SOLUTION: In the figure ABCD is a rectangle with AB=12cm and BC=7 cm. Inside ABCD there is another tilted rectangle KLMN. where AK= BL= CM= DN= x cm. If the area of KLMN is 54 cm2 find x.

Question 28455: In the figure ABCD is a rectangle with AB=12cm and BC=7 cm. Inside ABCD there is another tilted rectangle KLMN. where AK= BL= CM= DN= x cm. If the area of KLMN is 54 cm2 find x.
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
IF K,L,M,N ARE ON AB,BC,CD AND DA THEN,WE GET A DIFFERENT ANSWER.
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.
FROM YOUR ANSWER OF X=2,YOU CAN CHECK OUT.SO THE PROBLEM HAS TO BE REWORDED.
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 ONSIDERED.
SO X=2

SO GOT YOUR ANSWER?.BUT CORRECT THE PROBLEM GIVING THE SKETCH TO GIVE US A PROPER SETUP.
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In the figure ABCD is a rectangle with AB=12cm and BC=7 cm. Inside ABCD there is another tilted rectangle KLMN. where AK= BL= CM= DN= x cm. If the area of KLMN is 54 cm2 find x.
KLMN IS SITUATED SYMMETRICALLY W.R.T ABCD.SO THEIR SIDES SHALL BE IN THE RATIO OF SQUARE ROOT OF THEIR AREAS THAT IS (54/84)^0.5.LET KP AND KQ BE PERPENDICULARS TO AB AND AC RESPECTIVELY.HENCE

KL=12*(54/84)^0.5
KQ=(12-12*(54/84)^0.5)/2
KQ^2=((12-12*(54/84)^0.5)/2)^2
==1.41442889
KN=7*(54/84)^0.5
KP=(7-7*(54/84)^0.5)/2
KP^2=((7-7*(54/84)^0.5)/2)^2
=0.481298719
SO X^2=KP^2+KQ^2
X=(1.414^2+0.4813^2)^0.5
=1.493668534
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