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3 years ago

What is the area of the rectangle?

Mathematics

1 answer:

den301095 [7]3 years ago

7 0

Answer:

$50\ units^{2}$

Step-by-step explanation:

Plot the figure to better understand the problem

see the attached figure

we know that

If the figure is a rectangle

then

$AB=CD \\AD=BC$

The area of the rectangle is equal to

$A=B*h$

where

B is the base

h is the height

the base B is equal to the distance AB

the height h is equal to the distance AD

Step 1

Find the distance AB

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

$A(-5,5)\\B(0,-5)$

substitute the values

$d=\sqrt{(-5-5)^{2}+(0+5)^{2}}\\d=\sqrt{(-10)^{2}+(5)^{2}}\\dAB=\sqrt{125}\ units$

Step 2

Find the distance AD

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

$A(-5,5)\\D(-1,7)$

substitute the values

$d=\sqrt{(7-5)^{2}+(-1+5)^{2}}\\d=\sqrt{(2)^{2}+(4)^{2}}\\dAD=\sqrt{20}\ units$

Step 3

Find the area of the rectangle

$A=AB*AD$

we have

$dAB=\sqrt{125}\ units\\dAD=\sqrt{20}\ units$

substitute

$A=\sqrt{125}*\sqrt{20}\\A=\sqrt{2,500}\\A=50\ units^{2}$

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The integral along the surface of the cylinder (with outward/positive orientation) is then

$\displaystyle\iint_S(x^2+y^2+z^2)\,\mathrm dS=\left\{\iint_{T_1}+\iint_{T_2}+\iint_{T_3}\right\}(x^2+y^2+z^2)\,\mathrm dS$
$=\displaystyle\int_{u=0}^{u=2\pi}\int_{v=0}^{v=3}((2\cos u)^2+(2\sin u)^2+v^2)\left\|{{\mathbf r}_1}_u\times{{\mathbf r}_2}_v\right\|\,\mathrm dv\,\mathrm du+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}((r\cos\theta)^2+(r\sin\theta)^2+0^2)\left\|{{\mathbf r}_2}_r\times{{\mathbf r}_2}_\theta\right\|\,\mathrm d\theta\,\mathrm dr+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}((r\cos\theta)^2+(r\sin\theta)^2+3^2)\left\|{{\mathbf r}_3}_r\times{{\mathbf r}_3}_\theta\right\|\,\mathrm d\theta\,\mathrm dr$
$=\displaystyle2\int_{u=0}^{u=2\pi}\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv\,\mathrm du+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}r^3\,\mathrm d\theta\,\mathrm dr+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}r(r^2+9)\,\mathrm d\theta\,\mathrm dr$
$=\displaystyle4\pi\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv+2\pi\int_{r=0}^{r=2}r^3\,\mathrm dr+2\pi\int_{r=0}^{r=2}r(r^2+9)\,\mathrm dr$
$=136\pi$