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2 years ago

Helllp please Find the area of The following figures. Dimensions are in centimeters.

Mathematics

2 answers:

lukranit [14]2 years ago

5 0

Answer:

$\huge\mathfrak\red{Answer...} \\ \\ area \: of \: square \: = side {}^{2} \\ \\ side = \: 6 \: cm \\ \\ area \: of \: square \: = 6 \times 6 \\ \\ area \: of \: square \: = 36 \: cm.sq \\ \\ area \: of \: semicircle \: = (1 \div 2)\pi \: r {}^{2} \\ \\ radius = 2 \: cm \\ \\ area \: of \: semicircle = 6.28 \: cm.sq \\ \\ area \: of \: whole \: figure \: = 36 + 6.28 \\ \\ area = 42.28 \: cm.sq$

AleksAgata [21]2 years ago

3 0

Answer:

Seperate the shapes!

Step-by-step explanation:

CIRCLE

First we want to find the area of the circle which would be:

\pi r^{2}

3.14159265359 x 2^{2}

3.14159265359 x 4cm

= 12.5663706144cm^{2} /2 (must divide by two because it is half of a circle)

= 6.28318530718cm^{2}

RECTANGLE/SQUARE

Find the area of the square.

AREA = L x W

Length = 2cm + ?

Width = 6cm

To find the length we must use the diameter of the circle (which is the radius but times it by 2)

To find diameter: r x 2

= 2cm x 2

= 4cm

Length = 4 + 2 = 6cm

Width = 6 cm

Area (Length x Width) = 6cm x 6cm

= 36cm^{2}

TOTAL AREA:

Just add the area of the circle and the area of the square together.

Area total = 36cm^{2} + 6.28318530718cm^{2}

= 42.2831853072cm^{2}

* Also round your total answer to the nearest tenth or hundreth if needed to do so.

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Step-by-step explanation:

It has something to do with the Weierstrass substitution, 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

$\sin(2x)= \dfrac{2\tan(x)}{1+\tan^2(x)}$

we know $\sin(x)=\dfrac{2t}{1+t^2}$ , but sure, we can derive this formula considering the double angle identity

$\sin(x)= 2\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2}\right)$

Recall

$\sin \arctan t = \dfrac{t}{\sqrt{1 + t^2}} \text{ and } \cos \arctan t = \dfrac{1}{\sqrt{1 + t^2}}$

Thus,

$\sin(x)= 2 \left(\dfrac{t}{\sqrt{1 + t^2}}\right) \left(\dfrac{1}{\sqrt{1 + t^2}}\right) = \dfrac{2t}{1 + t^2}$

Similarly for cosine, consider the double angle identity

Thus,

$\cos(x)= \left(\dfrac{1}{\sqrt{1 + t^2}}\right)^2- \left(\dfrac{t}{\sqrt{1 + t^2}}\right)^2 = \dfrac{1}{t^2+1}-\dfrac{t^2}{t^2+1} =\dfrac{1-t^2}{1+t^2}$