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1 year 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]1 year 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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Find m 1 and m 2 in the diagram

Marysya12 [62]

Answers:
M for 1 is: 114
M for 2 is: 34

Why?
First you need to find out what the angle is for the corner to the left of angle 1 to do so

180-135 which equals 45

then to find on you take 180 (the sun of all interior angles in a triangle equal 180) and subtract all the angles give...
180-21-45= 114 (which makes since because angle one is obtuse and so is it)
Now for angle to you need to find the corner that is next to angle 1 to do so you do...

180-114 (because angle equals 114) this equals 66

Then you do 180-66-80 to get angle 2 which equals 34

Hope this helps :D

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Write each expression as an algebraic (nontrigonometric) expression in u, u > 0.

max2010maxim [7]

Answer:

$\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)=\frac{20\sqrt{u^2-100}}{u^2}\text{ where } u>0$

Step-by-step explanation:

We want to write the trignometric expression:

$\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)\text{ where } u>0$

As an algebraic equation.

First, we can focus on the inner expression. Let θ equal the expression:

$\displaystyle \theta=\sec^{-1}\left(\frac{u}{10}\right)$

Take the secant of both sides:

$\displaystyle \sec(\theta)=\frac{u}{10}$

Since secant is the ratio of the hypotenuse side to the adjacent side, this means that the opposite side is:

$\displaystyle o=\sqrt{u^2-10^2}=\sqrt{u^2-100}$

By substitutition:

$\displaystyle= \sin(2\theta)$

Using an double-angle identity:

$=2\sin(\theta)\cos(\theta)$

We know that the opposite side is √(u² -100), the adjacent side is 10, and the hypotenuse is u. Therefore:

$\displaystyle =2\left(\frac{\sqrt{u^2-100}}{u}\right)\left(\frac{10}{u}\right)$

Simplify. Therefore:

$\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)=\frac{20\sqrt{u^2-100}}{u^2}\text{ where } u>0$

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