A square mirror has sides measuring 2 ft. less than the sides of a square painting. If the difference between their areas is 32 ft^2, find the lengths of the sides of the mirror and the painting?
Square mirror= x-2
I don't know where to go from there...
The dimensions of a HP flat panel monitor are such that its length is 3in more than its width. If the length were doubled and if the width were decreased by 1in. the area would be increased by 150 in^2. What are the length and the width?
I don't know where to start....
Math - Reiny, Sunday, July 13, 2008 at 7:09pm
translate the English into math.
<A square mirror has sides measuring 2 ft. less than the sides of a square painting>
let the sides of the square painting be x ft.
Then the sides of the mirror are each x-2 ft.
area of mirror = (x-2)(x-2)
area of painting = (x)(x)
<the difference in their areas is 32 ft^2
----> x^2 - (x-2)^2 = 32
x^2 - (x^2 - 4x + 4 = 32
x^2 - x^2 + 4x - 4 = 32
4x = 36
x = 9
old width : x
old length : x+3
old area : x(x+3)
new length : 2(x+3)
new width : x+3 - 1 or x+2
new area : 2(x+3)(x+2)
<area is increased by 150 in^s> or
the difference in their areas is 150
2(x+3)(x+2) - x(x+3) = 150
Can you take it from there?
Let me know what you got.
CORRECTION Math - Reiny, Sunday, July 13, 2008 at 7:17pm
new width should be x-1
new area should be 2(x+3)(x-1)
equation should be
2(x+3)(x-1) - x(x+3) = 150