Question Number 135243 by benjo_mathlover last updated on 11/Mar/21
Commented by benjo_mathlover last updated on 11/Mar/21
Answered by liberty last updated on 13/Mar/21
![since AB∣∣CD triangles ABP and CDP are similar with a side ratio of ∣((AB)/(CD))∣ = (2/5). let ∣AP∣=2x & ∣BP∣=2y Then ∣CP∣=5x and ∣DP∣=5y since ∠APB=90° , [ ABCD] = (1/2)∣AC∣.∣BD∣ = ((49xy)/2) let α =∠ADP and β=∠BCP tan α=((2x)/(5y)) and tan β=((2y)/(5x)) note α+β=45° ⇒1=((tan α+tan β)/(1−tan α.tan β)) ⇒xy = ((10(x^2 +y^2 ))/(21)). We have ∣AB∣^2 =∣AP∣^2 +∣BP∣^2 → x^2 +y^2 =4 so xy=((40)/(21)) and [ ABCD ]=((49)/2).((40)/(21))=((140)/3)](https://www.tinkutara.com/question/Q135416.png)
$${since}\:{AB}\mid\mid{CD}\:{triangles}\:{ABP}\:{and} \\ $$$${CDP}\:{are}\:{similar}\:{with}\:{a}\:{side}\:{ratio} \\ $$$${of}\:\mid\frac{{AB}}{{CD}}\mid\:=\:\frac{\mathrm{2}}{\mathrm{5}}.\:{let}\:\mid{AP}\mid=\mathrm{2}{x}\:\&\:\mid{BP}\mid=\mathrm{2}{y} \\ $$$${Then}\:\mid{CP}\mid=\mathrm{5}{x}\:{and}\:\mid{DP}\mid=\mathrm{5}{y} \\ $$$${since}\:\angle{APB}=\mathrm{90}°\:,\:\left[\:{ABCD}\right]\:= \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mid{AC}\mid.\mid{BD}\mid\:=\:\frac{\mathrm{49}{xy}}{\mathrm{2}} \\ $$$${let}\:\alpha\:=\angle{ADP}\:{and}\:\beta=\angle{BCP} \\ $$$$\mathrm{tan}\:\alpha=\frac{\mathrm{2}{x}}{\mathrm{5}{y}}\:{and}\:\mathrm{tan}\:\beta=\frac{\mathrm{2}{y}}{\mathrm{5}{x}} \\ $$$${note}\:\alpha+\beta=\mathrm{45}°\:\Rightarrow\mathrm{1}=\frac{\mathrm{tan}\:\alpha+\mathrm{tan}\:\beta}{\mathrm{1}−\mathrm{tan}\:\alpha.\mathrm{tan}\:\beta} \\ $$$$\Rightarrow{xy}\:=\:\frac{\mathrm{10}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)}{\mathrm{21}}.\:{We}\:{have} \\ $$$$\mid{AB}\mid^{\mathrm{2}} =\mid{AP}\mid^{\mathrm{2}} +\mid{BP}\mid^{\mathrm{2}} \rightarrow\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{4} \\ $$$${so}\:{xy}=\frac{\mathrm{40}}{\mathrm{21}}\:{and}\:\left[\:{ABCD}\:\right]=\frac{\mathrm{49}}{\mathrm{2}}.\frac{\mathrm{40}}{\mathrm{21}}=\frac{\mathrm{140}}{\mathrm{3}} \\ $$$$ \\ $$