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3 years ago

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Jackie cut a piece of paper along its diagonal, as shown below, forming two triangles. A rectangle is cut diagonally to form 2 e

qual triangles. How do the height and base length of the two triangles compare to the height and base length of the original piece of paper?

1 answer:

Zanzabum3 years ago

8 0

Answer:

The two dimensions are the same in the triangles as they were in the rectangle.

Step-by-step explanation:

Height of a triangle is the perpendicular distance from the one vertex of the triangle to the one side of the triangle( that side is called base of the triangle).

Let the length of the given rectangle is l unit and width of the rectangle is b unit,

When the rectangle is cut along its diagonal,

Then we found two right triangles,

In which both having dimensions,

height = b and base = l,

Thus, the dimension of the triangles are same as the rectangles.

⇒ First option is correct.

Read more on Brainly.com - brainly.com/question/12150549#readmore

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The only solution is x=7
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MaRussiya [10]

Answer:

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Step-by-step explanation:

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3 years ago

Please solve these equations much appreciated

Reptile [31]

Answer:

Part c) $V=16,500\ m^3$

Part d) $V=93,312\pi\ mm^3$ or $V=292,999.68\ mm^3$

Step-by-step explanation:

Part c) Find the volume of the inverted rectangular pyramid

The volume of the rectangular pyramid is equal to

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zavuch27 [327]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Which of the following are true statements.

Mariulka [41]

Answer:

Second statement is true.

The lengths 7, 40 and 41 can not be sides of a right triangle. The lengths 12, 16, and 20 can be sides of a right triangle.

Step-by-step explanation:

for first part of statement

The lengths 7, 40 and 41 can not be sides of a right triangle.

If the square of long side is equal to the sum of square of other two sides

then the given length can be sides of a right triangle.

Check the given length by Pythagoras Theorem.

$c^{2} =a^{2} +b^{2}$ ----------(1)

Let $c=41$ and $a = 7$ and $b=40$

Put all the value in equation 1.

$41^{2} =7^{2} +40^{2}$

$1681=49+1600$

$1681=1649$

Therefore, the square of long side is not equal to the sum of square of other two sides, So given lengths 7, 40 and 41 can not be sides of a right triangle.

for second part of statement.

The lengths 12, 16, and 20 can be sides of a right triangle.

Check the given length by Pythagoras Theorem.

Let $c=20$ and $a = 12$ and $b=16$

$20^{2} =12^{2} +16^{2}$

$400=144+256$

$400=400$

Therefore, the square of long side is equal to the sum of square of other two sides, So given the lengths 12, 16, and 20 can be sides of a right triangle.

Therefore, The lengths 7, 40 and 41 can not be sides of a right triangle. The lengths 12, 16, and 20 can be sides of a right triangle.

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