Week 1

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Day 1

Hexagon Area

Module 2 Day 1 Challenge Part 1 Module 2 Day 1 Challenge Part 2 Module 2 Day 1 Challenge Part 3 Module 2 Day 1 Challenge Part 4 Module 2 Day 1 Your Turn Part 1 Module 2 Day 1 Your Turn Part 2

10
Topics

34
Posts

Π

@The-Blade-Dancer (long leg)^2 = (3^2) - (3 / 2)^2 which simplifies to 3/2√3
Day 2

Octagon Area

Module 2 Day 2 Challenge Part 1 Module 2 Day 2 Challenge Part 2 Module 2 Day 2 Challenge Part 3 Module 2 Day 2 Challenge Part 4 Module 2 Day 2 Challenge Part 5 Module 2 Day 2 Challenge Part 6 Module 2 Day 2 Your Turn Part 1 Module 2 Day 2 Your Turn Part 2

10
Topics

28
Posts

Π

RIP its been 2 years and its still on debbie's todo list
Day 3

Round Area

Module 2 Day 3 Challenge Part 1 Module 2 Day 3 Challenge Part 2 Module 2 Day 3 Challenge Part 3 Module 2 Day 3 Challenge Part 4 Module 2 Day 3 Your Turn Part 1 Module 2 Day 3 Your Turn Part 2 Module 2 Day 3 Your Turn Part 3

17
Topics

71
Posts

Π

@debbie he's done 1,000 exams? That's a sheesh moment
Day 4

Diagonal Area

Module 2 Day 4 Challenge Part 1 Module 2 Day 4 Challenge Part 2 Module 2 Day 4 Challenge Part 3 Module 2 Day 4 Your Turn Part 1 Module 2 Day 4 Your Turn Part 2 Module 2 Day 4 Your Turn Part 3

13
Topics

45
Posts

S

I found another way to solve this question that is simpler in my opinion (note: I have taken a high school geometry class before)

First solve for the base angles of the lower triangle. The lower triangle must be isosceles because it is in an isosceles trapezoid so the base angles must be equal. Since angles in a triangle add up to 180, the sum of these angles is 180-120=60. Each angle is 30 because 30=30 and 30+30=60.
45c19703-3daf-4ae3-ab74-1c18a38e91d3-image.png

Next draw a 30-60-90 triangle with one of the 30 degree angles and by breaking up a right angle on the top base into 30 and 60. Notice the diagonal of this 30-60-90 triangle is the same as the diagonal of the isosceles trapezoid.
66d086d8-aa32-4112-8bee-216348b96709-image.pngg)

Then move the triangle to the right of the 30-60-90 to the right of the trapezoid like this to create a rectangle of the same area:
32ce6a08-f0ba-4d5b-9b77-08fe1de4816b-image.png

(optional) To prove why this works you could draw a rectangle inside the isosceles trapezoid. The angles with one red mark are equal to each other. The angles with two red marks are also equal to each other. An angle with one red mark and an angle with two red marks adds up to 90 (or 180-90) because they are acute angles in a right triangle.
3c32cc8d-7479-44cf-ada6-52a8321fe1fd-image.png

Next, let x be the length of the diagonal. This is the hypotenuse of the 30-60-90 triangle! Because of this, we can identify the lengths of the rectangle as x/2 and x(√3)/2 using the ratio of 1, √3, 2. To find the area of the rectangle (which is equal to the area of the isosceles trapezoid), simply multiply the sides x/2 and x(√3)/2 which equals x²(√3)/4, the formula found in the video.
0eee0f16-c905-4e86-b419-ecd4d224feb6-image.png

Finally, set x²(√3)/4 equal to 36(√3).
Divide both sides by √3 to get
x²/4 = 36
Multiply both sides by 4 to get
x² = 144
Take the square root to get
x = 12

Since we set x as the diagonal of the trapezoid, the diagonal of the trapezoid is 12.

R

FUN 🥳

• • RZ923

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LOL because i joined but didn't do any math 😁
T

Why does a parallelogram with two diagonals end up making four triangles of the same area?

• • Tylenol

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@CoherentMango

M2W1D4-ch-why-same-area-all-triangles-in-parallelogram.png

This is a really interesting and surprising characteristic of parallelograms, that the four triangles formed from the intersection of their diagonals have the same area! I'm glad you asked about it.

Many facts go into this concept:
Fact 1: Diagonals of a parallelogram bisect each other
M2W1D4-ch-why-same-area-congruent-angles.png

Using the parallel lines of the parallelogram, we get the above congruent angles. To give the technical term for these congruent angles, the \( \textcolor{blue}{\text{ blue }} \) and \(\textcolor{purple}{\text{ purple }}\) ones are called "alternate interior angles," and the \( \textcolor{orange}{\text{yellow}}\) ones are called "vertical angles."

M2W1D4-ch-why-same-area-why-same-base.png

These two triangles have the same angles, so they are congruent, which means that the \(\textcolor{red}{\text{ red }}\) side and \(\textcolor{green}{\text{ green }}\) sides are the same length. So, the diagonals of a parallelogram cut each other in half (bisect each other)!

In the same way, the other two diagonal segments are also the same length.

M2W1D4-ch-why-same-area-why-same-base2.png
Fact 2: Each of the three sides of a triangle can be thought of a base
There are three ways of thinking of a base and height for a given triangle. This is because there are three sides of a triangle! In the diagram below, I have drawn the three different heights for the \(\textcolor{purple}{\text{ purple }}\) triangle, as well as the three different heights for the \(\textcolor{orange}{\text{ yellow }}\) triangle. The heights are color-coded with their matching bases, so the \(\textcolor{red}{\text{ red }}\) base goes with the \(\textcolor{red}{\text{ red }}\) altitude, the \(\textcolor{green}{\text{ green }}\) base goes with the \(\textcolor{green}{\text{ green }}\) altitude, and the \(\textcolor{purple}{\text{ purple }}\) base goes with the \(\textcolor{purple}{\text{ purple }}\) altitude.

M2W1D4-ch-why-same-area-three-different-bases.png

An altitude must be perpendicular \( \perp \) to the base. That means an altitude might have to lie outside the triangle, which is true for obtuse triangles.

Now, looking at the triangles one pair at a time, we see that they all have the same base and height.

M2W1D4-ch-why-same-area-pair1.png

M2W1D4-ch-why-same-area-pair2.png

M2W1D4-ch-why-same-area-pair3.png

Thus the four triangles created by drawing diagonals of any parallelogram each have the same area!
Even if the bases aren't the same length, you can apply this idea to other triangles
You can generalize this idea in order to find the ratio of areas of triangles which have a common vertex and common base.

In the picture below, the two \( \textcolor{purple}{\text{ purple }} \) triangles have the same altitude, which is the perpendicular \( \perp \) line from the top vertex to the base line.

M2W1-ratio-of-bases-ratio-of-areas.png

Thus to find their relative areas, just look at the relative lengths of their bases. In this example, one base is \( \frac{3}{2} \times\) the other, and the ratio of their areas is \( 2 : 3.\)

🙂

Day 1

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Why does a parallelogram with two diagonals end up making four triangles of the same area?