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3 years ago

Which pair of triangles can be proved congruent by the SAS Postulate?

Mathematics

1 answer:

Brrunno [24]3 years ago

5 0

This is another one for my "impossible geometry" file.

If X is the midpoint of both AD and BE, then ABDE is a parallelogram, and point C cannot exist.

The congruent segments are
BX ≅ EX
AX ≅ DX
Of course, vertical angles AXB and DXE are congruent.
Then, by SAS, the congruent triangles are
ΔAXB ≅ ΔDXE
or
ΔABX ≅ ΔDEX

Selection C is close, but has points D and E swapped. The figure is misleading, which may be a contributor to the problem.

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A piece of wire 11 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral tria

forsale [732]

Answer:

11 meters

Step-by-step explanation:

First, we can say that the square has a side length of x. The perimeter of the square is 4x, and that is how much wire goes into the square. To maximize the area, we should use all the wire possible, so the remaining wire goes into the triangle, or (11-4x).

The area of the square is x², and the area of an equilateral triangle with side length a is (√3/4)a². Next, 11-4x is equal to the perimeter of the triangle, and since it is equilateral, each side has (11-4x)/3 length. Plugging that in for a, we get the area of the equilateral triangle is

(√3/4)((11-4x)/3)²

= (√3/4)(11/3 - 4x/3)²

= (√3/4)(121/9 - 88x/9 + 16x²/9)

= (16√3/36)x² - (88√3/36)x + (121√3/36)

The total area is then

(16√3/36)x² - (88√3/36)x + (121√3/36) + x²

= (16√3/36 + 1)x² - (88√3/36)x + (121√3/36)

Because the coefficient for x² is positive, the parabola would open up and the derivative of the parabola would be the local minimum. Therefore, to find the maximum area, we need to go to the absolute minimum/maximum points of x (x=0 or x=2.75)

When x=0, each side of the triangle is 11/3 meters long and its area is

(√3/4)a² ≈ 5.82

When x=2.75, each side of the square is 2.75 meters long and its area is

2.75² = 7.5625

Therefore, a maximum is reached when x=2.75, or the wire used for the square is 2.75 * 4 = 11 meters

3 0

3 years ago

A student is asked to find the length of the hypotenuse of a right triangle. The length of one leg is 34

VikaD [51]

Answer:

41.6 centimeters

Step-by-step explanation:

The right triangle formula: c^2 = a^2 + b^2

c is the hypotenuse side of the triangle

a and b are the legs of the triangle.

Find the hypotenuse : c^2 = (24)^2 + (34)^2

c = square root of ( (24)^2 + (34)^2)

c = square root of ( 576 + 1,156)

c = 41.6173

5 0

3 years ago

Who already learned algebra ?

beks73 [17]

I did im in algebra 1 now

3 0

4 years ago

Read 2 more answers

Need helpp fastt plss will give brainliest to who is correct plss helpp

katen-ka-za [31]

Answer:

352

Step-by-step explanation:

(216-128)*(4)

(88)*(4)

352

4 0

3 years ago

A function is expressed by the equation y=0.3x−6. For what value of the independent variable will the value of the function be e

Anna11 [10]

Answer:

0; 10; 20

Step-by-step explanation:

x is the independent variable

y is the dependent variable

y is dependent on x

a) For what value of the independent variable will the value of the function be equal to −6

y=0.3x−6

-6 = 0.3x-6

0=0.3x

x = 0

Therefore, if the independent variable is 0, the value of the function will be -6.

b) For what value of the independent variable will the value of the function be equal to −3

y=0.3x−6

-3 = 0.3x-6

0.3x = -3+6

0.3x = 3

x = 3/0.3

x = 10

Therefore, if the independent variable is 10, the value of the function will be -3.

c) For what value of the independent variable will the value of the function be equal to 0.

y=0.3x−6

0=0.3x-6

6 = 0.3x

x = 6/0.3

x = 20

Therefore, if the independent variable is 20, the value of the function will be 0.

7 0

3 years ago