Can you find the area of the Green shaded region? | Two Squares | (Olympiad Math) |
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- Опубліковано 14 жов 2024
- Learn how to find the area of the Green shaded region. Two squares are given. Yellow and Blue areas are 250 and 150 respectively. Important Geometry skills are also explained: area of the square formula; area of the triangle formula; Pythagorean Theorem; similar triangles. Step-by-step tutorial by PreMath.com
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Thanks, Professor, for some weekend fun!❤😀🥂
You are so welcome! ❤️🌹
Thanks for your feedback! Cheers! 😀
You are awesome. Keep it up 👍
What a good feeling on Sunday, I solved it by myself.😊
Maths is one of my passions while retirement since this year. Please feed us! Thanks Sir and greetings from old Germany.🙋🏼
for calculating the area of BEP we also can use rapport of lenghts, for the area we can use the square of the rapport, for the triangle, the hypothenuse has a rapport of 1/5, for calculating the area of BEP we can calculate the area of CFP and multiply by 1/5 of square which is equal of 1/25, it gives 6 cm square
Thanks for your feedback! Cheers! 😀
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Area of minor square
A₁= 250+150=400 cm²
Side of minor square:
s = √A₁ = 20 cm
Height of blue right triangle
h = 2A/b = 150/20 = 15 cm
Angle of rotation:
tan α = 15/20
α = 36,8699°
Diagonal of minor square:
d = √(2A₁) = 20√2 cm
Side of mayor square:
S = d cos (45°-α)
S = 28 cm
Green shaded area =
Area of trapezoid - Area of isosceles right triangle
A = A₂ - ½A₁
A = ½(b+b)h - ½A₁
A = ½[S+S-d.sin(45°-α)].S - ½A₁
A= ½(2.28-4).28 - 400/2
A = 728 - 200
A = 528 cm² ( Solved √ )
Good job 👌👍🌹
Thanks for your feedback! Cheers! 😀
You are awesome. Keep it up 👍
Square CFEG:
A = s²
(250+150) = s²
s² = 400
s = √400 = 20
Trapezoid EGCP:
A = h(a+b)/2
250 = 20(EP+20)/2
250 = 10EP + 200
10EP = 50
EP = 5
Triangle ∆CFP:
FP² + CF² = PC²
15² + 20² = PC²
PC² = 225 + 400 = 625
PC = √625 = 25
As ∆PBE and ∆CFP share angles at P and are right triangles, they are similar.
Triangle ∆PBE:
PB/EP = FP/PC
PB/5 = 15/25
PB = (5)3/5 = 3
BE/PB = CF/FP
BE/3 = 20/15
BE = (3)4/3 = 4
A = bh/2 = 4(3)/2 = 6
The green area equals the area of DCBA minus EGCP (250) minus ∆PBE (6).
A = (25+3)² - 250 - 6
A = 784 - 256 = 528 cm²
thanks Professor.....
Very Very useful video sir
Thanks dear ❤️
Let a, b be the sides of squares, then b=sqrt(400)=20, and 15=300/20=PF, and PE=20-15=5, and thus PB=5×3/5=3, and PC=25 clearly, so a=25+3=28, therefore the answer is 28^2-250-3×4/2=528.😊
Excellent!
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
each side of the small square = ✓(250 cm^2 + 150 cm^2) = 20 cm
the three sides of the blue triangle => 20 cm, 150x2 cm^2/20 cm = 15 cm, ✓((20^2 + 15^2) cm^2) = ✓(625 cm^2) = 25 cm
the hypotenuse of the white right-angled triangle = (250 cm^2 - 150 cm^2) / 20 cm = 5 cm
the white triangle and the blue triangle are similar
the remaining two sides of the white right-angled triangle: 5 cm (20/25) = 4 cm and 5 cm (15/25) = 3 cm
each side of the large square = the hypotenuse of the blue triangle + the short side of the white triangle = 25 cm + 3 cm = 28 cm
area of the white triangle = (4 cm)(3 cm)/2 = 6 cm^2
green area = 28x28 cm^2 - 250 cm^2 - 6 cm^2 = 784 cm^2 - 256 cm^2 = 528 cm^2
⇒A⬠AEGCD=528
Given: A⏢GCPE=250, A△PFC=150, □GCFE and □ABCD are at different angles and overlap, with corners of □GCFE hitting sides of □ABCD at points E and P
Use A□=s²: for □GCFE, A=250+150, A=400; s²=400 ∴for □GCFE, s=20 Line GE=20, GC=20, EF=20
Set point J: Draw line ∥ FC, from point P. line intersects line GC at point J
ⒸA△PFC=150, △PJC≅△PFC ∴△PJC∾△PFC, A△PJC=150
A⏢GCPE=250, A△PJC=150 ∴▭GJPE=100
Use A▭=bh; A▭GJPE=100 100=20•GJ; ∴line GJ=5
∴Line JC=15 (NB could have figured this earlier)
Use a²+b²=c² WHERE △ABC is right triangle
15²+20²=Line CP² 225+400=CP² 625=CP² ∴CP=25
For ∠CPF, ∠CPF∾∠EPB Adj/Opp(COT)=3/4
∴line PE=5
EB²+BP²=PE 3²+4²=5² ∴PB=3, A△EPB=6
CP+PB=28, A□ABCD=28² ∴A□ABCD=784
A⬠AEGCD=784-6-250
⇒A⬠AEGCD=528
I was going to use letters and numbers with circles to point back to earlier conclusions but as I am only writing answer to ensure I leave a comment (algorithm purposes) I changed my mind.
My list of symbols I now use for math proofs now include (copy as desired):
⇒√∛∜±≈∉∈≡•÷±≥≤≠≟≝¯⁰¹²³⁴⁵⁶⁷⁸⁹ˣᵐⁿᵃᵇ⁺⁻ᐟ⸍⁼⁽⁾≮≯≨≩≅≆≌≝∾∾∿≈≉≣≢≡|∥∢∡∠△□⏢▭▱⬠°⌠⋅¼⅓½⅔¾∞αβπδΩθ°⌠µ!∑∪∫∶∷∝∴∵∋①②③④⑤⑥⑦⑧⑨⑩ⒶⒷⒸⒹⒺⒻⒼ
Pretty much the same way here: I called the blue triangle a congruent 5x magnification of the white triangle making white a 3-4-5, so large square has sides of 28cm. I actually did this complete one in my head - no calculator, no pen and paper. 28^2 - 250 - 6.
The little white triangle is a 1/5-scale of the blue triangle' so the white triangle has area (1/5)^2 = 1/25th that of the blue triangle, or area of 6.
@@timeonly1401 Sure thing: 3-4-5 gives an area of 6 as does 150/25.
Excellent!
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
Yay! I solved the problem.
Super ❤️
You are awesome. Keep it up 👍
250+150=400=20²
20*PF/2=150 PF=15 EP=5
PC=√[15²+20²]=25
BP/5 = 15/25 BP=3
BP+PC=3+25=28
EB/3 = 20/15 EB=4
EBP=3*4/2=6
area of Green Region : 28*28 - (250+6) = 784 - 256 = 528cm²
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
The area of the small square is 250+150=400=20^2, so thhe side length of the small square is 20. the blue triangle has the area of 150 and one leg of length 20, so the other leg of this right triangle must have length 15. (because the area is 150). So the hypotenuse of thhis triangle mmust be 25 (pythagoras). The small white triangle is siilar to the green triangle and hypotenuse of 5, so the lengt of the legs are 4 and 5 and therefor the are is 4*3/2=6. The side length of the big sqare is 25+3=28, so the areaof the big square is 28^2=784. The green area is the area of the big square minus area of the white triangle minus 250, i.e 784-6-250=528.
Excellent!
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
@@PreMathThanks. i first mixed up the length of the sides of the blue triangle and calculated with 23^2 as the area of the big square, but i saw my mistake and corrected the error ...
FC=√(250+150) =20 → PF=2*150/20=15 → EP=20-15=5 →→ 5*3=15 y 5*4=20 → PC=5*5=25 →→ s= EP/PC =5/25=1/5 → BP=15s=15/5=3 → BC=3+25=28 →→ EBP=PFCs²=150*(1/25)=6 →→→ Área verde =28² - 250 - 6 =528
Gracias y un saludo cordial.
Excellent! ❤️
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
We don’t need to figure out EB because we know it’s a 3-4-5 right angle triangle.
I converted cm to parsecs- it was a so much bigger problem to solve.
😀
I solved the problem and I by my calculation it is also 528 cm².
Excellent!
You are awesome. Keep it up 👍
....,L=28....Ah=28^2-250-4*3/2=784-250-6=528
Excellent! ❤️
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
😎
❤️🌹
We cannot like assume EFCG is a perfect square unless it is given
Easy
❤️
You are awesome. Keep it up 👍
@@PreMath thank s
my teacher .
528
Sir i think you have an error in the Sol
While us😢ratio of similar triangles
It should be
BP/5=20/25
And then continue
So ans would i guess
29²-250-6=585cm²
No, there’s no mistake; the ratio of the shortest side and hypotenuse of the small triangle (BP/EP) is the same as the ratio of the shortest side and hypotenuse of the big triangle (FP/CP). That is:
BP/EP = BP/5 = FP/CP = 15/25 = 3/5
You can tell this is correct by recognising that the big triangle is a 3,4,5 special triangle (15,20,25 -> 5*3,5*4,5*5), and as the big and little triangles are similar triangles, you know it’s also a 3,4,5 special triangle, and given the hypotenuse is 5, you know the other two sides must be 3 and 4.
@@boneistt thank you for pointing it out bro
A=528
Excellent! ❤️
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
THE AREA OF THE BLUE TRIANGLE IS IRELEVANT
Don't see how you can get the area of the angled square without it. You could rotate the angled square clockwise or counterclockwise and keep the overlap area at 250cm² and it would change both the area of the angled square and the larger square, not to mention the green area. There's nothing which sets what percentage of the angled square overlaps other than the differences in the areas. You can certainly determine all the dimensions of the problem from the yellow trapezoid alone (even the blue triangle, by dropping a perpendicular to GC from P), but not without knowing the full area of the angled square or at least it's side length, which you can't get from the trapezoid alone. Given just the yellow area alone, since the diagram is not necessarily to scale, the angled square side length could be anywhere from 16 to 22, and that's just the integer values (it's actually between 5√10 and 10√5).