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2 years ago

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A rectangular 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?

1 answer:

snow_tiger [21]2 years ago

7 0

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 $=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ , we get

$=\frac{-60\pm \sqrt{60^2-4\cdot \:1\left(-875\right)}}{2\cdot \:1}$

$x=\frac{-60+\sqrt{60^2-4\cdot \:1\left(-875\right)}}{2\cdot \:1}=5\left(\sqrt{71}-6\right)$

$x=\frac{-60-\sqrt{60^2-4\cdot \:1\left(-875\right)}}{2\cdot \:1}:\quad -5\left(6+\sqrt{71}\right)$

We cant take negative value.

So, $x=5\left(\sqrt{71}-6\right)=12.13$

We could take approximately 12 cm.

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