SOLUTION: A picture has dimensions 20cm by 30 cm. It is surrounded by a frame of uniform width whose outer edge is a rectangle with an area of 1800 cm squared. To the nearest .1 cm, find the

Algebra ->  Rectangles -> SOLUTION: A picture has dimensions 20cm by 30 cm. It is surrounded by a frame of uniform width whose outer edge is a rectangle with an area of 1800 cm squared. To the nearest .1 cm, find the      Log On


   

Question 871846: A picture has dimensions 20cm by 30 cm. It is surrounded by a frame of uniform width whose outer edge is a rectangle with an area of 1800 cm squared. To the nearest .1 cm, find the width of the frame.
Found 2 solutions by josgarithmetic, mananth:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Let w be the width of the frame around the picture.

Picture+Frame=1800, the entire area. may be more straightforward.
%2820%2B2w%29%2830%2B2w%29=1800.
2%2A2%2810%2Bw%29%2815%2Bw%29=18%2A10%2A10
%2810%2Bw%29%2815%2Bw%29=18%2A5%2A5
150%2B15w%2B10w%2Bw%5E2=450
w%5E2%2B25w=450-150
w%5E2%2B25w-300=0------you can finish this quadratic equation from here; just pick the value which works.

Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
Let the width be x
..
Length of picture 30 cm 54
width of picture 20 cm 44

Area = 600 m^2
Area of frame 1800 m^2
length of frame & plot 30 + 2 x
width of frame & plot 20 + 2 x

( 30 + 2 x ) ( 20 + 2 x ) + 0 = 1,800

600 + 60 x + 40 x + 4 X^2 + 0 = 1,800
4 X^2 + 100 x + -1,800 = 0
Find the roots of the equation by quadratic formula
a= 4 b= 100 c= -1,800
b^2-4ac= 10,000 - -28,800
b^2-4ac= 38,800 sqrt%28%0938%2C800%09%29= 197
x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29
x1=%28-b%2Bsqrt%28b%5E2-4ac%29%29%2F%282a%29
x1=( -100 + 197 )/ 8
x1= 12.122
x2=( -100 -197 ) / 8
x2= -15.500
Ignore negative value
width = 12.1 cm