Calculate area of the Green shaded Quadrilateral EFGH | Think outside the box | Fun Olympiad

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Calculate area of the Green shaded Quadrilateral EFGH | Think outside the box | Fun Olympiad

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[Waldlaeufer]

I took a quite risky approach since I was not sure whether it is correct at first:
White Area = 1/2(a + b)(56 - 1/2(x + y)) + 1/2(x + y)(56 - 1/2(a + b))
White Area = 1/2(35)(56 - 10.5) + 1/2(21)(56 - 17.5)
White Area = 1/2(35)(45.5) + 1/2(21)(38.5) = 1200.5
Square Area = 56² = 3136
Green Area = Square Area - White Area = 3136 - 1200.5 = 1935.5 square units

My reasoning:
Add two bases and take the average height in both directions (horizontally and vertically) to get the areas of the triangles. You can also think of the formulas as formulas for the area of rectangles: The average width and the average height of the rectangles are used.

[miguelgnievesl]

I think that it is not necessary to divide the green area to solve this geometric problem. It is enough to solve each of the 4 triangles and add them. We subtract this result from the area of the square and problem solved.

[robertberg]

Thanks for an elegant solution. I just added together the Area of the for triangles ((56-a)X + (56-X)b + (56-b)y + (56-y))/2 and simplified and a, b, X and Y turned out to cancel each other out. The Area turned out to 1200.5 which I subtracted from the Area of the rektangel.

[Abby-hisf]

Amazing short cut! no need to calculate the four small triangles individually!

[NASIRable]

Amazing way to conclude, just coming out of the box. Thanks for sharing nice stuff.

[barttemolder]

The difference between the white (w) and green (g) area is the rectangle of size 21 by 35 (r).
Together they have the area of the big square (s).
so g = w + r and g = s - w => 2g = s + w + r - w => 2g = s + w => g = (s + w) / 2 => g = (3136 + 735) / 2 => g = 1935.5

[ludmilaivanova]

4 white areas are: (56-a)x/2
(56-b)y/2
(56-x)b/2
(56-y)a/2. we need to add them up and subtract the sum from 56^2

[umutkargili]

Lovely. You said let's think out of the box but dig your way inside the box(56x56)😉

[sie_khoentjoeng]

We can make a hollow rectangular (A=3136-735=2401) which the green area is half of this i.e. 1200.5, and then add the inner rectangular (735).
The total area of green is 1200.5+735=1935.5 unit

[bernardocademia]

If S² = Area of the square and Ai, i = 1 to 3 is area of white triangles then, the of shaded region =S²-(A1+A2+A3+A4) by principle of inclusion and exclusion (PIE). But,
A1+A2+A3+A4 =28(x+y+a+b)-(1/2)(a+b)(x+y). Given that x+y =21, and a+b=35, therefore
A1+A2+A3+A4
=28(56)-(1/2)(35)(21)
Then the area of shaded region is
As=56²-(28)(56)+(35)(10.5)
As =1935.5 unit²

[nivan]

Hello Sir. I tried another way and got the answer 👍

[ybodoN]

We have enough information to find that:
• EG = √(56² + 21²) = 7√73
• FH = √(56² + 35²) = 7√89
But we can't use the formula K = (pq / 2) sin θ where the lengths of the diagonals are p and q and the angle between them is θ.
So, let's translate EG and FH to obtain a parallelogram.
Find the length of the diagonals in this parallelogram:
• √[(56 + 21)² + (56 - 35)²] = 7√130
• √[(56 + 35)² + (56 - 21)²] = 7√194
Now, thanks to Heron's formula:
• triangle with sides 7√73, 7√89, 7√130 has an area of 1935.5
• triangle with sides 7√73, 7√89, 7√194 has an area of 1935.5
Both ways, 1935.5 is the area of the green shaded quadrilateral.
As a bonus, we can also calculate that sin θ = 79 / √6497.

[J-yv]

Please me show how to solve for a, b, x, and y. Use Calculus?

[santiagoarosam]

If we transform the four peripheral right triangles into rectangles whose diagonals are the respective hypotenuses, we obtain a rectangle inside the original square whose dimensions are: Base=56-(a+b)=56-35=21 → Height=56-( x+y)=56-21=35 → Interior rectangle area= R =21*35=735 →→ Area of peripheral rectangles= P = 56² - R= 3136 - 735= 2401 →→ Area of peripheral right triangles = P/2= 2401/2=1200.50 →→ Quadrilateral area EFGH = 56² - (P/2)= 3136 – 1200.50 = 1935.50
Regards

[easternbrown]

Another way: a, b, x and y are not precisely defined by the problem, so you can assume the limiting case that both a and x are zero. The answer is then the area of the square less the two resulting white triangles which turns out to simplify neatly down to 56^2 - 7^4/2 or 1935.5.

[yashveersingh]

Hello sir Can you please make some videos on basic calculus