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@PreMath

Ай бұрын

Many thanks dear 🙏❤️

Fantastic 👍, thank you teacher 🌹🙏.

@PreMath

Ай бұрын

Always welcome🌹❤️

All are similar right-angle triangles, BE=3×4/5=12/5, AE=3×3/5=9/5, RD=12/5×4/5=48/25, DC=48/25×4/3=64/25, therefore the answer is 1/2×(12/5×9/5+48/25×64/25)=1/2×(108/25+3072/625)=1/2×5772/625=2886/625.😊

Another good problem, solved by proportions. Thank you!

The large triangle has area 6, and the white triangle is a scaled version, with an area reduction factor of (BE/5)^2 = 0.2304. Hence 6*(1-0.2304) = 2886/625

Area of a 3k,4k,5k triangle is 6k². With k=1, Area of ABC=6. BE=h is height of ABC with base AC=5, then 5h/2=6, h=12/5. h=5k (another k) then 12/5=5k, k=12/25. Area of BDE=6k²=6(12/25)²=864/625. Finally Red area=6-(864/625)=2886/625=4.6176.

First: BC = 4 (evident), Then we notice that the 4 triangles ABC, AEB, BDE and EDC are all similar (common angles). The side lengthes of AEB are 3/5 of the side lengthes of ABC (as AB/AC = 3/5), so BE = (3/5).BC =(3/5).4 = 12/5 and AE = (3/5).AB = (3/5).3 = 9/5. The side lengthes of BDE are 4/5 of the side lengthes of AEB (as BE/AB = (12/5)/3 = 4/5), so BD = (4/5).AE = (4/5).(9/5) = 36/25 and DE = (4/5).BE = (4/5).((12/5) =48/25. So the area of BDE is (1/2).BD.DE = (1/2).((36/25).(48/25) = 864/625. The area of ABC is (1/2).AB.BC = (1/2).3.4 = 6, so by difference the red shaded area is 6 - 864/625 = 2886/625

As CA = 5 and AB = 3, ∆ABC is a 3-4-5 Pythagorean triple right triangle, and BC = 4. If ∠A = α and ∠C = β, where β = 90°- α, then by complementary angles, ∠EBA is β and ∠CED = α, and thus ∠DBE = α and ∠DEB = β. Therefore ∆EDC, ∆BDE, and ∆BEA are all similar to ∆ABC. BE/AB =BC/CA BE/3 = 4/5 BE = (3)4/5 = 12/5 BD/EB = AB/CA BD/(12/5) = 3/5 BD = (3/5)(12/5) = 36/25 DE/EB = BC/CA DE/(12/5) = 4/5 DE = (4/5)(12/5) = 48/25 Triangle ∆ABC: A = bh/2 = 4(3)/2 = 12/2 = 6 Triangle ∆BDE: A = bh/2 = (36/25)(48/25)/2 A = 36(24)/625 = 864/625 Total red area: A = 6 - 864/625 A = (3750-864)/625 A = 2886/625 = 4.6176

h=3,b=4...White triangol..b=3a,h=4a..Ared=3*4/2-3a*4a/2=6-6a^2...triangoli simili 3/4=4a/(4-3a)..a=12/25..Ared=6-6*144/625=(3750-864)/625=2886/625

Many similar triangles. BC = 4, from 3,4, 5 triangle. 5/4 = 3/BE. BE = 12/5. BD = 12/5 x 3/5. DE = 12/5 x 4/5.. Area of white triangle = 1/2 x BD x DE. 1/2 x 12/5 x 3/5 x 12/5 x 4/5 = 1728/1250 = 1.3824. Area of triangle ABC = 6. Area of red triangles = 6 - 1.3824. 4.6176.

BC=√[5^2-3^2]=√16=4 3＊4/2=5＊BE/2 BE=12/5 DE=9.6/5 BD=7.2/5 Red region area : 3＊4/2 - 96/50＊72/50/2 =6 - 6912/5000 = 23088/5000 = 5772/1250 = 2886/625

S=4,6176≈4,62😊

4 similar triangles giving 12 equations. What a workout! ( Only needed to use 3 though)

Pythagorean Triples (3, 4, 5) So, BC = 4. Now to think outside the box (probably)! The altitude of a right triangle is visible. By the Right Triangle Similarity Theorem, △ABC ~ △AEB ~ △BEC. Apply the Geometric Mean (Leg) Theorem. Label AE = x. AB² = AE * AC 3² = x * 5 5x = 9 x = 9/5 So, AE = 9/5 = 1.8. Therefore, CE = 5 - 1.8 = 3.2. Both △ABC & △EDC share ∠C. So, △ABC ~ △EDC by AA. DE/AB = CE/AC DE/3 = 3.2/5 DE/3 = 16/25 DE = 48/25 = 1.92 CD/BC = CE/AC CD/4 = 3.2/5 CD/4 = 16/25 CD = 64/25 = 2.56 So, BD = 4 - 2.56 = 1.44 Red region area = △ABC Area - △BDE Area A = (bh)/2 = (3 * 4)/2 = 12/2 = 6 A = (1.44 * 1.92)/2 = 2.7648/2 = 1.3824 = 13824/10000 = 864/625 Red region area = 6 - 1.3824 = 4.6176 = 3750/625 - 864/625 = 2886/625 So, the area of the red shaded region is 2886/625 square units, or 4.6176 square units.

This is Cool

S = 1.8 × 2.4/2 + 6 × (3.2/5)^2 S = 4.6176

My way of solution is ▶ let ∠ BAC= α ∠ ABC= 90° ⇒ ∠ BCA= 90-α ∠ BCA= β ⇒ α + β = 90° if we write this for all these four triangles, we can say that : ΔABC ~ ΔAEB ~ ΔBDE ~ ΔEDC According to the Pythagorean theorem, for the ΔABC : AB²+BC²= AC² 3²+BC²= 5² BC= 4 length units ΔABC ~ ΔAEB BC/ EB= AC/AB= AB/AE BC= 4 AC= 5 AB= 3 4/EB= 5/3= 3/AE ⇒ EB= 12/5 AE= 9/5 ΔAEB ~ ΔBDE EB/ ED= AB/EB= AE/BD EB= 12/5 AB= 3 AE= 9/5 (12/5)/ED= 3/(12/5)= (9/5)/BD ⇒ 3*ED= 144/25 ED= 48/25 3*BD= (12/5)*(9/5) BD= 36/25 A(ΔABC)= 3*4/2 A(ΔABC)= 6 square units A(ΔEBD)= BD*ED/2 A(ΔEBD)= (36/25)*(48/25)/2 A(ΔEBD)= 864/625 Ared= A(ΔABC) - A(ΔEBD) Ared= 6- 864/625 Ared= 2886/625 square units b) Second way of solution ▶ for the ΔABC: cos(α)= AB/AC AB= 3 AC= 5 α= arccos(AB/AC) α= arccos(3/5) α= 53,13° β= 90-α β= 36,87° for the ΔABE: sin(α)= EB/AB AB= 3 sin(53,13°)= EB/3 EB= 2,4 length units for the ΔEBD: sin(α)= ED/EB sin(53,13°)= ED/2,4 ED= 1,92 length units cos(α)= BD/EB cos(53,13°)= BD/2,4 BD= 1,44 length units A(ΔABC)= 3*4/2 A(ΔABC)= 6 square units A(ΔEBD)= BD*ED/2 A(ΔEBD)= 1,44*1,92/2 A(ΔEBD)= 1,3824 Ared= A(ΔABC) - A(ΔEBD) Ared= 6- 1,3824 Ared= 4,6176 square units ( = 2886/625)

Shaded area = 4.6176 sq.units

ED=x BD=y Area of triangle CDE+Area of trpezium ABDE=½×3×4 ½x(4-y)+½(x+3)y=12 x(4-y)+y(x+3)=12 4x-xy+xy+3y=12 4x+3y=12 ...(1) y/x=¾ y=¾x ...(2) From(1),(2): x=48/25 y=36/25 Shaded area=6-½xy=6-½(48/25)(36/25)=2886/625

شكرا لكم على المجهودات يمكن استعمال S(ABC)=6 AEB et CAB semblables et AE/AC =3/5 S(AEB)=(3/5)^2 . S(ABC) EDC et ACB semblables et DC/CB =16/25 S(EDC)=(16/25)^2 . S(ABC) S(ADB)+S(EDC)=2886/625

1) 5^2 - 3^2 = BC^2 ; BC^2 = 25 - 9 ; BC^2 = 16 ; BC = 4 2) Area [ABC] = (3 * 4) / 2 ; A = 12 / 2 ; A = 6 3) BE = 12/5 ; BE = 2,4 4) AE^2 = 9 - 5,76 ; AE^2 = 3,24 ; AE = 1,8 5) Area [ABE] = 1,8 * 2,4 / 2 = 2,16 6) EC = 5 - 1,8 = 3,2 7) 3 / 5 = ED / 3,2 ; ED = 1,92 8) 3 / 4 = 1,92 / DC ; DC = 2,56 9) Area [CDE] = (1,92 * 2,56) / 2 = 2,4576 10) Red Region Area = 2,16 + 2,4576 = 4,6176 11) Answer : Red Region Area is equal to 4,6176 Square Units.

answer=40 isit

Let's find the area: . .. ... .... ..... The triangle ABC is obviously a right (3,4,5) triangle. The triangles within ABC are also right triangles and maybe all these triangles are similar. Let's check it: ABC: ∠ABC = 90° ∠BAC = α ∠ACB = β ABE: ∠AEB = 90° ∠BAE = α ∠ABE = β BDE: ∠BDE = 90° ∠DBE = α ∠BED = β CDE: ∠CDE = 90° ∠CED = α ∠DCE = β So indeed all four triangles are similar. Since BE is the height of the triangle ABC according to the base AC, we can calculate the length of BE in the following way: A(ABC) = (1/2)*AB*h(AB) = (1/2)*AC*h(AC) AB*h(AB) = AC*h(AC) AB*BC = AC*BE ⇒ BE = AB*BC/AC = 3*4/5 = 12/5 Since the right triangles ABC and BDE are similar, we can conclude: BD/BE = AB/AC ⇒ BD = BE*AB/AC = (12/5)*3/5 = 36/25 DE/BE = BC/AC ⇒ DE = BE*BC/AC = (12/5)*4/5 = 48/25 Now we are able to calculate the size of the red area: A(red) = A(ABC) − A(BDE) = (1/2)*AB*BC − (1/2)*BD*DE = (1/2)*3*4 − (1/2)*(36/25)*(48/25) = 6 − 864/625 = 3750/625 − 864/625 = 2886/625 = 4.6176 Best regards from Germany