When a rectangle is dilated, how do the perimeter and area of the rectangle change?

1 answer:

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A transformation of a figure in which all of the dimensions of the figure are multiplied by the same scale factor is called a dilation.

Effect of Dilation on Perimeter

Whenever a figure is dilated by a scale factor, the perimeter of the figure changes according to the same scale factor.

Effect of Dilation on Area

When a figure is dilated by a scale factor of $\frac{a}{b}$ , the area of the figure is dilated by a scale factor of $\frac{a^{2}}{b^{2}}$

Example-

Lets imagine a rectangle with length 10 cm and width 8 cm

So area becomes = 10*8 = 80 square cm

Lets suppose the rectangle is reduced by a scale factor of $\frac{1}{2}$ to produce a new rectangle.

So we will find the square of scale factor = $(\frac{1}{2})^{2}=\frac{1}{4}$

Now to find the area of the new rectangle(dilated one) we will multiply the area of the original rectangle by 1/4

= $80*\frac{1}{4}=20$

Hence, area becomes 20 square cm.

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Calculus 3 chapter 16

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Evaluate $\vec F$ at $\vec r$ :

$\vec F(x,y,z) = x\,\vec\imath + y\,\vec\jmath + xy\,\vec k \\\\ \implies \vec F(\vec r(t)) = \vec F(\cos(t), \sin(t), t) = \cos(t)\,\vec\imath + \sin(t)\,\vec\jmath + \sin(t)\cos(t)\,\vec k$

Compute the line element $d\vec r$ :

$d\vec r = \dfrac{d\vec r}{dt} dt = \left(-\sin(t)\,\vec\imath+\cos(t)\,\vec\jmath+\vec k\bigg) \, dt$

Simplifying the integrand, we have

$\vec F\cdot d\vec r = \bigg(-\cos(t)\sin(t) + \sin(t)\cos(t) + \sin(t)\cos(t)\bigg) \, dt \\ ~~~~~~~~= \sin(t)\cos(t) \, dt \\\\ ~~~~~~~~= \dfrac12 \sin(2t) \, dt$

Then the line integral evaluates to

$\displaystyle \int_C \vec F\cdot d\vec r = \int_0^\pi \frac12\sin(2t)\,dt \\\\ ~~~~~~~~ = -\frac14\cos(2t) \bigg|_{t=0}^{t=\pi} \\\\ ~~~~~~~~ = -\frac14(\cos(2\pi)-\cos(0)) = \boxed{0}$