I have been watching videos on how to build the base for a shed. The builders use lumber for two sides of the base (which are the same length) and different sized lumber for the other two sides. To ensure the base is a rectangle, they measure the diagonals. If the diagonals are equal, the base is rectangular. I am trying to prove this mathematically. Here’s what I have so far: Given: $ AB = CD $ $ AD = BC $ Proof: If $ AC = BD $, then triangle $ \triangle ACD $ is congruent to triangle $ \triangle BCD )$. And here I am stuck. Any ideas on how to prove this ..read more

I'd like to ask about the following type of problem: in a regular polygon a ball is launched from one of the edges. When the ball collides, it rotates clockwise by the interior angle of the polygon. Now, what work has been done relating to this problem is it the case that the trajectory is periodic as long as the slope of the initial direction is less then the interior angle of the polygon ..read more

In the figure, if MN = 40 and NQ = 9, Calculate PQ. AB and MB are diameters.(Answer:21) *Note: The question does not mention whether the diameters are equal. I don't know if it would be mandatory to get the result I try: Let AB = 2R and MB=2r $\triangle MQB \sim \triangle MKN \implies \frac{BQ}{KN} = \frac{MQ}{KN} = \frac{2r}{MN}$ $\therefore \frac{BQ}{KN} = \frac{49}{KN} = \frac{2r}{40}=\frac{1}{20}$ $\triangle ABK\sim \triangle ABQ\implies \frac{2R}{2r} = \frac{KB}{KN} = \frac{AQ}{MQ}$ $\therefore \frac{R}{r} = \frac{KB}{KN} = \frac{AQ}{49}$ $\triangle OPQ: PQ^2 = R^2 - OQ^2 \implies PQ ..read more

I'd like to ask about the following type of problem: in a regular polygon a ball is launched from one of the edges. When the ball collides, it rotates clockwise by the interior angle of the polygon. Now, when is the trajectory periodic? what work has been done relating to this problem is it the case that the trajectory is periodic as long as the slope of the initial direction is less then the interior angle of the polygon ..read more

Consider an equilateral triangle $ABC$ with points $D$, $E$, and $F$ being the midpoints of sides $AB$, $AC$, and $BC$ respectively. Let $M $ be a point on the line $ BC $. If we rotate the line segment $ MD $ about $M $ by $60°$ clockwise to get $MD' $, how can we prove that $ D' $ always lies on the line $ EF $? Details and Assumptions: $ ABC $ is an equilateral triangle. $ D $, $ E$, and $ F $ are midpoints of sides $AB $, $ AC $, and $BC $, respectively. $ M $ is a point on the line $ BC $. Line segment $ MD $ is rotated by $60°$ clockwise about point $ M$ to obtain $MD' $. Objective: Pr ..read more

In the figure, if MN = 40 and NQ = 9, Calculate PQ. AB and MB are diameters.(Answer:21) *Note: The question does not mention whether the diameters are equal. I don't know if it would be mandatory to get the result I try: Let AB = 2R and MB=2r $\triangle MQB \sim \triangle MKN \implies \frac{BQ}{KN} = \frac{MQ}{KN} = \frac{2r}{MN}$ $\therefore \frac{BQ}{KN} = \frac{49}{KN} = \frac{2r}{40}=\frac{1}{20}$ $\triangle ABK\sim \triangle ABQ\implies \frac{2R}{2r} = \frac{KB}{KN} = \frac{AQ}{MQ}$ $\therefore \frac{R}{r} = \frac{KB}{KN} = \frac{AQ}{49}$ $\triangle OPQ: PQ^2 = R^2 - OQ^2 \implies PQ ..read more

Here is a polygon with a dot inside an edge, and a dot outside another edge. How do you calculate the $x$ and $y$ position of any dot (whether it's inside or outside of the line's edge) positioned along an edge of a regular polygon (ranging from 3 to 20 sides)? The dot has radius $r$ (like 1 to 20px) The dot is placed an offset $o$ from the line's edge. The position is relative to the center of the polygon. You say where you want the dot to go using a fraction/percent from 0 to 1, along an edge, where edges are numbered from 1-5 (1-n for arbitrary polygon) going like a clock. I am a novice ..read more

In an isosceles trapezoid, is the area equal to: $A=lh$, here $l$ is the leg of the trapezoid, $h$ is the height and $A$ is the area ..read more

In an isosceles trapezoid, is the area equal to:$$A=lh$$ Where $l$ is the leg of the trapezoid, $h$ is the height and $A$ is the area ..read more

In the figure, $OP = 1$ and $PF = 3$. Calculate $r$. $DC$ and $DE$ are tangents. (Answer:$r=2$) Could this issue be resolved only with this data? I try: $Point Theorem: FA.FB = FC.FE \implies (3-(r-1).(4+r) = FC.(FC+FE)$ $DC=DE$ $\triangle AEB: AE^2+EB^2 =(2r)^2 = 4r^2 ..read more