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

]]>