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3 years ago

What is the height of a rectangle whose base is 40 inches and whose diagonal is 41 inches?

Mathematics

2 answers:

maw [93]3 years ago

8 0

Answer: 9 inches

Step-by-step explanation:

romanna [79]3 years ago

7 0

Answer:

You must draw a right triangle and label the hypotenuse as 41 and the longest leg (horizontal in this case, since we are to determine the height) as 40. We do not know the height so call it h.

Now the Pythagorean (sp /) Theorem states that the square of the hyp. is equal to the sum of the squares of the two legs.

so, 41 ^ 2 = 40 ^ 2 + h ^ 2

41 ^ 2 = 1681 40 ^ 2 = 1600 Therefore h ^2 must be equal to 81. This means that h = 9.

The height is 9 inches.

Step-by-step explanation:

mark me as brainliest pls :)

~Alex

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timama [110]

Answer:

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Step-by-step explanation:

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kiruha [24]

Answer:

V ≈ 1847.26

Step-by-step explanation:

<u>Circular Cone Formulas in terms of radius r and height h:</u>

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OverLord2011 [107]

Answer:

Part A) $sin(A)=\frac{2\sqrt{42}}{23}$

Part B) $cos(A)=\frac{19}{23}$

Part C) $tan(A)=\frac{2\sqrt{42}}{19}$

Step-by-step explanation:

Part A) we know that

In the right triangle ABC of the figure the sine of angle A is equal to divide the opposite side angle A by the hypotenuse

$sin(A)=\frac{BC}{AB}$

substitute the values

$sin(A)=\frac{2\sqrt{42}}{23}$

Part B) we know that

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In the right triangle ABC of the figure the tangent of angle A is equal to divide the opposite side angle A by the adjacent side angle A

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substitute the values

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