Are-give-two-parallel-lines-and-a-point-A-is-given-between-them-and-the-distance-from-A-to-the-lines-are-a-and-b-Determine-the-cathetus-of-right-angled-triangle-in-A-knowing-that-the-other-vertices

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Question Number 76386 by Maclaurin Stickker last updated on 27/Dec/19

$$Aregivetwoparallellinesandapoint$$$$Aisgivenbetweenthem,andthe$$$$distancefromAtothelinesare$$$$\mathit{a}and\mathit{b}.Determinethecathetusof$$$$right-angledtriangleinAknowing$$$$thattheotherverticesbelongto$$$$theparallellinesandtheareaof$$$$thetriangleisequalto{\mathit{k}}^2.$$

Commented by Maclaurin Stickker last updated on 27/Dec/19

$$thankyou$$

Commented by MJS last updated on 27/Dec/19

$$\mathrm{no}.\mathrm{the}\mathrm{distance}\mathrm{between}\mathrm{a}\mathrm{point}\mathrm{and}\mathrm{a}\mathrm{line}$$$$\mathrm{is}\mathrm{the}\mathrm{minimum}\mathrm{distance}\Rightarrow a\mathrm{and}b\mathrm{are}\mathrm{part}$$$$\mathrm{of}\mathrm{a}\mathrm{line}\mathrm{right}\mathrm{angled}\mathrm{to}\mathrm{the}2\mathrm{parallels}$$

Answered by mr W last updated on 27/Dec/19

$$\frac{x}{a}=\frac{b}{y}$$$$\Rightarrow y=\frac{ab}{x}$$$$AC=\sqrt{a^2+x^2}=a\sqrt{1+\frac{x^2}{a^2}}$$$$AB=\sqrt{b^2+y^2}=\sqrt{b^2+\frac{a^2{b}^2}{x^2}}=b\sqrt{1+\frac{a^2}{x^2}}$$$$\frac{1}{2}\sqrt{(a^2+x^2)(b^2+\frac{a^2{b}^2}{x^2})}=k^2=area$$$$(1+\frac{x^2}{a^2})(1+\frac{a^2}{x^2})={\left(\frac{2k^2}{ab}\right)}^2$$$$let\xi =\frac{x^2}{a^2},\lambda ={\left(\frac{2k^2}{ab}\right)}^2$$$$(1+\xi )(1+\frac{1}{\xi })=\lambda$$$${\xi }^2-(\lambda -2)\xi +1=0$$$$\Rightarrow \xi =\frac{\lambda -2\pm \sqrt{\lambda (\lambda -4)}}{2}$$$$\Rightarrow x=a\sqrt{\frac{\lambda -2\pm \sqrt{\lambda (\lambda -4)}}{2}}$$$$weseethereisalwayssolution\left(s\right)$$$$if\lambda ⩾4,i.e.k^2⩾ab.$$$$$$$$cathetiofthesearchedtriangle:$$$$AC=a\sqrt{1+\xi }$$$$AB=b\sqrt{1+\frac{1}{\xi }}$$

Commented by mr W last updated on 27/Dec/19

Commented by john santu last updated on 27/Dec/19

$$greatsir$$

Commented by john santu last updated on 27/Dec/19

$$sirwhy\measuredangle A={90}^o?$$

Commented by mr W last updated on 27/Dec/19

$$itisgiveninthequestionthatthe$$$$rightangleisatpointA.$$

Commented by john santu last updated on 27/Dec/19

$$ohhyes.thanksyousir$$

Commented by john santu last updated on 27/Dec/19

$$howtudrawingthistriangle$$$$inyourphonesir?$$

Commented by mr W last updated on 27/Dec/19

$$i{don}^{\prime }tuseanyspecialapp.ijust$$$$usetheappwhichispreinstalled$$$$inmyhuaweismartphone.$$

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