Question Number 157379 by quvonch3737 last updated on 22/Oct/21
Commented by quvonch3737 last updated on 22/Oct/21
$${prove}\:{help} \\ $$
Answered by mr W last updated on 22/Oct/21
![R=((3+4)/( (√2)))=(7/( (√2))) tan α=((4−3.5)/(3.5))=(1/7) ⇒cos α=(7/( (√(50)))) A_(yellow) =(R^2 /2)[sin (45+α)+sin (45−α)]−(R^2 /2) A_(yellow) =(R^2 /2)((√2) cos α−1) A_(yellow) =(R^2 /2)((√2) ×(7/( (√(50))))−1) A_(yellow) =(R^2 /5) i.e. shaded area is 20% of the square. A_(yellow) =(1/5)×((7/( (√2))))^2 =((49)/(10))](https://www.tinkutara.com/question/Q157394.png)
$${R}=\frac{\mathrm{3}+\mathrm{4}}{\:\sqrt{\mathrm{2}}}=\frac{\mathrm{7}}{\:\sqrt{\mathrm{2}}} \\ $$$$\mathrm{tan}\:\alpha=\frac{\mathrm{4}−\mathrm{3}.\mathrm{5}}{\mathrm{3}.\mathrm{5}}=\frac{\mathrm{1}}{\mathrm{7}}\:\Rightarrow\mathrm{cos}\:\alpha=\frac{\mathrm{7}}{\:\sqrt{\mathrm{50}}} \\ $$$${A}_{{yellow}} =\frac{{R}^{\mathrm{2}} }{\mathrm{2}}\left[\mathrm{sin}\:\left(\mathrm{45}+\alpha\right)+\mathrm{sin}\:\left(\mathrm{45}−\alpha\right)\right]−\frac{{R}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${A}_{{yellow}} =\frac{{R}^{\mathrm{2}} }{\mathrm{2}}\left(\sqrt{\mathrm{2}}\:\mathrm{cos}\:\alpha−\mathrm{1}\right) \\ $$$${A}_{{yellow}} =\frac{{R}^{\mathrm{2}} }{\mathrm{2}}\left(\sqrt{\mathrm{2}}\:×\frac{\mathrm{7}}{\:\sqrt{\mathrm{50}}}−\mathrm{1}\right) \\ $$$${A}_{{yellow}} =\frac{{R}^{\mathrm{2}} }{\mathrm{5}} \\ $$$${i}.{e}.\:{shaded}\:{area}\:{is}\:\mathrm{20\%}\:{of}\:{the}\:{square}. \\ $$$${A}_{{yellow}} =\frac{\mathrm{1}}{\mathrm{5}}×\left(\frac{\mathrm{7}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} =\frac{\mathrm{49}}{\mathrm{10}} \\ $$
Commented by mr W last updated on 22/Oct/21
Commented by Tawa11 last updated on 23/Oct/21
$$\mathrm{Great}\:\mathrm{sirs} \\ $$
Answered by ajfour last updated on 22/Oct/21
Commented by ajfour last updated on 22/Oct/21
![lets call square side s. diagonal s(√2)=3+4 ⇒ s=(7/( (√2))) D[((3s)/7), ((4s)/7)] slope of AD tan θ=((s−((4s)/7))/(s−((3s)/7)))=(3/4) A[s(1−cos θ), s(1−sin θ)] A[(s/5), ((2s)/5)] Area △ABC =(s^2 /2)−(s/2)(x_A +y_A ) =(s^2 /5) ★](https://www.tinkutara.com/question/Q157404.png)
$${lets}\:{call}\:{square}\:{side}\:{s}. \\ $$$${diagonal}\:\:\:{s}\sqrt{\mathrm{2}}=\mathrm{3}+\mathrm{4} \\ $$$$\Rightarrow\:\:{s}=\frac{\mathrm{7}}{\:\sqrt{\mathrm{2}}} \\ $$$${D}\left[\frac{\mathrm{3}{s}}{\mathrm{7}},\:\frac{\mathrm{4}{s}}{\mathrm{7}}\right]\:\:\: \\ $$$${slope}\:{of}\:{AD}\: \\ $$$$\:\mathrm{tan}\:\theta=\frac{{s}−\frac{\mathrm{4}{s}}{\mathrm{7}}}{{s}−\frac{\mathrm{3}{s}}{\mathrm{7}}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${A}\left[{s}\left(\mathrm{1}−\mathrm{cos}\:\theta\right),\:{s}\left(\mathrm{1}−\mathrm{sin}\:\theta\right)\right] \\ $$$${A}\left[\frac{{s}}{\mathrm{5}},\:\frac{\mathrm{2}{s}}{\mathrm{5}}\right] \\ $$$${Area}\:\bigtriangleup{ABC}\: \\ $$$$\:\:=\frac{{s}^{\mathrm{2}} }{\mathrm{2}}−\frac{{s}}{\mathrm{2}}\left({x}_{{A}} +{y}_{{A}} \right) \\ $$$$\:\:=\frac{{s}^{\mathrm{2}} }{\mathrm{5}}\:\:\bigstar \\ $$$$ \\ $$