document.write( "Question 288724: A picture that measures 10cm by 8cm is to be surrounded by a mat before framing. The width of the mat is to be the same on all sides of the picture. The area of the mat is equal to the area of the picture. What is the width of the mat, to the nearest tenth of a centimetre? \n" ); document.write( "

Algebra.Com's Answer #209255 by checkley77(12844) $\"\"$ $\"About$

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10*8=80 cm^2 is the area of the picture.
\n" ); document.write( "(10+2x)(8+2x)-80=80
\n" ); document.write( "80+16x+20x+4x^2-160=0
\n" ); document.write( "4x^2+36x-80=0
\n" ); document.write( "4(x^2+9x-20)=0
\n" ); document.write( " $\"x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\"$
\n" ); document.write( "x=(-9+-sqrt[9^2-4*1*-20])/2*1
\n" ); document.write( "x=(-9+-sqrt[81+80])/2
\n" ); document.write( "x=(-9+-sqrt161)/2
\n" ); document.write( "x=(-9+-12.69)/2
\n" ); document.write( "x=(-9+-12.69)/2
\n" ); document.write( "x=(-9+12.69)/2
\n" ); document.write( "x=3.69/2
\n" ); document.write( "x=1.8 cm. is the width of the mat.
\n" ); document.write( "Proof:
\n" ); document.write( "(10+2*1.8)(8+2*1.8)-80=80
\n" ); document.write( "(10+3.6)(8+3.6)=160
\n" ); document.write( "13.8*11.6=160
\n" ); document.write( "160~160 \n" ); document.write( "

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