Can you find the area of the Green shaded region? | Two Squares | (Olympiad Math) |

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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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27 окт 2023

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Комментарии : 52

@bigm383 8 месяцев назад

Thanks, Professor, for some weekend fun!❤😀🥂

@PreMath 8 месяцев назад

You are so welcome! ❤️🌹 Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

@ganymed1236 8 месяцев назад

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.🙋🏼

@francismoles9852 8 месяцев назад

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

@PreMath 8 месяцев назад

Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

@Ibrahimfamilyvlog2097l 8 месяцев назад

Good job 👌👍🌹

@PreMath 8 месяцев назад

Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

@3LLT33 8 месяцев назад

We don’t need to figure out EB because we know it’s a 3-4-5 right angle triangle.

@masudi67 8 месяцев назад

thanks Professor.....

@marioalb9726 8 месяцев назад

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 √ )

@mathbynisharsir5586 8 месяцев назад

Very Very useful video sir

@PreMath 8 месяцев назад

Thanks dear ❤️

@quigonkenny 3 месяца назад

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

@michaelgarrow3239 8 месяцев назад

I converted cm to parsecs- it was a so much bigger problem to solve.

@PreMath 8 месяцев назад

😀

@danieldennis9831 8 месяцев назад

⇒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ⒶⒷⒸⒹⒺⒻⒼ

@misterenter-iz7rz 8 месяцев назад

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.😊

@PreMath 8 месяцев назад

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

@cyruschang1904 8 месяцев назад

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

@MrPaulc222 8 месяцев назад

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.

@timeonly1401 8 месяцев назад

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.

@MrPaulc222 8 месяцев назад

@@timeonly1401 Sure thing: 3-4-5 gives an area of 6 as does 150/25.

@PreMath 8 месяцев назад

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

@Copernicusfreud 8 месяцев назад

Yay! I solved the problem.

@PreMath 8 месяцев назад

Super ❤️ You are awesome. Keep it up 👍

@juergenilse3259 8 месяцев назад

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.

@PreMath 8 месяцев назад

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

@juergenilse3259 8 месяцев назад

@@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 ...

@wackojacko3962 8 месяцев назад

😎

@PreMath 8 месяцев назад

❤️🌹

@santiagoarosam430 8 месяцев назад

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.

@PreMath 8 месяцев назад

Excellent! ❤️ Thanks for sharing! Cheers! You are awesome. Keep it up 👍

@timurkodzov718 8 месяцев назад

I solved the problem and I by my calculation it is also 528 cm².

@PreMath 8 месяцев назад

Excellent! You are awesome. Keep it up 👍

@himo3485 8 месяцев назад

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

@PreMath 8 месяцев назад

Thanks for sharing! Cheers! You are awesome. Keep it up 👍

@killing_gaming0973 8 месяцев назад

We cannot like assume EFCG is a perfect square unless it is given

@giuseppemalaguti435 8 месяцев назад

....,L=28....Ah=28^2-250-4*3/2=784-250-6=528

@PreMath 8 месяцев назад

Excellent! ❤️ Thanks for sharing! Cheers! You are awesome. Keep it up 👍

@DB-lg5sq 8 месяцев назад

528

@abdulkadirbuyuksoy2076 8 месяцев назад

Easy

@PreMath 8 месяцев назад

❤️ You are awesome. Keep it up 👍

@abdulkadirbuyuksoy2076 8 месяцев назад

@@PreMath thank s my teacher .

@prossvay8744 8 месяцев назад

A=528

@PreMath 8 месяцев назад

Excellent! ❤️ Thanks for sharing! Cheers! You are awesome. Keep it up 👍

@user-yk8hf2hb2x 8 месяцев назад

THE AREA OF THE BLUE TRIANGLE IS IRELEVANT

@quigonkenny 3 месяца назад

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).

@trishanuagarwal9220 8 месяцев назад

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

@boneistt 8 месяцев назад

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.

@trishanuagarwal9220 8 месяцев назад

@@boneistt thank you for pointing it out bro