SOLUTION: A rectancle JKLM is inscribed in an another rectangle ABCD such that the vertices J,K,L and M touches the side AB,BC, CD and DA respectively.where AB = 12cm and BC = 7cm. Also A

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Question 28582: A rectancle JKLM is inscribed in an another rectangle ABCD such that the vertices J,K,L and M touches the side AB,BC, CD and DA respectively.where AB = 12cm and BC = 7cm. Also AJ = BK= CL= DM= x.The area of rectangle JKLM is 54cm2. Find x?
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
SEE DRAWING BELOW
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LET AK=BL=CM=DN=X
AB=CD=12 AND BC=DA=7.
FIRSTLY KLMN WILL NOT BE A RECTANGLE.IT WILL BE PARALLELOGRAM AS YOU CAN SEE BELOW.
LET DC BE X AXIS AND DA BE Y AXIS WITH D AS ORIGIN.HENCE COORDINATES OF DIFFERENT POINTS ARE
D...(0,0) M..(10,0) C…(12,0) L…(12,5) B….(12,7) K….(2,7) A….(0,7) N….(0,2)
SLOPE OF MN =(0-2)/(10-0)= -1/5
SLOPE OF NK =(7-2)/(2-0)= 5/2
PRODUCT OF SLOPES =(-1/5)*((5/2)=-1/2…..SO THEY ARE NOT PERPENDICULAR.
SO KLMN IS NOT A RECTANGLE.
THEN FORMULA FOR AREA ALSO CHANGES.
LET US FIND AREAS OF 4 TRIANGLES AKN,KBL,LCM AND MDN ALL OF WHICH ARE RIGHT ANGLED TRIANGLES.
SUM OF THEM IS 84-54=30..AS THIS IS EQUAL TO AREA OF ABCD(84) - AREA OF KLMN (54)
SUM OF AREAS OF 4 TRIANGLES AKN,KBL,LCM AND MDN ARE
(1/2){X(12-X)+X(7-X)+X(12-X)+X(7-X)}=X(12-X)+X(7-X)=X(12-X+7-X)=X(19-2X)=19X-2X^2=84-54=30
2X^2-19X+30=0
2X^2-4X-15X+30=0
2X(X-2)-15(X-2)=0
(X-2)(2X-15)=0
X-2=0….OR……X=2
2X-15=0….OR….X=7.5…WHICH IS BEYOND SIDES BC AND DA..SO NOT TO BE CONSIDERED.
SO X=2