Question-221132

Question Number 221132 by fantastic last updated on 25/May/25

Commented by fantastic last updated on 25/May/25

$${can}\:{someone}\:{give}\:{me}\:{a}\:{DETAILED}\:{and}\:{EASY}\:{SOLUTION}?\: \\ $$$${i}\:{solved}\:{it}\:{in}\:{very}\:{lengthy}\:{and}\:{hard}\:{method} \\ $$

Commented by mr W last updated on 25/May/25

$${please}\:{crop}\:{and}\:{rotate}\:{the}\:{image} \\ $$$${properly}\:{when}\:{posting}\:{it}! \\ $$

Commented by fantastic last updated on 25/May/25

$${Sometimes}\:{it}\:{becomes}\:{blurred}\:{I}\:{do}\:{not}\:{for}\:{what} \\ $$

Commented by fantastic last updated on 25/May/25

Commented by fantastic last updated on 25/May/25

$${oh}\:!\:{i}\:{have}\:{found}\:{one} \\ $$

Answered by Frix last updated on 25/May/25

Rectangle ABCD A= ((0),(0) ) B= ((p),(0) ) C= ((p),(q) ) D= ((0),(q) ) E= ((0),(h) ) ∣BE∣=(√(h^2 +p^2 ))=L is given Circle through B andE with center ((m),(0) ) and radius q (x−m)^2 +y^2 =q^2 Inserting B, E ⇒ m=((p^2 −h^2 )/(2p)) q=((h^2 +p^2 )/(2p))=(L^2 /(2p)) Area of rectangle =pq=(L^2 /2)

$$\mathrm{Rectangle}\:{ABCD} \\ $$$${A}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\:{B}=\begin{pmatrix}{{p}}\\{\mathrm{0}}\end{pmatrix}\:\:{C}=\begin{pmatrix}{{p}}\\{{q}}\end{pmatrix}\:\:{D}=\begin{pmatrix}{\mathrm{0}}\\{{q}}\end{pmatrix} \\ $$$${E}=\begin{pmatrix}{\mathrm{0}}\\{{h}}\end{pmatrix} \\ $$$$\mid{BE}\mid=\sqrt{{h}^{\mathrm{2}} +{p}^{\mathrm{2}} }={L}\:\mathrm{is}\:\mathrm{given} \\ $$$$\mathrm{Circle}\:\mathrm{through}\:{B}\:\mathrm{and}{E}\:\mathrm{with}\:\mathrm{center}\:\begin{pmatrix}{{m}}\\{\mathrm{0}}\end{pmatrix} \\ $$$$\mathrm{and}\:\mathrm{radius}\:{q} \\ $$$$\left({x}−{m}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={q}^{\mathrm{2}} \\ $$$$\mathrm{Inserting}\:{B},\:{E}\:\Rightarrow \\ $$$${m}=\frac{{p}^{\mathrm{2}} −{h}^{\mathrm{2}} }{\mathrm{2}{p}} \\ $$$${q}=\frac{{h}^{\mathrm{2}} +{p}^{\mathrm{2}} }{\mathrm{2}{p}}=\frac{{L}^{\mathrm{2}} }{\mathrm{2}{p}} \\ $$$$\mathrm{Area}\:\mathrm{of}\:\mathrm{rectangle}\:={pq}=\frac{{L}^{\mathrm{2}} }{\mathrm{2}} \\ $$

Answered by mr W last updated on 25/May/25

Commented by mr W last updated on 25/May/25

ΔCDB ∼ ΔACB (b/a)=(a/(2R)) ⇒bR=(a^2 /2) area of rectangle =bR=(a^2 /2)=(6^2 /2)=18

$$\Delta{CDB}\:\sim\:\Delta{ACB} \\ $$$$\frac{{b}}{{a}}=\frac{{a}}{\mathrm{2}{R}}\:\Rightarrow{bR}=\frac{{a}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${area}\:{of}\:{rectangle}\:={bR}=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{6}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{18} \\ $$

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