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Question Number 74882 by ajfour last updated on 03/Dec/19

Commented by ajfour last updated on 03/Dec/19

If each small blue triangles have the same area, and each small quadrilaterals have the same area, find ratio of length to breadth of the outer rectangle.

$${If}\:{each}\:{small}\:{blue}\:{triangles} \\ $$$${have}\:{the}\:{same}\:{area},\:{and}\:{each} \\ $$$${small}\:{quadrilaterals}\:{have}\:{the} \\ $$$${same}\:{area},\:{find}\:{ratio}\:{of}\:{length} \\ $$$${to}\:{breadth}\:{of}\:{the}\:{outer}\:{rectangle}. \\ $$

Commented by mind is power last updated on 03/Dec/19

Op.a=1.CS⇒CS=a.op OA.AQ=CB.CS⇒QA=CS=a.op (BQ)//(OS),QA=CS⇒OS=BQ...E E⇒OSBQ parallelogram⇒BS=OQ ⇒∠QOP=∠RBS sinceSame area⇒BG=OE withe same angle ⇒RG=EP Same idea ⇒CRAP parellograme 1conclusion OP^→ =RB^→ , PA^→ =CR^→ OS^→ =Q^→ B,BG^→ =EO^→ ,HS^→ =QF^→ P(p,0) A(1,0), S(0,s), C(0,a) B(1,a) AQ^→ =SC^→ ⇒Q(1,a−s) R(1−p,a) (OQ) equatin :y=(a−s)x (SB) y=(a−s)x+s (CP):y=−(a/p)x+a (RA)=−(a/p)x+(a/p) E=(OQ)∩(CP)⇒(a−s)x=−(a/p)x+a ⇒x=((ap)/(ap+a−ps)),y=((ap(a−s))/(ap+a−ps)) F=(OQ)∩(RA)⇒(a−s)x=((−ax+a)/p) ⇒x=(a/(ap+a−sp)),y=((a(a−s))/(ap+a−sp)) H=(SB)∩(CP)⇒(a−s)x+s=((−a)/p)x+a ⇒((p(a−s))/(pa+a−sp))=x,y=((−a(a−s)+a(ap+a−sp))/(pa+a−sp)) =((a(s+ap−sp))/(pa+a−sp)) G=(RA)∩(SB)⇒((−ax)/p)+(a/p)=(a−s)x+s ⇒x=((a−ps)/(ap+a−sp)),y=((−a^2 +aps+a^2 p+a^2 −aps)/(p(ap−sp+a)))=(a^2 /((ap−sp+a))) (1/2)Det(QF^→ ,QA^→ )=(1/2)Det(OP^→ ,OE^→ ) QF^→ (((−ap+sp)/(ap+a−sp)),(((a−s)(ps−ap))/(ap+a−sp))) QA^→ =(0,s−a) ⇒((p(s−a)^2 )/(ap−sp+a))=(((ap^2 (a−s))/(ap−sp+a))) ⇒(s−a)=−ap ⇒p=((a−s)/a) too bee continued 2 other equation to use det(pA^→ ,PE^→ )=det(BQ^→ ,BG^→ )=det(GH^→ ,GF^→ )

$$\mathrm{Op}.\mathrm{a}=\mathrm{1}.\mathrm{CS}\Rightarrow\mathrm{CS}=\mathrm{a}.\mathrm{op} \\ $$$$\mathrm{OA}.\mathrm{AQ}=\mathrm{CB}.\mathrm{CS}\Rightarrow\mathrm{QA}=\mathrm{CS}=\mathrm{a}.\mathrm{op} \\ $$$$\left(\mathrm{BQ}\right)//\left(\mathrm{OS}\right),\mathrm{QA}=\mathrm{CS}\Rightarrow\mathrm{OS}=\mathrm{BQ}...\mathrm{E} \\ $$$$\mathrm{E}\Rightarrow\mathrm{OSBQ}\:\mathrm{parallelogram}\Rightarrow\mathrm{BS}=\mathrm{OQ} \\ $$$$\Rightarrow\angle\mathrm{QOP}=\angle\mathrm{RBS} \\ $$$$\mathrm{sinceSame}\:\mathrm{area}\Rightarrow\mathrm{BG}=\mathrm{OE}\:\mathrm{withe}\:\mathrm{same}\:\mathrm{angle} \\ $$$$\Rightarrow\mathrm{RG}=\mathrm{EP} \\ $$$$\mathrm{Same}\:\mathrm{idea}\:\Rightarrow\mathrm{CRAP}\:\mathrm{parellograme} \\ $$$$\mathrm{1conclusion} \\ $$$$\mathrm{O}\overset{\rightarrow} {\mathrm{P}}=\mathrm{R}\overset{\rightarrow} {\mathrm{B}},\:\:\mathrm{P}\overset{\rightarrow} {\mathrm{A}}=\mathrm{C}\overset{\rightarrow} {\mathrm{R}} \\ $$$$\mathrm{O}\overset{\rightarrow} {\mathrm{S}}=\overset{\rightarrow} {\mathrm{Q}B},\mathrm{B}\overset{\rightarrow} {\mathrm{G}}=\mathrm{E}\overset{\rightarrow} {\mathrm{O}},\mathrm{H}\overset{\rightarrow} {\mathrm{S}}=\mathrm{Q}\overset{\rightarrow} {\mathrm{F}} \\ $$$$\mathrm{P}\left(\mathrm{p},\mathrm{0}\right)\:\:\:\:\mathrm{A}\left(\mathrm{1},\mathrm{0}\right),\:\mathrm{S}\left(\mathrm{0},\mathrm{s}\right),\:\mathrm{C}\left(\mathrm{0},\mathrm{a}\right) \\ $$$$\mathrm{B}\left(\mathrm{1},\mathrm{a}\right) \\ $$$$\mathrm{A}\overset{\rightarrow} {\mathrm{Q}}=\mathrm{S}\overset{\rightarrow} {\mathrm{C}}\Rightarrow\mathrm{Q}\left(\mathrm{1},\mathrm{a}−\mathrm{s}\right) \\ $$$$\mathrm{R}\left(\mathrm{1}−\mathrm{p},\mathrm{a}\right) \\ $$$$\left(\mathrm{OQ}\right)\:\mathrm{equatin}\::\mathrm{y}=\left(\mathrm{a}−\mathrm{s}\right)\mathrm{x} \\ $$$$\left(\mathrm{SB}\right)\:\mathrm{y}=\left(\mathrm{a}−\mathrm{s}\right)\mathrm{x}+\mathrm{s} \\ $$$$\left(\mathrm{CP}\right):\mathrm{y}=−\frac{\mathrm{a}}{\mathrm{p}}\mathrm{x}+\mathrm{a} \\ $$$$\left(\mathrm{RA}\right)=−\frac{\mathrm{a}}{\mathrm{p}}\mathrm{x}+\frac{\mathrm{a}}{\mathrm{p}} \\ $$$$\mathrm{E}=\left(\mathrm{OQ}\right)\cap\left(\mathrm{CP}\right)\Rightarrow\left(\mathrm{a}−\mathrm{s}\right)\mathrm{x}=−\frac{\mathrm{a}}{\mathrm{p}}\mathrm{x}+\mathrm{a} \\ $$$$\Rightarrow\mathrm{x}=\frac{\mathrm{ap}}{\mathrm{ap}+\mathrm{a}−\mathrm{ps}},\mathrm{y}=\frac{\mathrm{ap}\left(\mathrm{a}−\mathrm{s}\right)}{\mathrm{ap}+\mathrm{a}−\mathrm{ps}} \\ $$$$\mathrm{F}=\left(\mathrm{OQ}\right)\cap\left(\mathrm{RA}\right)\Rightarrow\left(\mathrm{a}−\mathrm{s}\right)\mathrm{x}=\frac{−\mathrm{ax}+\mathrm{a}}{\mathrm{p}} \\ $$$$\Rightarrow\mathrm{x}=\frac{\mathrm{a}}{\mathrm{ap}+\mathrm{a}−\mathrm{sp}},\mathrm{y}=\frac{\mathrm{a}\left(\mathrm{a}−\mathrm{s}\right)}{\mathrm{ap}+\mathrm{a}−\mathrm{sp}} \\ $$$$\mathrm{H}=\left(\mathrm{SB}\right)\cap\left(\mathrm{CP}\right)\Rightarrow\left(\mathrm{a}−\mathrm{s}\right)\mathrm{x}+\mathrm{s}=\frac{−\mathrm{a}}{\mathrm{p}}\mathrm{x}+\mathrm{a} \\ $$$$\Rightarrow\frac{\mathrm{p}\left(\mathrm{a}−\mathrm{s}\right)}{\mathrm{pa}+\mathrm{a}−\mathrm{sp}}=\mathrm{x},\mathrm{y}=\frac{−\mathrm{a}\left(\mathrm{a}−\mathrm{s}\right)+\mathrm{a}\left(\mathrm{ap}+\mathrm{a}−\mathrm{sp}\right)}{\mathrm{pa}+\mathrm{a}−\mathrm{sp}} \\ $$$$=\frac{\mathrm{a}\left(\mathrm{s}+\mathrm{ap}−\mathrm{sp}\right)}{\mathrm{pa}+\mathrm{a}−\mathrm{sp}} \\ $$$$\mathrm{G}=\left(\mathrm{RA}\right)\cap\left(\mathrm{SB}\right)\Rightarrow\frac{−\mathrm{ax}}{\mathrm{p}}+\frac{\mathrm{a}}{\mathrm{p}}=\left(\mathrm{a}−\mathrm{s}\right)\mathrm{x}+\mathrm{s} \\ $$$$\Rightarrow\mathrm{x}=\frac{\mathrm{a}−\mathrm{ps}}{\mathrm{ap}+\mathrm{a}−\mathrm{sp}},\mathrm{y}=\frac{−\mathrm{a}^{\mathrm{2}} +\mathrm{aps}+\mathrm{a}^{\mathrm{2}} \mathrm{p}+\mathrm{a}^{\mathrm{2}} −\mathrm{aps}}{\mathrm{p}\left(\mathrm{ap}−\mathrm{sp}+\mathrm{a}\right)}=\frac{\mathrm{a}^{\mathrm{2}} }{\left(\mathrm{ap}−\mathrm{sp}+\mathrm{a}\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Det}\left(\mathrm{Q}\overset{\rightarrow} {\mathrm{F}},\mathrm{Q}\overset{\rightarrow} {\mathrm{A}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Det}\left(\mathrm{O}\overset{\rightarrow} {\mathrm{P}},\mathrm{O}\overset{\rightarrow} {\mathrm{E}}\right) \\ $$$$\mathrm{Q}\overset{\rightarrow} {\mathrm{F}}\left(\frac{−\mathrm{ap}+\mathrm{sp}}{\mathrm{ap}+\mathrm{a}−\mathrm{sp}},\frac{\left(\mathrm{a}−\mathrm{s}\right)\left(\mathrm{ps}−\mathrm{ap}\right)}{\mathrm{ap}+\mathrm{a}−\mathrm{sp}}\right) \\ $$$$\mathrm{Q}\overset{\rightarrow} {\mathrm{A}}=\left(\mathrm{0},\mathrm{s}−\mathrm{a}\right) \\ $$$$\Rightarrow\frac{\mathrm{p}\left(\mathrm{s}−\mathrm{a}\right)^{\mathrm{2}} }{\mathrm{ap}−\mathrm{sp}+\mathrm{a}}=\left(\frac{\mathrm{ap}^{\mathrm{2}} \left(\mathrm{a}−\mathrm{s}\right)}{\mathrm{ap}−\mathrm{sp}+\mathrm{a}}\right) \\ $$$$\Rightarrow\left(\mathrm{s}−\mathrm{a}\right)=−\mathrm{ap} \\ $$$$\Rightarrow\mathrm{p}=\frac{\mathrm{a}−\mathrm{s}}{\mathrm{a}} \\ $$$$\mathrm{too}\:\mathrm{bee}\:\mathrm{continued} \\ $$$$\mathrm{2}\:\mathrm{other}\:\mathrm{equation} \\ $$$$\mathrm{to}\:\mathrm{use} \\ $$$$\mathrm{det}\left(\mathrm{p}\overset{\rightarrow} {\mathrm{A}},\mathrm{P}\overset{\rightarrow} {\mathrm{E}}\right)=\mathrm{det}\left(\mathrm{B}\overset{\rightarrow} {\mathrm{Q}},\mathrm{B}\overset{\rightarrow} {\mathrm{G}}\right)=\mathrm{det}\left(\mathrm{G}\overset{\rightarrow} {\mathrm{H}},\mathrm{G}\overset{\rightarrow} {\mathrm{F}}\right) \\ $$

Commented by ajfour last updated on 03/Dec/19

Answered by mr W last updated on 03/Dec/19

Commented by mr W last updated on 03/Dec/19

(f/e)=(c/(a+b)) af=c×(c/(c+d))×e ⇒(f/e)=(c^2 /(a(c+d))) (c/(a+b))=(c^2 /(a(c+d))) ⇒(a/(a+b))=(c/(c+d)) ⇒(a/b)=(c/d)=k (1/2)[(((a+b)^2 f)/a)−af]=(((a+b)d)/3) ⇒f=((2(a+b)ad)/(3(2a+b)b))=((2k(k+1)d)/(3(2k+1))) ⇒e=((2(k+1)^2 b)/(3(2k+1))) (f/(a−e))=((c+d)/a) ⇒(((2k)/(3(2k+1)))/(k−((2(k+1)^2 )/(3(2k+1)))))=(1/k) ⇒2k^2 −k−2=0 ⇒k=((1+(√(17)))/4)≈1.28 (k/(k+1))=((1+(√(17)))/(5+(√(17))))=(((√(17))−3)/2)≈0.56 i.e. the ratio of side lengthes of the outer rectangle can be of any value. the condition is only (a/b)=(c/d)=((1+(√(17)))/4) or (a/(a+b))=(c/(c+d))=(((√(17))−3)/2)

$$\frac{{f}}{{e}}=\frac{{c}}{{a}+{b}} \\ $$$${af}={c}×\frac{{c}}{{c}+{d}}×{e} \\ $$$$\Rightarrow\frac{{f}}{{e}}=\frac{{c}^{\mathrm{2}} }{{a}\left({c}+{d}\right)} \\ $$$$\frac{{c}}{{a}+{b}}=\frac{{c}^{\mathrm{2}} }{{a}\left({c}+{d}\right)} \\ $$$$\Rightarrow\frac{{a}}{{a}+{b}}=\frac{{c}}{{c}+{d}} \\ $$$$\Rightarrow\frac{{a}}{{b}}=\frac{{c}}{{d}}={k} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\left({a}+{b}\right)^{\mathrm{2}} {f}}{{a}}−{af}\right]=\frac{\left({a}+{b}\right){d}}{\mathrm{3}} \\ $$$$\Rightarrow{f}=\frac{\mathrm{2}\left({a}+{b}\right){ad}}{\mathrm{3}\left(\mathrm{2}{a}+{b}\right){b}}=\frac{\mathrm{2}{k}\left({k}+\mathrm{1}\right){d}}{\mathrm{3}\left(\mathrm{2}{k}+\mathrm{1}\right)} \\ $$$$\Rightarrow{e}=\frac{\mathrm{2}\left({k}+\mathrm{1}\right)^{\mathrm{2}} {b}}{\mathrm{3}\left(\mathrm{2}{k}+\mathrm{1}\right)} \\ $$$$\frac{{f}}{{a}−{e}}=\frac{{c}+{d}}{{a}} \\ $$$$\Rightarrow\frac{\frac{\mathrm{2}{k}}{\mathrm{3}\left(\mathrm{2}{k}+\mathrm{1}\right)}}{{k}−\frac{\mathrm{2}\left({k}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{3}\left(\mathrm{2}{k}+\mathrm{1}\right)}}=\frac{\mathrm{1}}{{k}} \\ $$$$\Rightarrow\mathrm{2}{k}^{\mathrm{2}} −{k}−\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow{k}=\frac{\mathrm{1}+\sqrt{\mathrm{17}}}{\mathrm{4}}\approx\mathrm{1}.\mathrm{28} \\ $$$$\frac{{k}}{{k}+\mathrm{1}}=\frac{\mathrm{1}+\sqrt{\mathrm{17}}}{\mathrm{5}+\sqrt{\mathrm{17}}}=\frac{\sqrt{\mathrm{17}}−\mathrm{3}}{\mathrm{2}}\approx\mathrm{0}.\mathrm{56} \\ $$$${i}.{e}.\:{the}\:{ratio}\:{of}\:{side}\:{lengthes}\:{of}\:{the} \\ $$$${outer}\:{rectangle}\:{can}\:{be}\:{of}\:{any}\:{value}. \\ $$$${the}\:{condition}\:{is}\:{only} \\ $$$$\frac{{a}}{{b}}=\frac{{c}}{{d}}=\frac{\mathrm{1}+\sqrt{\mathrm{17}}}{\mathrm{4}} \\ $$$${or} \\ $$$$\frac{{a}}{{a}+{b}}=\frac{{c}}{{c}+{d}}=\frac{\sqrt{\mathrm{17}}−\mathrm{3}}{\mathrm{2}} \\ $$

Commented by ajfour last updated on 03/Dec/19

sir could i make the question clear ? the small blue triangles have the same area. The four pink quadrilaterals and the central one have an equal other area. What is the ratio of length of the outer rectangle to its breadth?

$${sir}\:{could}\:{i}\:{make}\:{the}\:{question} \\ $$$${clear}\:?\:\:{the}\:{small}\:{blue}\:{triangles} \\ $$$${have}\:{the}\:{same}\:{area}. \\ $$$${The}\:{four}\:{pink}\:{quadrilaterals} \\ $$$${and}\:{the}\:{central}\:{one}\:{have}\:{an} \\ $$$${equal}\:{other}\:{area}. \\ $$$${What}\:{is}\:{the}\:{ratio}\:{of}\:{length}\:{of} \\ $$$${the}\:{outer}\:{rectangle}\:{to}\:{its}\:{breadth}? \\ $$$$ \\ $$

Commented by mr W last updated on 04/Dec/19

thank you sir for clarifying! i have modified my answer. the ratio of lengthes of the rectangle can be of any value. but the sides must be divided with a certain ratio.

$${thank}\:{you}\:{sir}\:{for}\:{clarifying}! \\ $$$${i}\:{have}\:{modified}\:{my}\:{answer}. \\ $$$${the}\:{ratio}\:{of}\:{lengthes}\:{of}\:{the}\:{rectangle} \\ $$$${can}\:{be}\:{of}\:{any}\:{value}.\:{but}\:{the}\:{sides}\: \\ $$$${must}\:{be}\:{divided}\:{with}\:{a}\:{certain}\:{ratio}. \\ $$

Commented by mr W last updated on 04/Dec/19

A_(red) =(((a+b)d)/3)=(((k+1)bd)/3) A_(blue) =((af)/2)=(k^2 /((2k+1)))×(((k+1)bd)/3) ⇒(A_(red) /A_(blue) )=((2k+1)/k^2 )=((3(√(17))+5)/8)≈2.17 i.e. the ratio of area of a blue triangle to that of a red zone is independent from the ratio of side lengthes of the outer rectangle. it′s not possible that a triangle area is equal to that of a red area.

$${A}_{{red}} =\frac{\left({a}+{b}\right){d}}{\mathrm{3}}=\frac{\left({k}+\mathrm{1}\right){bd}}{\mathrm{3}} \\ $$$${A}_{{blue}} =\frac{{af}}{\mathrm{2}}=\frac{{k}^{\mathrm{2}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)}×\frac{\left({k}+\mathrm{1}\right){bd}}{\mathrm{3}} \\ $$$$\Rightarrow\frac{{A}_{{red}} }{{A}_{{blue}} }=\frac{\mathrm{2}{k}+\mathrm{1}}{{k}^{\mathrm{2}} }=\frac{\mathrm{3}\sqrt{\mathrm{17}}+\mathrm{5}}{\mathrm{8}}\approx\mathrm{2}.\mathrm{17} \\ $$$${i}.{e}.\:{the}\:{ratio}\:{of}\:{area}\:{of}\:{a}\:{blue}\:{triangle} \\ $$$${to}\:{that}\:{of}\:{a}\:{red}\:{zone}\:{is}\:{independent} \\ $$$${from}\:{the}\:{ratio}\:{of}\:{side}\:{lengthes}\:{of}\:{the} \\ $$$${outer}\:{rectangle}.\:\:{it}'{s}\:{not}\:{possible}\:{that} \\ $$$${a}\:{triangle}\:{area}\:{is}\:{equal}\:{to}\:{that}\:{of}\:{a} \\ $$$${red}\:{area}. \\ $$

Commented by mr W last updated on 04/Dec/19

Commented by mr W last updated on 04/Dec/19

Commented by mr W last updated on 04/Dec/19

Commented by mr W last updated on 04/Dec/19

ξ=(((√(17))−3)/2) in each case with different (L/B), blue triangles have same area, red and yellow areas are equal.

$$\xi=\frac{\sqrt{\mathrm{17}}−\mathrm{3}}{\mathrm{2}} \\ $$$${in}\:{each}\:{case}\:{with}\:{different}\:\frac{{L}}{{B}}, \\ $$$${blue}\:{triangles}\:{have}\:{same}\:{area}, \\ $$$${red}\:{and}\:{yellow}\:{areas}\:{are}\:{equal}. \\ $$

Commented by ajfour last updated on 04/Dec/19

Thanks sir, totally convinced. can the blue, yellow, and red areas all be equal for certain side ratio of outer rectangle, Sir?

$${Thanks}\:{sir},\:{totally}\:{convinced}. \\ $$$${can}\:{the}\:{blue},\:{yellow},\:{and}\:{red} \\ $$$${areas}\:{all}\:{be}\:{equal}\:{for}\:{certain} \\ $$$${side}\:{ratio}\:{of}\:{outer}\:{rectangle},\:{Sir}? \\ $$

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