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?
2 Answers | Add Yours
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
By quadratic formula
Since length is never negative so another value rejected.
Thus dimensions of painting= 21.41 x 31.41
Join to answer this question
Join a community of thousands of dedicated teachers and students.Join eNotes