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If-the-area-of-a-convex-quadrilateral-is-2k-2-and-the-sum-of-its-diagonals-is-4k-2-then-show-that-this-quadrilateral-is-an-orthodiagonal-one-

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Question Number 151907 by mathdanisur last updated on 24/Aug/21
If the area of a convex quadrilateral is 2k^2 and the sum of its diagonals is 4k^2 , then show that this quadrilateral is an orthodiagonal one.
$$\mathrm{If}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{convex}\:\mathrm{quadrilateral} \\ $$$$\mathrm{is}\:\mathrm{2}\boldsymbol{\mathrm{k}}^{\mathrm{2}} \:\mathrm{and}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{its}\:\mathrm{diagonals} \\ $$$$\mathrm{is}\:\mathrm{4}\boldsymbol{\mathrm{k}}^{\mathrm{2}} ,\:\mathrm{then}\:\mathrm{show}\:\mathrm{that}\:\mathrm{this}\:\mathrm{quadrilateral} \\ $$$$\mathrm{is}\:\mathrm{an}\:\mathrm{orthodiagonal}\:\mathrm{one}. \\ $$
Commented by mr W last updated on 24/Aug/21
please check the question! you can not compare the area with the sum of diagonals! as you can not say 2 cm^2 is equal to 2 cm.
$${please}\:{check}\:{the}\:{question}! \\ $$$${you}\:{can}\:{not}\:{compare}\:{the}\:{area}\:{with} \\ $$$${the}\:{sum}\:{of}\:{diagonals}!\:{as}\:{you}\:{can}\:{not} \\ $$$${say}\:\mathrm{2}\:{cm}^{\mathrm{2}} \:{is}\:{equal}\:{to}\:\mathrm{2}\:{cm}. \\ $$
Commented by mathdanisur last updated on 24/Aug/21
Sorry Ser, 4k
$$\mathrm{Sorry}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{4}\boldsymbol{\mathrm{k}} \\ $$
Answered by mr W last updated on 24/Aug/21
Commented by mr W last updated on 25/Aug/21
say AC=a, BD=b area=((ah_1 )/2)+((ah_2 )/2)=((ab sin θ)/2)=2k^2 ⇒ab=((4k^2 )/(sin θ)) a+b=4k a,b are roots of: x^2 −4kx+((4k^2 )/(sin θ))=0 Δ=16k^2 −((16k^2 )/(sin^2 θ))≥0 1≥(1/(sin^2 θ)) ⇒sin^2 θ≥1 since sin θ≤1, i.e. sin^2 θ≤1 ⇒the only possibility is sin θ=1, i.e. θ=90°
$${say}\:{AC}={a},\:{BD}={b} \\ $$$${area}=\frac{{ah}_{\mathrm{1}} }{\mathrm{2}}+\frac{{ah}_{\mathrm{2}} }{\mathrm{2}}=\frac{{ab}\:\mathrm{sin}\:\theta}{\mathrm{2}}=\mathrm{2}{k}^{\mathrm{2}} \\ $$$$\Rightarrow{ab}=\frac{\mathrm{4}{k}^{\mathrm{2}} }{\mathrm{sin}\:\theta} \\ $$$${a}+{b}=\mathrm{4}{k} \\ $$$${a},{b}\:{are}\:{roots}\:{of}: \\ $$$${x}^{\mathrm{2}} −\mathrm{4}{kx}+\frac{\mathrm{4}{k}^{\mathrm{2}} }{\mathrm{sin}\:\theta}=\mathrm{0} \\ $$$$\Delta=\mathrm{16}{k}^{\mathrm{2}} −\frac{\mathrm{16}{k}^{\mathrm{2}} }{\mathrm{sin}^{\mathrm{2}} \:\theta}\geqslant\mathrm{0} \\ $$$$\mathrm{1}\geqslant\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\theta} \\ $$$$\Rightarrow\mathrm{sin}^{\mathrm{2}} \:\theta\geqslant\mathrm{1} \\ $$$${since}\:\mathrm{sin}\:\theta\leqslant\mathrm{1},\:{i}.{e}.\:\mathrm{sin}^{\mathrm{2}} \:\theta\leqslant\mathrm{1} \\ $$$$\Rightarrow{the}\:{only}\:{possibility}\:{is}\:\mathrm{sin}\:\theta=\mathrm{1}, \\ $$$${i}.{e}.\:\theta=\mathrm{90}° \\ $$
Commented by mathdanisur last updated on 25/Aug/21
Thank you Ser
$${Thank}\:{you}\:{Ser} \\ $$

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