That explanation sounded a whole lot better than the KZread 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.

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

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

Easy . Nice solution.

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.

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.

Shaded area=1/2(√29)^2=29/2

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.

Сколько исписать на такую простую задачу ?

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

29 күн бұрын

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

@LuisdeBritoCamacho

29 күн бұрын

@@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

29 күн бұрын

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

@rabotaakk-nw9nm

28 күн бұрын

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

@rabotaakk-nw9nm

28 күн бұрын

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