Problem 1

ABCD is a square with area 625 square meters and AEFD is a rhombus with area 500 square meters. What is the area of the shaded region ABCGEA?

Problem 2

Thanks to Tasnim for the suggestion! Three squares are shown. The large square has a side length of 60, and the two smaller squares are congruent and each has a side length of 20. What is the area of triangle ABC?

Problem 3

I was emailed this problem and told it comes from an Iranian Mathematical Olympiad. A piece of paper measuring 8 by 6 is folded several times, as shown. What is the area of the final shape? (Some smart aleck will point out that folding does not technically change the area of the paper, so imagine the final shape is a separate 2d object in the plane and calculate the area enclosed by its boundary.)

As usual, watch the video for solutions.

Find The Shaded Areas – 3 Problems

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“All will be well if you use your mind for your decisions, and mind only your decisions.” Since 2007, I have devoted my life to sharing the joy of game theory and mathematics. MindYourDecisions now has over 1,000 free articles with no ads thanks to community support! Help out and get early access to posts with a pledge on Patreon.

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Answer To Square And Rhombus Shaded 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).

Problem 1

Let s = AB = BC = CD = AD be the side length of the square. Then s² = 625, so s = 25.

All sides of a rhombus have the same length, so AE = AD = 25.

Construct the height EH upon AD. Then (EH)(AD) = 500, so (EH)25 = 500 and EH = 20.

Construct a perpendicular EI upon AB. Then AI = EH = 20, so BI = AB – AI = 25 – 20 = 5.

Triangle AIE is a right triangles, so

IE² + 20² = 25²
IE = √(25² – 20²)
IE = 15

Now the shaded region is the sum of the rectangle IBCG and the right triangle AIE, so we have:

IBCG + AIE
= 25(5) + 15(20)/2
= 125 + 150
= 275 square meters

Problem 2

There are many ways to solve it. But here’s one method I like. Imagine completing the square by adding 3 rectangles to the corners. The shaded triangle is then the area of the square minus the area of 3 right triangles.

Let S(b, h) denote the area of a triangle with base b and height h.

shaded triangle
= square(80) – S(80, 60) – S(20, 40) – S(40, 80)
= 80² – 80×60/2 – 20×40/2 – 40×80/2
= 6400 – 2400 – 400 – 1600
= 2000

Problem 3

We will take the area of the original paper and subtract out the triangles that are folded. The original folds are right triangles whose horizontal leg is half of 6. As the height of a triangle is then folded to the horizontal, the height equals the horizonal leg. So both legs are 3. The first two folds are of isosceles right triangles with legs of 3. The next folds are of two congruent triangles. One side is the hypotenuse of the isosceles right triangle, so it is 3√2. Each triangle is congruent to the triangle it is folded into, and each of those obviously has a height of 3 to a base of 3√2.

So we have:

area final shape
= 8×6 rectangle – 2(isosceles triangles legs 3) – 2(height 3, base 3√2)
= 48 – 2(3×3)/2 – 2(3×3√2)/2
= 48 – 9 – 9√2
= 39 – 9√2
≈ 26.272

Reference

Problem 1 @Standupmathsmks
https://youtu.be/gCaDguHz7Hw

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