document.write( "Question 378676: A picture that measures 10 cm by 5 cm is to be surrounded by a mat before being framed. The width of the mat is to be the same on all sides of the picture. The area of the mat is to be twice the area of the picture. What is the width of the mat? \n" ); document.write( "
Algebra.Com's Answer #268970 by mananth(16946)\"\" \"About 
You can put this solution on YOUR website!
A picture that measures 10 cm by 5 cm is to be surrounded by a mat before being framed. The width of the mat is to be the same on all sides of the picture. The area of the mat is to be twice the area of the picture.
\n" ); document.write( "...
\n" ); document.write( "Area of picture = 10*5 = 50 cm^2
\n" ); document.write( "Area of mat = 2*50 = 100 cm^2
\n" ); document.write( "...
\n" ); document.write( "let width of mat be x
\n" ); document.write( "width of frame including mat = 2x+5cm ( x on each side)
\n" ); document.write( "length = 2x+10
\n" ); document.write( "..
\n" ); document.write( "Total Area = (2x+5)(2x+10)
\n" ); document.write( "area of picture = 50 cm^2
\n" ); document.write( "area of mat = 100 cm^2
\n" ); document.write( "..
\n" ); document.write( "(2x+5)(2x+10)-50 =100
\n" ); document.write( "4x^2+30x+50-50=100
\n" ); document.write( "4x^2+30x=100
\n" ); document.write( "/2
\n" ); document.write( "2x^2+15x=50
\n" ); document.write( "-75
\n" ); document.write( "2x^2+15x-50=0
\n" ); document.write( "solve using quadratic formula
\n" ); document.write( "x1= (-15+sqrt(225+200))/4
\n" ); document.write( "x1=(-15+(sqrt(625))/4
\n" ); document.write( "x1= 2.5 cm the width of the mat
\n" ); document.write( "x2 is negative so ignore
\n" ); document.write( "...
\n" ); document.write( "m.ananth@hotmail.ca \n" ); document.write( "
\n" );