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3 years ago

Find the missing length of the triangle 20 cm 12 cm

Mathematics

1 answer:

larisa [96]3 years ago

3 0

The shape on the left is a square with side length of 12 cm

This means the height of the triangle is 12cm.

Knowing the height and the hypotenuse using the Pythagorean you can solve for the base.

20^2 = 12^2 + x^2

400 = 144 + x^2

Subtract 144 from both sides:

256 = x^2

Take the square root of both sides

X = sqrt(256)

X = 16

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E(5,3) and F (2,-1) are two vertices of a square EFGH and H is in the x-axis .Find the coordinates of H and G. Please need answe

timurjin [86]

Answer:

coordinates of H = (1, 0)

Coordinates of G = ( - 3.6, -2) or (5.6, -2)

Step-by-step explanation:

E(5,3) and F (2,-1) are two vertices of a square EFGH and H is in the x-axis.

Let the coordinates of H is (x, 0) and G is (a, b).

The length of side EF is

$EF = \sqrt{(5 -2)^2 + (3 +1)^2} = 5$

So,

$EH = \sqrt{(5 -x)^2 + (3 -0)^2} = 5\\\\(5 -x)^2+ 9 = 25\\\\5 - x = 4\\\\x = 1$

And

$GH = \sqrt{(a -x)^2+ b^2} = 5\\\\(a -1)^2+ b^2 = 25\\\\a^2 + b^2 + 1 - 2 a = 25\\\\a^2 + b^2 - 2a = 24 .... (1)$

Now

$FG = \sqrt{(a -2)^2+ (b +1)^2} = 5\\\\(a -2)^2+ (b+1)^2 = 25\\\\a^2 + b^2 + 2b - 2 a = 20\\ ..... (2)$

Solving (1) and (2)

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Check the picture below.

so, the maximum height is reached at the parabola's vertex, where x = how many seconds it took, y = how hight it went.

$\bf \textit{ vertex of a vertical parabola, using coefficients}\\\\
\begin{array}{lccclll}
h = &{{ -16}}t^2&{{ +36}}t&{{ +10}}\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad 
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)$

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\\\\\\
\left( \cfrac{81}{2}~~,~~\cfrac{121}{4} \right)\implies \left(\stackrel{\textit{how many seconds}}{40\frac{1}{2}}~~,~~\stackrel{\textit{how many feet up}}{30\frac{1}{4}} \right)$

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