asinus

\(A_{oct}=(\sqrt{\frac{(LE)^2}{2}}+EF+\sqrt{\frac{(FG)^2}{2}})\cdot (\sqrt{\frac{(FG)^2}{2}}+GH+\sqrt{\frac{(HI)^2}{2}})-\frac{4(LE)^2}{4}\\ A_{oct}=(1+1+1)\cdot (1+1+1)-2\\ \color{blue}A_{oct}=7\)

 !

Where is the diagram below? Where is "down"?

 !

Given a right triangle CAB with the right angle at B and angle bisector AD, let AD = AC.

Solution:

The question is nonsensical.

The length of the angle bisector of an acute angle in a right triangle is not equal to the length of the hypotenuse.

 !

With AB = AC = 1 and angle BAC = 60°, it is an equilateral triangle with side length 1. D lies on AB, E lies on the arc BC. The condition BD = DE is only met if E is identical to C and D is identical to A. F lies on DB, B lies on the arc DE. The condition DF = FG is only met if F is identical to B and G is identical to C.

Thus, the length DF = AB = 1.

  !

The maximum possible distance between their endpoints is the diameter of the circle, namely 2r.

 !

What is DE?

The lines extended from BD and CE form triangle ABC.

AB=13, AC=19, BC=3.

The following holds: |AC-AB| < BC.

Therefore, there is no triangle with side lengths {13, 19, 3}.

The question does not yield a sine answer.

  !

Find the set of all possible values of a/b.

\(a+b=1\\ a=1-b\\ b=1-a\\ \dfrac{a}{b}=\dfrac{a}{1-a}\)

\(\color{blue}Possibility_{\frac{a}{b}}= \{0<\mathbb R <∞\}\)

  !

Let k be a positive real number and find  k, so that the line is tangent to the circle.

\(y=\sqrt{k-x^2}\\ y=-x+3+k\)

\(\color{blue}k\notin \{+\mathbb{R}\}\)

\(y=\sqrt{k-x^2}\\ y=-x+3+k \)

\(y=\sqrt{{\color{blue}1.365}-x^2}\\ y=-x+3\color{blue}-1.365\)

\(\color{blue}k\in \{ \approx 1.365,\ \approx-1.365\}\)   determined with graphics. 

If +k is replaced by -k in the problem, it is solvable. 

 !

Where is the circle sector shown??

What is the radius of the circle?

\(A_{sector}=\dfrac{\pi r^2\alpha}{360°}\\ A_{sector}=\dfrac{60}{\pi}\\ r_{sector}=\sqrt{\dfrac{60\cdot 360°}{\pi \alpha}\cdot \dfrac{2\pi}{360°}}\\ \color{blue}r_{sector}=\sqrt{\dfrac{120}{\alpha }}\)

 !

What is the area of rectangle ZWXY?

Let ZW be the baseline and WX be the height of the rectangle ZWXY, and ZW be x.

Then:

\(\overline{WX}=\dfrac{x-2}{tan\ 60°}\\ \overline{AX}=\dfrac{tan\ 30°(x-2)}{tan\ 60°}=\dfrac{x-2}{3}\\ \)

\(\triangle ZAY\\ (x-\overline{AX})^2+\overline{WX}^2=4^2 \\ (x-\dfrac{x-2}{3})^2+(\dfrac{x-2}{tan\ 60°})^2=16\\ (\dfrac{2x}{3}+\dfrac{2}{3})^2+(\dfrac{x}{tan\ 60°}-\dfrac{2}{\tan\ 60°})^2=16\\ x\in \{-4,\dfrac{32}{7}\}\ \color{blue}WolframAlpha\)

\(A_{ZWXY}=x\cdot \dfrac{x-2}{tan\ 60°}=\dfrac{32}{7}\cdot \dfrac{\frac{32}{7}-2}{tan\ 60°}\\ \color{blue}A_{ZWXY}=6.7868\)

The area of rectangle ZWXY is 6.7868.

  !