All categories

4 years ago

How are rigid transformations used to justify the SAS congruence theorem?

Mathematics

2 answers:

Yuliya22 [10]4 years ago

7 0

Rigid transformations preserve segment lengths and angle measures. If you can find a rigid transformation, or a combination of rigid transformations, to map one triangle onto the other, then the triangles are congruent. To prove SAS, we started with two distinct triangles that had a pair of congruent corresponding sides and a congruent corresponding included angle. Then we performed a translation, followed by a rotation, followed by a reflection, to map one triangle onto the other, proving the SAS congruence theorem.

Katena32 [7]4 years ago

3 0

(Refer to the diagram in the attached photo.)

In this diagram there is defined two Sides and an Angle in between (SAS).

Notice how there are two loose endpoints in the diagram. A single line segment can be drawn connecting the endpoints, forming a unique triangle.

If the original diagram is transformed, the dimensions and angles formed by the third line segment remains the same relative to the first two.

Therefore, if two triangles have the same SAS configuration, then they are congruent.

You might be interested in

2x/7 divides by 4y/14?

Sonbull [250]

Remember that $\frac{\frac{a}{b}}{\frac{c}{d}}=(\frac{a}{b})(\frac{d}{c})=\frac{ad}{bc}$

try and simpliy each first
4y/14=(2/2)(2y/7)=2y/7

so 2x/7 divided by 2y/7
$\frac{\frac{2x}{7}}{\frac{2y}{7}}=(\frac{2x}{7})(\frac{7}{2y})=\frac{(2x)(7)}{(2y)(7)}$ = $\frac{14x}{14y}=\frac{x}{y}$

7 0

3 years ago

Hdc produces microcomputer hard drives at four different production facilities (f1, f2, f3, and f4) hard drive production at f1,

OLEGan [10]

Let $D$ denote the event that an HD is defective, and $F_i$ the event that a particular HD was produced at facility $i$ .

You are asked to compute

$\mathbb P(F_2\mid D)$
$\mathbb P(F_4\mid D)$
$\mathbb P(D^C)$

From the definition of conditional probabilities, the first two will require that you first find $\mathbb P(D)$ . Once you have this, part (c) is trivial.

I'll demonstrate the computation for part (a). Part (b) is nearly identical.

(a)
$\mathbb P(F_2\mid D)=\dfrac{\mathbb P(F_2\cap D)}{\mathbb P(D)}$

Presumably, the facility responsible for producing a given HD is independent of whether the HD is defective or not, so $\mathbb P(F_2\cap D)=\mathbb P(F_2)\mathbb P(D)=0.20\times0.015$ .

Use the law of total probability to determine the value of the denominator:

$\mathbb P(D)=\mathbb P(D\mid F_1)+\mathbb P(D\mid F_2)+\mathbb P(D\mid F_3)+\mathbb P(D\mid F_4)$

We know each of the component probabilities because they are given explicitly: 0.015, 0.02, 0.01, and 0.03, respectively. So

$\mathbb P(D)=0.015+0.02+0.01+0.03=0.075$

and thus

$\mathbb P(F_2\mid D)=\dfrac{0.2\times0.015}{0.075}=0.04$

(b) Similarly,
$\mathbb P(F_4\mid D)=\dfrac{0.4\times0.03}{0.075}=0.16$

(c)
$\mathbb P(D^C)=1-\mathbb P(D)=0.925$

6 0

3 years ago

What is 1/5 converted to a percent

uranmaximum [27]

You would just do 1÷5 so it would equal 20 or 20%.

8 0

3 years ago

Read 2 more answers

874×1 = 874×10 = 874×100 = 874×1,000 =

PtichkaEL [24]

Answer:

874×1 = 874

874×10 = 8,740

874×100 = 87,400

874×1,000 = 874,000

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Suppose a triangle has sides a, b, and c, and let theta be opposite the side of length a. If cos theta < 0, what must be true

katrin2010 [14]

If $\cos \theta\ \textless \ 0$ , then angle $\theta$ is obtuse and triangle with sides a, b, c is obtuse triangle.
In an arbitrary triangle can be only one obtuse angle, and the side which lies opposite to the largest angle is the largest. Then since angle $\theta$ is opposite the side of length a you can conclude that a>c and a>b.

6 0

4 years ago