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2 years ago

Can yall please help lol

Mathematics

2 answers:

const2013 [10]2 years ago

4 0

Answer:

1.) Triangle ABC is congruent to Triangle CDA because of the SAS theorem

2.) Triangle JHG is congruent to Triangle LKH because of the SSS theorem

Step-by-step explanation:

Alright. Let's start with the 1st figure. How do we prove that triangles ABC and CDA (they are named properly) are congruent? First, we can see that segments BC and AD have congruent markings, so that can help us. We also see a parallel marking for those segments as well, meaning that the diagonal AC is also a transversal for those parallel segments. That means we can say that angle CAD is congruent to angle ACB because of the alternate interior angles theorem. Then, the 2 triangles also share the side AC (reflexive property).

So, we have 2 congruent sides and 1 congruent angle for each triangle. And in the way they are listed, this makes the triangles congruent by the SAS theorem since the angle is adjacent to the 2 sides that are congruent.

The second figure is way easier. As you can clearly see by the congruent markings on the diagram, all the sides on one triangle are congruent to the other. So, since there are 3 sides congruent, we can say the triangles JHG and LKH are congruent by the SSS theorem.

svetoff [14.1K]2 years ago

3 0

Step-by-step explanation:

a)ASA .

if any 2 angles and a side of a triangle are equal to the corresponding the angles and side of the other triangle , then the two triangles are congruent.

BC : AD .

b) SSS

IF the 3 sides of a triangle are equal to the 3 corresponding sides of the other triangle ,then the two triangles are congruent.

JG:HL

JH :LK

HG:HK

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3 years ago

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<h3>Is the triangle a right triangle?</h3>

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where a = length

b = base

c = hypotenuse

12 ² + 24²

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To learn more about Pythagoras theorem, please check: brainly.com/question/14580675

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1 year ago

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2 years ago

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2 years ago

Write an equation of the line that passes through the given point and is (a) parallel and (b) perpendicular to the given line.

Harrizon [31]

Answer:

Equation of the line that passes through the given point and is (a) parallel is: $\mathbf{y=-2x+11}$

Equation of the line that passes through the given point and is (b) perpendicular is: $\mathbf{y=\frac{1}{2}x+1}$

Step-by-step explanation:

We need to Write an equation of the line that passes through the given point and is (a) parallel and (b) perpendicular to the given line.

First we will find slope of the line given in graph

The slope can be found using formula: $Slope=\frac{y_2-y_1}{x_2-x_1}$

We have points (1,6) and (2,2)

Slope is:

$Slope=\frac{y_2-y_1}{x_2-x_1}\\Slope=\frac{2-6}{2-1}\\Slope=\frac{-4}{2}\\Slope=-2$

Part a)

Write an equation of the line that passes through the given point and is (a) parallel

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$y=mx+b\\3=-2(4)+b\\3=-8+b\\b=3+8\\b=11$

The equation of line will be:

$y=mx+b\\y=-2x+11$

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Part b)

Write an equation of the line that passes through the given point and is (b) perpendicular

When the lines are perpendicular they have opposite slope. So, slope of required line: m = 1/2

Using slope m =1/2 and point(4,3) we can find y-intercept b

$y=mx+b\\3=\frac{1}{2}(4)+b\\3=2+b\\b=3-2\\b=1$

The equation of line will be:

$y=mx+b\\y=\frac{1}{2}x+1$

Equation of the line that passes through the given point and is (b) perpendicular is: $\mathbf{y=\frac{1}{2}x+1}$

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2 years ago