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3 years ago

Can someone help me match these up? i know the given is number one I need help with the rest.

Mathematics

2 answers:

EleoNora [17]3 years ago

6 0

$C-1\\E-2\\B-3\\D-4\\A-5$

SSSSS [86.1K]3 years ago

6 0

Answer:

1-C, 2-E, 3-B, 4-D, 5-A

Step-by-step explanation:

Firstly, from the figure, it is shown that line RS is parallel to the line segment AB and ∠1=∠2. Therefore, 1 matches with C.

Now, from the definition of parallel lines, the corresponding angles are always equal, therefore ∠B=∠1. Therefore, 2 matches with E.

Since, we are given that the lines are parallel, therefore by property of parallel lines, alternate angles are always equal. Therefore, ∠A=∠2. Hence, 3 matches with B.

Now, we know that ∠B=∠1 and ∠A=∠2, also that ∠1=∠2, therefore from these equations, ∠A=∠B. Hence, 4 matches with D.

If two angles of the triangle are equal that is ∠A=∠B, then the sides opposite to equal angles are always equal, therefore, RA=RB. Hence, 5 matches with A.

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3 years ago

This figure is made up of a quadrilateral and a semicircle.

Arturiano [62]

Answer:

The correct option is B. The area of the figure is 40.4 units².

Step-by-step explanation:

The line AB divides the figure in two parts one is a rectangle and another is semicircle.

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$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

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$AB=\sqrt{(2+3)^2+(4-2)^2}=\sqrt{25+4}=\sqrt{29}$

The length of AD is

$AD=\sqrt{(-1+3)^2+(-3-2)^2}=\sqrt{4+25}=\sqrt{29}$

Since AB=AD, therefore ABCD is a square. The area of the of square is

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The area of square is 29 units².

The area of a semicircle is

$A_2=\frac{\pi}{2}r^2$

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$r=\frac{d}{2}=\frac{\sqrt{29}}{2}$

The area of the semicircle is

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3 years ago

Identify the correct corresponding parts

andrey2020 [161]

Answer:

The correct corresponding part is;

$\overline {CB}$ ≅ $\overline {CD}$

Step-by-step explanation:

The information given symbolically in the diagram are;

ΔCAB is congruent to ΔCED (ΔCAB ≅ ΔCED)

Segment $\overline {CA}$ is congruent to $\overline {CE}$ ( $\overline {CA}$ ≅ $\overline {CE}$ )

Segment $\overline {CB}$ is congruent to $\overline {CD}$ ( $\overline {CB}$ ≅ $\overline {CD}$ )

From which, we have;

∠A ≅ ∠E by Congruent Parts of Congruent Triangles are Congruent (CPCTC)

∠B ≅ ∠D by CPCTC

Segment $\overline {AB}$ is congruent to $\overline {DE}$ ( $\overline {AB}$ ≅ $\overline {DE}$ ) by CPCTC

Segment $\overline {AE}$ bisects $\overline {BD}$

Segment $\overline {BD}$ bisects $\overline {AE}$

Therefore, the correct option is $\overline {CB}$ ≅ $\overline {CD}$

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