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vova2212 [387]
2 years ago
7

I will give brainliest, These figures are similar. the perimeter and area of one is given. the perimeter of the other is also gi

ven, find its area and round to the nearest 10th

Mathematics
1 answer:
kkurt [141]2 years ago
3 0

Answer:

Step-by-step explanation:

step 1

Find the scale factor

we know that

If two figures are similar, then the ratio of its perimeters is equal to the scale factor

Let

z ----> the scale factor

x----> the perimeter of the larger figure

y ----> the perimeter of the smaller figure

we have

substitute

step 2

Find the area of the larger figure

we know that

If two figures are similar, then the ratio of its areas is equal to the scale factor squared

Let

z ----> the scale factor

x----> the area of the larger figure

y ----> the area of the smaller figure

we have

substitute

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Answer:

See below

Step-by-step explanation:

It has something to do with the<em> </em><u><em>Weierstrass substitution</em></u>, where we have

$\int\, f(\sin(x), \cos(x))dx = \int\, \dfrac{2}{1+t^2}f\left(\dfrac{2t}{1+t^2}, \dfrac{1-t^2}{1+t^2} \right)dt$

First, consider the double angle formula for tangent:

\tan(2x)= \dfrac{2\tan(x)}{1-\tan^2(x)}

Therefore,

\tan\left(2 \cdot\dfrac{x}{2}\right)= \dfrac{2\tan(x/2)}{1-\tan^2(x/2)} = \tan(x)=\dfrac{2t}{1-t^2}

Once the double angle identity for sine is

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======================================================

5\cos(x) =12\sin(x) +3, x \in [0, 2\pi ]

Solving

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\implies \dfrac{5-5t^2}{1+t^2}= \dfrac{24t}{1+t^2}+3 \implies  \dfrac{5-5t^2 -24t}{1+t^2}= 3

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t = \dfrac{-(-24)\pm \sqrt{(-24)^2-4(-8)\cdot 2}}{2(-8)} = \dfrac{24\pm 8\sqrt{10}}{-16} =  \dfrac{3\pm \sqrt{10}}{-2}

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