Pythagorean Tests & Worksheets - All Grades

They each have a length of $2c$ . Using the distance formula and looking at the difference of distances between the lengths just found and the other vertices of the major axes, one finds that $b = sqrt(a^2-c^2$ .
They each have a length of $c$ . Then, the right triangle formed between the vertices and the origin, and applying the Pythagorean theorem, results in $a^2 + b^2 = c^2$ .
They each have a length of $a$ . Looking at the right triangle formed by the origin, $F_1$ , and the vertex $(0,b)$ , and applying the Pythagorean theorem results in $a^2 = b^2 + c^2$ .
They each have a length of $a/2$ . Therefore, the sum of their lengths, $a$ , can be used as a value equal to the sum of the lengths of $b$ and $c$ .

Grade 12 Trigonometry CCSS: HSF-TF.C.9

Given that

$cos(2theta) = cos^2(theta)- sin^2(theta)$ , prove that

$cos(2theta) = 2cos^2(theta)-1$ .

Grade 11 Nonlinear Equations and Functions CCSS: HSG-GPE.A.3

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Letting

$b^2 x^2 - y^2 = b^2 a^2$

$(-c,0)$ and

$b^2 = c^2 - a^2$ come from?

Since $b$ usually appears in the equation for a hyperbola, it must be included. Using the Pythagorean theorem, $a^2 +b^2 = c^2$ , simply rearrange the equation.
It has to be done, to ensure the asymptotes are related to the equation. The equations of the asymptotes are $y= pm b/a$ , and knowing that $|c| > |a|$ , squaring and rearranging results in $b^2 = c^2 - a^2$ .
It's done to simply the equation. $b$ is not defined yet, and since $|c| > |a|$ , $c^2 > a^2$ , and so there must be a positive number, $b^2$ such that $b^2 = c^2 - a^2$ .
Knowing that $b$ is the length of the semi-minor axis, a right triangle can be formed with the center of the hyperbola and either foci, with $b$ as the length of one leg of this triangle. Applying the Pythagorean theorem results in $b^2 + c^2 = a^2$ , and simply rearrange.

Grade 11 Trigonometry CCSS: HSG-SRT.D.10

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Is this proof valid for all triangles? Explain your answer.

Grade 11 Trigonometry CCSS: HSG-SRT.D.10

Which of the following is true concerning the law of sines and right triangles?

The law of sines is not valid for right triangles.
The law of sines can only be proved for the acute angles of a right triangle.
The law of sines can be easily proved for a right triangle, using trig ratios and the fact that $sin(90°)=1$ .
The law of sines can proved for a right triangle, and the easiest proof involves the use of the Pythagorean Theorem and the formula for the area of a triangle.

Grade 11 Three Dimensional Shapes CCSS: HSG-SRT.C.8, HSG-MG.A.3

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Grade 11 Nonlinear Equations and Functions CCSS: HSG-GPE.A.3

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$x$ or

$b^2 = a^2-c^2$ be made?

Since $b$ is undefined so far, it can be defined as any value. Then, using the Pythagorean theorem, $a^2 + b^2 = c^2$ , simply rearrange to solve for $b^2$ .
Knowing that the semi-minor axis is $b$ units long, one can substitute the square of this value for $a^2-c^2$ .
Since $b$ is not yet defined, it can be used to simplify the equation by defining $b^2 = a^2 - c^2$ . A positive value for $a^2-c^2$ exists since $|a| > |c|$ .
Projecting $b$ , the length of the semi-minor axis, onto the semi-major axis it is seen that $b=a-c$ . Then, simply square both sides of the equation.

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