac{\mathrm{1}}{\mathrm{2}}\left[\left({a}−{b}\right)^{\mathrm{2}} −\left({a}−{c}\right)^{\mathrm{2}} \right]\frac{\left({d}−{b}\right)}{\:\sqrt{{a}^{\mathrm{2}} −\left({d}−{b}\overset{\mathrm{2}} {\right)}}}\:\right\}\:. \\ $$$$\:=\left({a}−{c}\right)\left({a}−{b}\right) \\ $$$$\:\:\:\:\:\:\:\:\:+\left({c}−{b}\right)\left({a}−\frac{{b}+{c}}{\mathrm{2}}\right)\frac{\left({d}−{b}\right)}{\:\sqrt{{a}^{\mathrm{2}} −\left({d}−{b}\right)^{\mathrm{2}} }}\:. \\ $$