Mathematics, 28.07.2019 02:30 darenl4478

When finding the length of the hypotenuse of a right triangle, why might you approximate the answer?

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Answer from: magemenproduct

In a right angled triangle, length of hypotenuse is calculated by using the formula :

$Hypotenuse^{2}= \sqrt{Base^{2} + Perpendicular^{2} }$

Now, there could be a possibility that the term under square root does not form a complete square, then an approximation of the root will make more sense in calculating the length of hypotenuse.

For real-world problems, you would want an approximation, so that you will have an estimate of the length you are looking to find out.

Answer from: paytonpaige22

If the square root is not a perfect square, then an approximation of the root often makes more sense in the context of the problem. For real-world problems, you would want an approximation, so you have an estimate of the length you are looking at.

Answer from: Quest

something that characterises a right triangle is the fact that we may use pythagoras' theorem to find any one of its sides given that we have the length of the other two sides. here, we have the length of the hypotenuse, √(87) yd, and one other side, √(23) yd, thus we can easily find the length of the remaining side using pythagoras' theorem as defined below:

c^2 = a^2 + b^2, where c is the length of the hypotenuse, and a and b are the lengths of the other two sides.

thus, given that c = √(87) and a = √(23), we may substitute these into the formula we defined above to get:

c^2 = a^2 + b^2

(√87)^2 = (√23)^2 + b^2 (substitute c = √87 and a = √23 into the equation)

87 = 23 + b^2 (evaluate (√87)^2 = 87 and (√23)^2 = 23 - remember that the square root of a number, squared will equal that very same number)

64 = b^2 (subtract 23 from both sides)

√64 = b (take the square root of both sides)

8 = b (evaluate √64 = 8)

therefor, the length of the unknown leg in the right triangle is 8 yd (first answer).

What is the length of the unknown leg in the right triangle ?

Answer from: Quest

$\begin {array}{c|c}\underline {\quad x\quad}& \underline{\quad y\quad}\\ -2& 4\\-1& 2\\0& 0\\1& 2\\ 2& 4\\\end{array}$

step-by-step explanation:

y = | -2(-2) | → y = | 4 | → y = 4 ⇒ (-2, 4)

y = | -2(-1) | → y = | 2 | → y = 2 ⇒ (-1, 2)

y = | -2(0) | → y = | 0 | → y = 0 ⇒ (0, 0)

y = | -2(1) | c → y = | -2 | → y = 2 ⇒ (1, 2)

y = | -2(2) | → y = | - 4 | → y = 4 ⇒ (2, 4)

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