5th grade math problem from china.

Question-
In the quadrilateral ABCD, angle A = 90°, angle ABD = 40°, angle BDC = 5°, angle C = 45°, and the length of AB is 6. Find the area of the quadrilateral ABCD.


                     
Answer-.
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I would have solved the problem using trigonometry, which is beyond what a 5th grader would have learned. Still, I will present this complicated solution to show it is possible. Then I will reveal the neat trick which is how 5th graders were expected to solve it.

First we will calculate some angles and lengths.

angle ADB = 90° – 40° = 50°
angle DBC = 180° – 45°- 5° = 130°

AD = 6 tan 40°
BD = 6/cos 40°

Then we can use the law of sines to calculate CB:

CB/BD = (sin 5°)/(sin 45°)

CB = (6/cos 40°)(sin 5°)/(sin 45°)

We can then calculate the areas of triangles ABD and CBD. Triangle ABD is a right triangle, so its area is half its base times its height. Triangle CBD can be solved using the formula (side 1)(side 2)(angle between)/2. So we get:

area ABD = 6(6 tan 40°)/2

area CBD = (6/cos 40°)[(sin 5°)/(sin 45°)](sin 130°)(1/2)

                        

area ABCD = 18

So now the question is, how would 5th graders have found the answer? Here’s the incredible method.

The trick: flip triangle CBD

If we flip triangle CBD, side BD remains in the same position, but we flip the other sides. Here is the shape we end up creating:

Now we can calculate some angles in this new figure:

angle ADC = 50° + 130° = 180°
angle ABC = 40° + 5° = 45°
angle ACB = 45°

Notice this is a right triangle with two equal 45 degree angles–so this is an isosceles right triangle!

The area of this shape equals quadrilateral ABCD (as we have only re-arranged the areas). So the other leg of this triangle AC also has a length of 6, and its area will be:

6 × 6 / 2 = 18

Amazing! This is unlike any math problem I’ve seen in class, a competition, or on a test. It’s quite a neat trick!

Thanks to Mr . Presh Talwalkar.