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Question-96032




Question Number 96032 by i jagooll last updated on 29/May/20
Commented by i jagooll last updated on 29/May/20
ABCD is a square with an area 240 cm^2   if DP = 3PC , then what is the   area of the shaded region ?
$$\mathrm{ABCD}\:\mathrm{is}\:\mathrm{a}\:\mathrm{square}\:\mathrm{with}\:\mathrm{an}\:\mathrm{area}\:\mathrm{240}\:\mathrm{cm}^{\mathrm{2}} \\ $$$$\mathrm{if}\:\mathrm{DP}\:=\:\mathrm{3PC}\:,\:\mathrm{then}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\: \\ $$$$\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{shaded}\:\mathrm{region}\:? \\ $$
Commented by mr W last updated on 29/May/20
say side length of square=a  a^2 =240    DP/PC=3/1  ⇒DP=(3/4)DC=((3a)/4)  ⇒PC=(a/4)=SD  A_(ΔDSP) =(1/2)×((3a)/4)×(a/4)=((3a^2 )/(32))  A_(ΔDSP) +A_(ΔOSP) =((square)/4)=(a^2 /4)  A_(ΔOSP) =(a^2 /4)−((3a^2 )/(32))=((5a^2 )/(32))  A_(shade) =2×A_(ΔOSP) =((5a^2 )/(16))=(5/(16))×square  =(5/(16))×240=75
$${say}\:{side}\:{length}\:{of}\:{square}={a} \\ $$$${a}^{\mathrm{2}} =\mathrm{240} \\ $$$$ \\ $$$${DP}/{PC}=\mathrm{3}/\mathrm{1} \\ $$$$\Rightarrow{DP}=\frac{\mathrm{3}}{\mathrm{4}}{DC}=\frac{\mathrm{3}{a}}{\mathrm{4}} \\ $$$$\Rightarrow{PC}=\frac{{a}}{\mathrm{4}}={SD} \\ $$$${A}_{\Delta{DSP}} =\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}{a}}{\mathrm{4}}×\frac{{a}}{\mathrm{4}}=\frac{\mathrm{3}{a}^{\mathrm{2}} }{\mathrm{32}} \\ $$$${A}_{\Delta{DSP}} +{A}_{\Delta{OSP}} =\frac{{square}}{\mathrm{4}}=\frac{{a}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${A}_{\Delta{OSP}} =\frac{{a}^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{3}{a}^{\mathrm{2}} }{\mathrm{32}}=\frac{\mathrm{5}{a}^{\mathrm{2}} }{\mathrm{32}} \\ $$$${A}_{{shade}} =\mathrm{2}×{A}_{\Delta{OSP}} =\frac{\mathrm{5}{a}^{\mathrm{2}} }{\mathrm{16}}=\frac{\mathrm{5}}{\mathrm{16}}×{square} \\ $$$$=\frac{\mathrm{5}}{\mathrm{16}}×\mathrm{240}=\mathrm{75} \\ $$
Commented by i jagooll last updated on 29/May/20
thank you both
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{both}\: \\ $$
Answered by john santu last updated on 29/May/20
Commented by john santu last updated on 29/May/20
area ABCD = 16a^2  = 240 ; a^2  =((240)/(16)) = 15  area of shaded region = (1/2)(a(√(10)))^2   =5a^2  = 5 ×15 cm^2  = 75 cm^2
$$\mathrm{area}\:\mathrm{ABCD}\:=\:\mathrm{16a}^{\mathrm{2}} \:=\:\mathrm{240}\:;\:\mathrm{a}^{\mathrm{2}} \:=\frac{\mathrm{240}}{\mathrm{16}}\:=\:\mathrm{15} \\ $$$$\mathrm{area}\:\mathrm{of}\:\mathrm{shaded}\:\mathrm{region}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{a}\sqrt{\mathrm{10}}\right)^{\mathrm{2}} \\ $$$$=\mathrm{5a}^{\mathrm{2}} \:=\:\mathrm{5}\:×\mathrm{15}\:\mathrm{cm}^{\mathrm{2}} \:=\:\mathrm{75}\:\mathrm{cm}^{\mathrm{2}} \: \\ $$

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