# A rectangular paintint has dimensions x and x+10. The painting is in a frame 2 in. wide. The total area of the picture and the frame is 900 in^2.What are the dimensions of the painting?

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the area of painting is x^2 + 10x, we have to add frame area,

two vertical rettangle of area 2(x+4) and two rettangles of area

2(x+10). So the frame area is 4(2x + 14)

Adding the area to the paint area we have:

x^2 + 10x + 8x +56 = 900

x^2 +18x -844 `Delta` = 324 - 4(-844)= 324 +3376= 3700

x 1= (-18 + 10 `sqrt(37)` )/2 = -9 + 5`sqrt(37)` =21,413

x2 = (-18 - 10`sqrt(37)` )/2 = -9 - 5 `sqrt(37)` = -39,413

We accept only positive solutione therfore the second one is to be negletted.

Thusthe dimension of hte paint is 21,413 In X 31,413 In

Dimensionsn of panting= x . (x+10)

dimensions of( paintng and frame )= (x+4).(x+14)

(Length of frame increase each side by 2 inch.)

Area of frame = 900 sq unit (i)

But area of frame= (x+4) X (x+14) (ii)

from (i) and (ii) ,we have

(x+4) x (x+14)=900

x^2+18x-56=900

x^2+18x-844=0

By quadratic formula

`x=(-18+-sqrt(18^2+4xx844))/2`

`x=(-18+-60.83)/2`

`x=21.41`

Since length is never negative so another value rejected.

Thus dimensions of painting= 21.41 x 31.41