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A rectangular paintint has dimensions x and x+10. The painting is in a frame 2 in....

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user1071113 | eNotes Newbie

Posted April 5, 2013 at 12:45 AM via web

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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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oldnick | (Level 1) Valedictorian

Posted April 5, 2013 at 2:15 AM (Answer #1)

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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  

 

 

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pramodpandey | College Teacher | (Level 3) Valedictorian

Posted April 5, 2013 at 3:44 AM (Answer #2)

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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

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