Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

$${Thank}\:{you}\:{Ser} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com