A photograph print measuring 24cm by 15cm is enclosed in a frame. A uniform space of width X cm is left between the edges of the photograph and the frame. If the area of the space is 270cm², find tje value of X

Answer 1

Area of photograph print:

24 ∙ 15 = 360 cm²

New dimension:

24 + 2 x and 15 + 2 x

Area of the space = ( 24 + 2 x ) ∙ ( 15 + 2 x ) - 24 ∙ 15 = 270

15 ∙ 24 + 15 ∙ 2 x + 2 x ∙ 24 + 2 x ∙ 2 x - 360 = 270

360 + 30 x + 48 x + 4 x² - 360 = 270

4 x² + 78 x = 270

4 x² + 78 x - 270 = 0

The solutions are:

x = - 45 / 2 and x = 3

x can't be negatve so x = 3 cm

Answer 2

P = 2*(x+24+x) + 2(x+15+x) = 8x + 78 = Perimeter = Total length(L).
A = L*W = L*X = 270,
(8x+78) * x = 270,
8x^2 + 78x - 270 = 0, Divide both sides by 2:
4x^2 + 39x - 135 = 0,

X = (-B +- sqrt(B^2-4AC))/2A,
X = (-39 +- sqrt(3681))/8,
X = (-39 +- 60.7)/8 = 2.7, and -12.5 cm.
X = 2.7 cm. Used positive value of X.