Given: △ABC is an isosceles triangle,

Mathematics

1 answer:

andre [41]3 years ago

8 0

Answer:

Perimeter of △BDC=Sum of all the sides=BD+DC+CB=15+15+15=45

Step-by-step explanation:

Given ΔABC is an isosceles triangle, thus, AB=BC=x and AC=12,

perimeter of △ABC is=42

⇒x+x+12=42

⇒2x=30

⇒x=15

Thus, AB=BC=15

Now, △BDC is an equilateral triangle therefore BD=DC=BC=x

Since, x=15, therefore BD=DC=BC=15

Now, Perimeter of △BDC=Sum of all the sides=BD+DC+CB=15+15+15=45

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∫c<br> x sin y ds, C is the line segment from (0, 1) to (3, 5)

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Parameterize the line segment $C$ by

$\mathbf r(t)=\langle x(t),y(t)\rangle=(1-t)\langle0,1\rangle+t\langle3,5\rangle=\langle3t,1+4t\rangle$

where $t\in[0,1]$ . Then

$\mathrm ds=\|\mathbf r'(t)\|\,\mathrm dt$
$\mathrm ds=\|\langle3,4\rangle\|\,\mathrm dt$
$\mathrm ds=5\,\mathrm dt$

So the integral is

$\displaystyle\int_Cx\sin y\,\mathrm ds=5\int_0^1x(t)\sin y(t)\,\mathrm dt$
$=\displaystyle5\int_0^13t\sin(1+4t)\,\mathrm dt$
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