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2 years ago

The tile pattern below is made up of trapezoids and rhombuses. What is the area of each figure in the pattern?

Mathematics

1 answer:

shutvik [7]2 years ago

6 0

The area of the rhombus and trapezoid from the figure are 2 square in and 5 square in respectively

<h3>How to find the area of a trapezoid and rhombus?</h3>

The given pattern consists of rhombus and trapezoids

The formula for calculating the area of rhombus is expressed as:

A = pq/2

Area of trapezoid = 0.5(a+b)h

Given the following

height = 2in

a = 2in

b = 3in

Ara of rhombus = 1(4)/2 = 2 square inches

Area of the trapezoid = 0.5(2+3) * 2

Area of the trapezoid = 5 square inches

Hence the area of the rhombus and trapezoid from the figure are 2 square in and 5 square in respectively

Learn more on area of rhombus and trapezoid here: brainly.com/question/2456096

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Evaluate the following integral using trigonometric substitution

serg [7]

Answer:

The result of the integral is:

$\arcsin{(\frac{x}{3})} + C$

Step-by-step explanation:

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$\int \frac{dx}{\sqrt{9-x^2}}$

Trigonometric substitution:

We have the term in the following format: $a^2 - x^2$ , in which a = 3.

In this case, the substitution is given by:

$x = a\sin{\theta}$

$dx = a\cos{\theta}d\theta$

In this question:

$a = 3$

$x = 3\sin{\theta}$

$dx = 3\cos{\theta}d\theta$

$\int \frac{3\cos{\theta}d\theta}{\sqrt{9-(3\sin{\theta})^2}} = \int \frac{3\cos{\theta}d\theta}{\sqrt{9 - 9\sin^{2}{\theta}}} = \int \frac{3\cos{\theta}d\theta}{\sqrt{9(1 - \sin^{\theta})}}$

We have the following trigonometric identity:

$\sin^{2}{\theta} + \cos^{2}{\theta} = 1$

$1 - \sin^{2}{\theta} = \cos^{2}{\theta}$

Replacing into the integral:

$\int \frac{3\cos{\theta}d\theta}{\sqrt{9(1 - \sin^{2}{\theta})}} = \int{\frac{3\cos{\theta}d\theta}{\sqrt{9\cos^{2}{\theta}}} = \int \frac{3\cos{\theta}d\theta}{3\cos{\theta}} = \int d\theta = \theta + C$

Coming back to x:

We have that:

$x = 3\sin{\theta}$

$\sin{\theta} = \frac{x}{3}$

Applying the arcsine(inverse sine) function to both sides, we get that:

$\theta = \arcsin{(\frac{x}{3})}$

The result of the integral is:

$\arcsin{(\frac{x}{3})} + C$

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