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find-the-Area-ABCD-




Question Number 198477 by a.lgnaoui last updated on 20/Oct/23
find the Area (ABCD)
findtheArea(ABCD)
Commented by a.lgnaoui last updated on 20/Oct/23
Commented by mr W last updated on 21/Oct/23
Area(ABCD)=((x^2 −y^2 )/4)
Area(ABCD)=x2y24
Answered by mr W last updated on 21/Oct/23
Commented by mr W last updated on 21/Oct/23
b^2 =d^2 +f^2    ...(i)  a^2 =c^2 +(f+c+d)^2 =c^2 +f^2 +c^2 +d^2 +2fc+2fd+2cd   ...(ii)  (ii)−(i):  a^2 −b^2 =2c^2 +2fc+2fd+2cd  ⇒c^2 +fc+fd+cd=((a^2 −b^2 )/2)  [ABCD]=(df/2)+((c(f+c+d))/2)                    =((c^2 +df+cf+cd)/2)=((a^2 −b^2 )/4) ✓
b2=d2+f2(i)a2=c2+(f+c+d)2=c2+f2+c2+d2+2fc+2fd+2cd(ii)(ii)(i):a2b2=2c2+2fc+2fd+2cdc2+fc+fd+cd=a2b22[ABCD]=df2+c(f+c+d)2=c2+df+cf+cd2=a2b24
Commented by a.lgnaoui last updated on 21/Oct/23
exactly  thanks
exactlythanks

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