**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.

**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.

g)

**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:

**(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.

**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.

**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.

]]>It's a really interesting question to consider the reverse line of logic. If two trapezoids have the same area, then do they have the same diagonal? Here let's look at a counterexample (an example that shows why this fact cannot be true):

The first trapezoid has bases of lengths 2 and 3, and a height of 5.

f712c5d4-0859-4d1e-aa73-d573f5fb9671-image.png

The segment labeled 2.5 runs from the base of the altitude to the bottom right corner.

The area of this trapezoid is

$$ \frac{1}{2} (2 + 3) \times 5 = \frac{25}{2} $$

Let's consider another trapezoid with the same area whose bases are almost equal in length, with height the same length as the bases. It would look almost like a square:

1edc8e83-7003-4e31-aa76-b999affe16e5-image.png

The diagonal of the square is equal to 5. However, the diagonal of the first trapezoid must be larger than 5, because one of the legs is already equal to 5, and the Pythagorean Theorem tells us that

$$ a^2 + b^2 = c^2 $$

where c is the hypotenuse. These two trapezoids have the same area but different length diagonals, so this shows that not all isosceles trapezoids with the same area have the same length diagonals.

Happy Learning!

The Daily Challenge Team

]]>You can see Prof. Loh talk about a cool way of solving quadratic equations which he came up with while developing Module 4: Algebra Tools. You can also read more about him and his quadratic method in this NYT article.

Happy Learning!

The Daily Challenge Team

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