Arectangular portrait measures 50cm by 70cm. it is surrounded by a rectangular frame of uniform width. if the area of the frame is the same as the area of the portrait, what is the approximate width of the frame?

Let us assume uniform width = x cm wide.

Length of rectangular portrait = 50cm and width of rectangular portrait = 70cm.

Therefore,  length of rectangle made by frame = 50 + x+x = (2x+50) cm.

And width of rectangle made by frame = 70+x+x = (2x+70) cm.

We know, the area of rectangular portrait = 50 × 70 = 3500 cm^2.

Total area of the rectangle made by frame would be =  (2x+50) * (2x+70)

We know,

Actual area of frame = Area of rectangle made by frame -  area of rectangular portrait.

We also given "the area of the frame is the same as the area of the portrait."

We can setup an equation now,

 3500 = (2x+50) * (2x+70) - 3500.

Subtracting 3500 from both sides, we get

3500-3500 = (2x+50) * (2x+70) - 3500-3500.

0 = (2x+50) * (2x+70) -7000.

FOIL (2x+50) * (2x+70), we get

0 = 2x*2x +2x*70 + 50*2x +50*70 - 7000.

0 = 4x^2 +140x +100x +3500 -7000.

4x^2 +240x -3500 = 0.

Dividing whole equation by 4, we get

x^2 +60x - 875 =0

Applying quadratic formula , we get

We cant take negative value.

So,

We could take approximately 12 cm.

depends on how many days you are talking about

step-by-step explanation:

step-by-step explanation:

hans did not square the 40-unit side; he merely multiplied it by 2.   incorrect.