SOLUTION: A square picture has a 1-centimeter border around it. Half of the total area of the picture and the border is made up by the border. Use a graph to determine the dimensions of the

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Question 66092: A square picture has a 1-centimeter border around it. Half of the total area of the picture and the border is made up by the border. Use a graph to determine the dimensions of the picture to the nearest tenth centimeter.
Choices:
A) 2.4cm by 2.4cm
B) 4.8cm by 4.8cm
C) 2cm by 2cm
D) 4.1 cm by 4.1cm

Found 2 solutions by josmiceli, venugopalramana:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The area of tyhe border is
B+=+1%2B1%2B1%2B1%2B4%28x%2A1%29 where x is the width of the square picture
B+=+4+%2B+4x
without the border
A = area of picture
A+=+x%5E2
B+=+A
4+%2B+4x+=+x%5E2
x%5E2+-+4x+-4+=+0
(B) is the correct choice
4.8%5E2+-+4%2A4.8+-+4+=+0
23.04+-+19.2+-+4+=+0
This is approximately correct

Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
A square picture has a 1-centimeter border around it. Half of the total area of the picture and the border is made up by the border. Use a graph to determine the dimensions of the picture to the nearest tenth centimeter.
Choices:
A) 2.4cm by 2.4cm
B) 4.8cm by 4.8cm
C) 2cm by 2cm
D) 4.1 cm by 4.1cm
BORDER WIDTH = 1 CM
LET PICTURE BE OF SIDE X CM.
AREA OF PICTURE = X^2
HENCE INCLUDING BORDER OF 1 CM ON ALL SIDES, THE SQUARE WILL BE
OF SIDE (X+1+1)=X+2
AREA OF PICTURE WITH BORDER =(X+2)^2
AREA OF BORDER=(X+2)^2-2X^2
WE GOT
(X+2)^2-X^2=[(X+2)^2]/2= Y SAY
WE CAN SOLVE THIS BY DRAWING GRAPHS OF
Y=2X^2...AND...........Y=(X+2)^2.........

WE FIND THAT THE 2 CURVES INTERSECT AT X=4.8 CM
HENCE THE PICTURE WITH BORDER HAS
6.8 CM SIDE AND PICTURE HAS 4.8 CM SIDE.