Which of these square numbers also happens to be the sum of two smaller square numbers?

A. 16
B. 25
C. 36
D. 49

The question was worth $16,000, but the contestant would drop down to just $1,000 for an incorrect answer. Can you figure it out?

As usual, watch the video for a solution.

When Not Knowing Math Could Cost You $15000

Or keep reading.

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Answer To When Not Knowing Math Could Cost You $15000

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

Every student should instantly recognize the correct answer of B. 25 from the famous right triangle triple:

3² + 4² = 5² = 25

The 3-4-5 triangle was first known to the ancient Egyptians who used it for surveying land and constructing pyraminds. This is so ancient that it pre-dates Greek mathematicians like Archimedes, Pythagoras and Euclid by more than 1000 years!

So every student should know the answer instinctively, and any doubt would reflect a failure of the education system.

What happened on the show

Sadly, our education system is failing students.

The contestant was having pressure under the spotlight, so he used a lifeline and asked the audience. The studio audience answered in the following proportions:

A. 16 (50%)
B. 25 (30%)
C. 36 (19%)
D. 49 (1%)

The online AOL community also chose option A with the highest percentage of replies.

A. 16 (45%)
B. 25 (22%)
C. 36 (23%)
D. 49 (10%)

The contestant went with the audience answer of (A) 16 and was eliminated from the show, leaving with just $1000 instead of advancing to $16,000.

The lesson is: always pay attention in math class!

Verifying other answers are wrong

Suppose some square is equal to the sum of two smaller squares. Then we would have:

c² = a² + b²
c² – a² = b²

To see if a perfect square is the sum of two smaller squares, we can subtract smaller square numbers, one by one, and see if the remainder is also a perfect square.

Let’s test 4² = 16. We can subtract the smaller squares and see that none of the resulting differences is a perfect square.

4² – 1² = 15
4² – 2² = 12
4² – 3² = 7

Now let’s analyze 6² = 36 in the same fashion.

6² – 1² = 35
6² – 2² = 32
6² – 3² = 27
6² – 4² = 20
6² – 5² = 11

Now let’s analyze 7² = 49 in the same fashion.

7² – 1² = 48
7² – 2² = 45
7² – 3² = 40
7² – 4² = 33
7² – 5² = 24
7² – 6² = 13

References

Egypt
https://commons.wikimedia.org/wiki/File:All_Gizah_Pyramids.jpg
Ricardo Liberato, CC BY-SA 2.0 , via Wikimedia Commons
https://commons.wikimedia.org/wiki/File:Corda-a-12-nodi.svg

Video clip
https://www.youtube.com/watch?v=JpGya7hKtng

Reddit
https://www.reddit.com/r/facepalm/comments/rprz3s/when_not_knowing_high_school_level_math_costs_you/

As usual, watch the video for a solution.

Missing Area In Square

Or keep reading.

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Answer To Missing Area In Square

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

Thanks Rolan for alerting me of several typos!

Triangles DHF and DAE are similar because each has a right angle, and they share a common angle ADE.

The ratio of their areas is the square of the ratio of corresponding sides:

[DHF]/[DAE] = 4/9
(HF)²/(AE)² = 4/9
(HF)/(AE)² = 2/3

Let HF = 2x so AF = 3x. Since we know the area of each right triangle and one leg, we can determine the value of the other leg in terms of x.

[DHF] = 4
(DH)(HF)/2 = 4
(DH)(2x)/2 = 4
DH = 4/x

[DAE] = 9
(AD)(AE)/2 = 4
(AD)(3x)/2 = 4
AD = 6/x

Since ABCD is a square CD = AD = 6/x.

Now construct GI parallel to CD.

Triangles GIF and DAE are congruent. They are similar because each has a right angle, and from the similarity of DAE and DHF the angles DFH and DEA are congruent. Since IG = AD the triangles GIF and DAE are congruent. Thus they have the same area of 9, and IF = AE = 3x.

We can now solve for the area of DIGC.

[DIGC]
= [DFGC] – [GIF]
= 4 + 11 – 9
= 6

But we have:

[DIGC] = (DC)(DI)
6 = (6/x)(DI)
x = DI

We now have the 3 sides of the right triangle DFH in terms of x, so we have:

(FH)² + (DH)² = (DF)²
(2x)² + (4/x)² = (4x)²
4x² + 16/x² = 16x²
16/x² = 12x²
4/3 = x⁴
2/√3 = x²

The area of the square ABCD is then:

(DC)²
= (6/x)²
= 36/x²
= 36/(2/√3)
= 18√3

But the area of the square is also the sum of 5, 4, 11, and the missing area. So we have:

5 + 4 + 11 + [missing] = 18√3
[missing] = 18√3 – 20

Thus the missing area is 18√3 – 20 ≈ 11.18 square meters.

Special thanks this month to:

Lee Redden
Kyle
Daniel Lewis

Thanks to all supporters on Patreon and YouTube!

Reference

Puzzle by Angel Silva Palacios, solution by Chew-Seong Cheong
https://www.facebook.com/photo/?fbid=1748313732795477&set=pb.100028506018835.-2207520000
https://www.pinterest.com/pin/713890978478397767/

A square is divided into 8 rectangles whose areas are equal. The height of one rectangle is shown as 8 cm.

What is the area of the entire square?

As usual, watch the video for a solution.

Area of Square From 8 Rectangles Equal Area

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Answer To Area of Square From 8 Rectangles Equal Area

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

The answer is 1225 cm². Here is how to derive it. Label the rectangles from B to I as shown, and let a be the width of rectangle B.

The area of B must be equal to 8a. All rectangles have the same area, so all have an area of 8a.

Consider the rectangle formed by B, C, and D. The total area is then 3(8a) = 24a. Since this rectangle width is also a, its height must be 24. That is the height of rectangle E.

Now iterate these same steps by joining rectangle E, which adds an area of 8a, to the previous rectangle. The new rectangle’s area is 32a, and its height is 24. So the width must be 32a/24 = 4a/3. This is the width of rectangle F.

Now append the rectangle F to increase the area by 8a. The new rectangle’s area is 40a, and its width is 4a/3. So the height must be 40a/(4a/3) = 30. This will be the height of rectangle G.

Next append the rectangle G to increase the area by 8a. The new rectangle’s area is 48a, and its height is 30. So the width must be 48a/30 = 8a/5. This will be the width of rectangle H.

Then append the rectangle H to increase the area by 8a. The new rectangle’s area is 56a, and its width is 8a/5. So the height must be 56a/(8a/5) = 35. This will be the height of rectangle I.

But the height of rectangle I is exactly the side of the square. So the square’s area is then 35² = 1225 cm².

For good measure we can easily solve for a. Append I to increase the area to 64a. But we know this equals 1225, so we must have a = 1225/64 cm.

From there one could solve for the side lengths of all the rectangles, if so desired.

What a fun problem!

References

Scientific American
https://www.scientificamerican.com/game/math-puzzle-great-lengths/

Previous puzzle 5 rectangles
https://www.youtube.com/watch?v=lrND8AeL23s
https://mindyourdecisions.com/blog/2020/12/20/5-rectangles-puzzle/