Chinese College Test Problem | You should be able to solve this!

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[@michaeldoerr5810]

That explanation sounded a whole lot better than the TH-cam comments for this video. And I guessed in my head correctly the relevant congruence postulate. I hope that this means that I do know how to solve this easily. Congruence postulates have to be applied to justify triangles being congruent regardless of rotation.

[@Uma-Bharat-India]

Points A, B, C are on a right angle tr. on a semi circle. Elaborate further.

[@jimlocke9320]

Once we have determined that ΔPEG is congruent to ΔCGF (5:15), we can determine that PG = CF = 2 and CG = EP = 5, so PC = PG + CG = 2 + 5 = 7. BE = PC because they are opposite sides of a rectangle, so BE = 7. By Pythagoras, EF = √(BE² + BF²) = √(7² + 3²) = √(49 + 9) = √(58). One formula for the area of an isosceles right triangle is hypotenuse squared divided by 4 = (√(58))²/4 = 58/4 = 29/2, as Math Booster also found.

[@Irtsak]

Easy . Nice solution.

[@santiagoarosam430]

H es la proyección ortogonal de E sobre DG→ EH=BC=3+2=5→ HG=FC=2→ EH²+HG²=EG²→ 5²+2²=29→ Área rosa EGF =EG²/2=29/2=14,5 ud².
Gracias y un saludo cordial.

[@prossvay8744]

Shaded area=1/2(√29)^2=29/2

[@jonpress6773]

Let x = EG = FG. Then EF = x√2. By Pythagoras, GC = √(x^2-2^2) = √(x^2-4) and EB = √(EF^2 - 3^2) = √(2x^2 - 9). As stated in the video, EP = BF+FC = 5. In ΔPGE, PG = √(x^2 - 5^2) = √(x^2 - 25). In parallelogram EBCP, EB = PC = PG+GC. so √(x^2 - 25) + √(x^2-4) = √(2x^2 - 9). Then do the algebra:
√(x^2 - 25) + √(x^2-4) = √(2x^2 - 9)
x^2 - 25 + x^2-4 + 2√[(x^2 - 25)(x^2-4)] = 2x^2 - 9
2√[(x^2 - 25)(x^2-4)] = 20
√[(x^2 - 25)(x^2-4)] = 10
(x^2 - 25)(x^2-4) = 100
x^4-29x^2+100 = 100
x^4-29x^2 = 0
x^2(x^2 - 29)
So x^2 = 0 or 29, but only 29 is reasonable.
Since the desired area is x^2/2, the answer is 29/2.

[@riccsin]

I solved system equations:
FG * cosβ = 2
sqrt(2) * FG * sin(β - π/4) = 3
Remembering :
sin(α - β) = sin α * cos β - cos α * sin β
sin² α + cos² α = 1

[@user-yr2jg7cm5r]

Сколько исписать на такую простую задачу ?

[@LuisdeBritoCamacho]

1) Draw a Parallel Line to CB passing Point E. The Point of interception with Line DG is E'
2) Line EE' = 5 lin un.
3) Line E'G = CF = 2 lin un, because the Triangle [EFG] is Isosceles.
4) EG^2 = 2^2 + 5^2 = 4 + 25 = 29
5) EF^2 = 29 + 29 ; EF^2 = 58
6) EB^2 = 58 - 9 ; EB^2 = 49 ; EB = 7 lin un
7) CE' = BE = 7 lin un
8) Area of Rectangle [BCE'E] = 5 * 7 = 35 sq un
9) Pink Area = 35 - (10 + 21/2) ; Pink Area = 35 - (10 + 21/2) ; Pink Area = 35 - 20,5 ; Pink Area = 14,5 Square Units.
10) Answer : Pink Area is equal to 14,5 Square Units or (EG^2 / 2) = 29/2 Square Units.

[@johndoe3092]

shaded area = EG^2 / 2 = 29 / 2 = 14.5
you have the answer after step (4) ! Why the rest ?

[@LuisdeBritoCamacho]

@@johndoe3092 , because I wanted to do it my way! I am using my Free Will. Did I made something wrong? My answer is wrong? No! What's your problem man? Don't you dare to tease me. Are we cleared? Stick the dodge in your arse hole!!

[@User-jr7vf]

@@johndoe3092 because he wants to complicate things as usual. This guy always comes up with these extra complicated solutions to sound smart.

[@rabotaakk-nw9nm]

@@LuisdeBritoCamacho I can see now:
2) EE'=5 ??? 🤔 Why?
3) E'G=2

[@rabotaakk-nw9nm]

@@LuisdeBritoCamacho I caB see:
@LuisdeBritoCamacho is smarter than @johndoe3092 and @User-jr7vf 🤣🤣🤣🤣🤣🤣