All categories

4 years ago

Parallelogram abcd is similar to parallelogram efgh. angle a is congruent to the measure of angle ___.

Mathematics

2 answers:

Vesna [10]4 years ago

8 0

Angle A is congruent to Angle E

RSB [31]4 years ago

4 0

Answer: angle e.

Step-by-step explanation:

Given: Parallelogram abcd is similar to parallelogram efgh.

We know that If any two polygons are similar then their corresponding angles are congruent and the measures of their corresponding sides are proportional.

Therefore, measure of angle a is congruent to the measure of angle e.

measure of angle b is congruent to the measure of angle f.

measure of angle c is congruent to the measure of angle g.

measure of angle d is congruent to the measure of angle h.

You might be interested in

Hey, babes! Here is a question for today-

Papessa [141]

Answer:

The equation has the form: y = a + b * x where a and b are constant numbers. The variable x is the independent variable, and y is the dependent variable. Typically, you choose a value to substitute for the independent variable and then solve for the dependent variable.

6 0

3 years ago

James bought a house that needs repairs in the 24 ft by 15 ft back yard. He

son4ous [18]

Answer:

below

Step-by-step explanation:

area to be repaired = 24*15=360ft^2

1 bag = 2ft^2

? 360ft^2

360/2 = 180 bags needed

cost of bags = 180*24.95=4491$

7 0

3 years ago

Please help me please !

Morgarella [4.7K]

Hi there!

»»————-　★　————-««

I believe your answer is:

Option C

»»————-　★　————-««

Here’s why:

⸻⸻⸻⸻

$\text{\underline{The Slope Formula Is:}}\\\\m=\frac{y_2-y_1}{x_2-x_1}\\\\(x_1,y_1)\text{ and } (x_2,y_2)\text{ are two points given.}\\\\\text{We are given the points: } (3,5) \text{ and } (9,2).\\\\\text{\underline{The formula for the points should be:}}\\\\m=\frac{5-2}{3-9}, \text{where } (9,2) \text{ is }(x_1,y_1)\text{ and } (3,5) \text{ is } (x_2,y_2).$

⸻⸻⸻⸻

»»————-　★　————-««

Hope this helps you. I apologize if it’s incorrect.

7 0

3 years ago

Plzzzz help mee plzzzzz plzzzzz

FromTheMoon [43]

Answer:

1. B

2. Seven Hundred thousand nine hundred and four

Step-by-step explanation:

3 0

3 years ago

Strain-displacement relationship) Consider a unit cube of a solid occupying the region 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1 After loa

Anastasy [175]

Answer:

please see answers are as in the explanation.

Step-by-step explanation:

As from the data of complete question,

$0\leq x\leq 1\\0\leq y\leq 1\\0\leq z\leq 1\\u= \alpha x\\v=\beta y\\w=0$

The question also has 3 parts given as

Part a: Sketch the deformed shape for α=0.03, β=-0.01 .

Solution

As w is 0 so the deflection is only in the x and y plane and thus can be sketched in xy plane.

the new points are calculated as follows

Point A(x=0,y=0)

Point A'(x+αx,y+βy)

Point A'(0+(0.03)(0),0+(-0.01)(0))

Point A'(0,0)

Point B(x=1,y=0)

Point B'(x+αx,y+βy)

Point B'(1+(0.03)(1),0+(-0.01)(0))

Point B'(1.03,0)

Point C(x=1,y=1)

Point C'(x+αx,y+βy)

Point C'(1+(0.03)(1),1+(-0.01)(1))

Point C'(1.03,0.99)

Point D(x=0,y=1)

Point D'(x+αx,y+βy)

Point D'(0+(0.03)(0),1+(-0.01)(1))

Point D'(0,0.99)

So the new points are A'(0,0), B'(1.03,0), C'(1.03,0.99) and D'(0,0.99)

The plot is attached with the solution.

Part b: Calculate the six strain components.

Solution

Normal Strain Components

$\epsilon_{xx}=\frac{\partial u}{\partial x}=\frac{\partial (\alpha x)}{\partial x}=\alpha =0.03\\\epsilon_{yy}=\frac{\partial v}{\partial y}=\frac{\partial ( \beta y)}{\partial y}=\beta =-0.01\\\epsilon_{zz}=\frac{\partial w}{\partial z}=\frac{\partial (0)}{\partial z}=0\\$

Shear Strain Components

$\gamma_{xy}=\gamma_{yx}=\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}=0\\\gamma_{xz}=\gamma_{zx}=\frac{\partial u}{\partial z}+\frac{\partial w}{\partial x}=0\\\gamma_{yz}=\gamma_{zy}=\frac{\partial w}{\partial y}+\frac{\partial v}{\partial z}=0$

Part c: Find the volume change

$\Delta V=(1.03 \times 0.99 \times 1)-(1 \times 1 \times 1)\\\Delta V=(1.0197)-(1)\\\Delta V=0.0197\\$

Also the change in volume is 0.0197

For the unit cube, the change in terms of strains is given as

$\Delta V={V_0}[(1+\epsilon_{xx})]\times[(1+\epsilon_{yy})]\times [(1+\epsilon_{zz})]-[1 \times 1 \times 1]\\\Delta V={V_0}[1+\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}+\epsilon_{xx}\epsilon_{yy}+\epsilon_{xx}\epsilon_{zz}+\epsilon_{yy}\epsilon_{zz}+\epsilon_{xx}\epsilon_{yy}\epsilon_{zz}-1]\\\Delta V={V_0}[\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}]\\$

As the strain values are small second and higher order values are ignored so

$\Delta V\approx {V_0}[\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}]\\ \Delta V\approx [\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}]\\$

As the initial volume of cube is unitary so this result can be proved.

5 0

3 years ago