This is an example easier than it looks and this is probably why AA similarity is way more useful than it is given credit for!!!. I better use that for practice!!!
∆DEC and ∆EBC share the same height, so DE/BC = 14/32 = 7/16. DE/BC = AD/AC = 7/16. AD/DC = 7/9. ∆ AED and ∆ DEC share the same height, so area of ∆ AED = 7/9 x area of ∆ DEC = (7/9) x 14 = 98/9.
Sobre el triángulo EBC construimos el rectángulo EBCF de área 2*32=64→ Área CFD=32-14=18. Si decimos que EB=4→ BC=16 ; ED=7 y DF=9→ Razón de semejanza entre AED y CFD, s=7/9→ s²=49/81→ Área AED=s²*18 =882/81 =98/9 =10,88 cm². Gracias y un saludo cordial.
Pause at 1:08. (I) “The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.” x/(x+14+32)=x/(x+46)=(AD/AC)^2. (2) "If two triangles share a height, then the ratio of their areas is equal to the ratio of their bases." Applying this to △EAD and △ECD, we get x/14=AD/CD, or x/(x+14)= AD/AC. Combine these 2 results and solve for x to get 18x=196 and x=98/9.
Solution: Initially, let's dimension the figure: AE = a ED = n EB = b BC = m Blue Triangle A = ½ base × height 32 = ½ m b m = 64/b Yellow Triangle A = ½ base × height 14 = ½ n b n = 28/b The Pink Triangle AED is similar to the Larger Triangle ABC The ratio between the areas of similar triangles is equal to the square of the similarity ratio. In this case, the bases "n" and "m" are similar Area AED = S Area ABC = S + 46 (14 + 32) Area AED/Area ABC = n²/m² S/(S + 46) = (28/b)²/(64/b)² S/(S + 46) = (28/b/64/b)² S/(S + 46) = (28/64)² S/(S + 46) = (7/16)² S/(S + 46) = 49/256 256S = 49S + 2254 207S = 2254 S = 2254/207 (÷23) S = 98/9 cm² ✅ S ~= 10,889 cm² ✅
Extend ED up to F, where CF is parallel to EB. As CF and EB are parallel and FE and BC are parallel, and ∠FEB = ∠EBC = 90°, then CFEB is a rectangle. As EC is the diagonal of CFEB, then the area of CFEB is twice the area of ∆EBC, or 2(32) = 64 cm². As the area of ∆CDE is 14 and the area of ∆EBC is 32, the area of triangle ∆CFD = 64-(32+14) = 64-46 = 18. As ∠CFD = ∠AED = 90° and ∠FDC and ∠EDA are veryical angles and thus congruent, then ∆CFD and ∆AED are similar triangles. As ∆CFD is 18 cm² and ∆CDE is 14 cm², then the ratio of their bases FD to DE is 18/14 = 9/7, as the triangles have the same height. As ∆CFD and ∆AED are similar, the ratio of all their sides are the same, so the ratio of their areas equals the square of the ratio of their sides. A = (7/9)²18 A = 49(18)/81 = 98/9 = 10.8̅ cm²
RUSSIA! Рассмотрим трапецию BCDE, её площадь равна (ВС+DЕ)*BE/2=46. BC=BE=8, DE=a; (8+а)*8/2=46; а=7/2; АЕ=х; исходя из подобия треугольников AED и ABC, составляем пропорцию 7/2х=8/(8+х), откуда находим х=56/9; теперь находим площадь розового треугольника, А=56*7/9*2*2=98/9.
BC =16x DE = 7x AE =7 p AB= 16p BE =( 16-7)p=9p Area of 🔺 CBE = 1/2*BC*BE=1/2 *16x *9p = 32 sq units xp =32/72=4/9 Area of 🔺 ADE = 1/2*7x *7p =49xp/2 =49*4/9*2=98/9 sq units
2.area of EDC = 28 = ED.EB and 2.area of EBC = BC.EB, so by division: ED/BC = 28/64 = 7/16 Triangles ADE and ACB are similar (same angles), so AE/AB = ED/BC = 7/16. Let's note AE = 7.k and AB = 16.k. Then by difference EB = 9.k 2.area of AED = AE.ED = 7.k.ED and 2.area of EDC = ED.EB = ED.9.k. So we have area of AED/ area of EDC = 7/9, and as area of EDC = 14, then we have that area of AED = (7/9).(14) = 98/9.
It’s tricky but fun😅 1/ Label the area of the purple area as A We have 1/2 DExEB= 14 (1) 1/2 BCxEB=32. (2) -> (1)/(2)-> DE/BC=14/32=7/16 The purple triangle and ABC triangle are similar so, DE/BC=AE/AB=7/16 -> A/(A+46) = sq(7/16)=49/256 -> A=46x49/207= 10.89 sq cm😅😅😅
Answer 10.88888 round to 10.89 Different approach Let's label the length of the blue A and the height C Let's label the base of the yellow B and the height C (the same height as the blue) then BC = 28 and AC = 64 Equation 1 B/A = 7/16 (divide BC by AC) Hence, B = 7/16 A (multiply both sides by A). This is the base of the purple Let's label the base of the purple P Since the purple is similar to the large triangle, then (P+C)/A = P/7/16 A (P+ C) = P/7/16 (multiply both sides by A) 7/16 P + 7/16 C = P (cross multiply) 7/16 P + 7/16 C= 16/16 P ( 16/16P = P) 7/16 C = 9/16 P 7/16 * 16/9 * C = P 7/9 C = P This is the height of the purple Hence, the base and height of the purple in terms of A and C are 7/9 C and 7 /16 A Hence, the area of the purple in terms of A and C is 7/9 C * 7/16 A * 1/2 = 49/144 * 1/2 * AC = 49/ 288 AC But recall AC =64 (see equation 1) Hence, the area of the purple is 49/288 * 64 = 3136/288 3136/288 = 10.888888889
Beginning @ 8:30 , For some odd reason the dimensions of stuff change for me standing on the South Pole or the Equator. Just can't trust myself. Sometimes I'm exact but not so precise. Maybe if I did all my measurements on the 45th Parallel North or South everything would be okay. 🙂
RESOLUTION PROPOSAL : 01) DE = b(ase) 02) BC = B(ase) 03) BE = h(eigth) 04) Trapezoid [BCDE] Area = (B + b) * (h/2) = 46 ; h*B + h*b = 92 05) b * h = 28 sq cm 06) B * h = 64 sq cm 07) h = 28/b and h = 64/B 08) 28/b = 64/B ; 28 * B = 64 * b ; 7B = 16b ; b = 7B/16 09) There are many infinite Solutions for this Equation : b = 7B/16 ; within the Variable Domain D = ]1 ; 14[ 10) Prime Factors of 28 = {2 ; 2 ; 7} 11) Prime Factors of 64 = {2 ; 2 ; 2 ; 2 ; 2 ; 2} 12) Then I saw that : 8 * 3,5 = 28. What's the right Solution? 13) Could it be (Possible Solution) BE = BC = 8 cm? And DE = 3,5 cm? I am not sure!! 14) Let's try. 15) 8/(8 + X) = 3,5/X ; 8X = 3,5 * (8 + X) ; 8X = 28 + 3,5X ; 4,5X = 28 ; X = 56/9 16) 2 * Purple Area = 35/10 * 36/9 = 1.260 / 90 = 126 / 9 = 14 17) PA = 14 / 2 ; PA = 7 Square Centimeters.
Let's assume that the area of the triangle AED is x, from which x/(x+14)=AD/AC=ED/BC=14/32, so x=98/9.
This is an example easier than it looks and this is probably why AA similarity is way more useful than it is given credit for!!!. I better use that for practice!!!
∆DEC and ∆EBC share the same height, so
DE/BC = 14/32 = 7/16.
DE/BC = AD/AC = 7/16.
AD/DC = 7/9.
∆ AED and ∆ DEC share the same height, so area of ∆ AED = 7/9 x area of ∆ DEC = (7/9) x 14 = 98/9.
Sobre el triángulo EBC construimos el rectángulo EBCF de área 2*32=64→ Área CFD=32-14=18.
Si decimos que EB=4→ BC=16 ; ED=7 y DF=9→ Razón de semejanza entre AED y CFD, s=7/9→ s²=49/81→ Área AED=s²*18 =882/81 =98/9 =10,88 cm².
Gracias y un saludo cordial.
Pause at 1:08.
(I) “The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.”
x/(x+14+32)=x/(x+46)=(AD/AC)^2.
(2) "If two triangles share a height, then the ratio of their areas is equal to the ratio of their bases." Applying this to △EAD and △ECD, we get
x/14=AD/CD, or x/(x+14)= AD/AC.
Combine these 2 results and solve for x to get 18x=196 and x=98/9.
I spent several hours to solve this tricky puzzle... If I was doing a test, I'd never solve this... Congrats professor! 👍
Solution:
Initially, let's dimension the figure:
AE = a
ED = n
EB = b
BC = m
Blue Triangle
A = ½ base × height
32 = ½ m b
m = 64/b
Yellow Triangle
A = ½ base × height
14 = ½ n b
n = 28/b
The Pink Triangle AED is similar to the Larger Triangle ABC
The ratio between the areas of similar triangles is equal to the square of the
similarity ratio. In this case, the bases "n" and "m" are similar
Area AED = S
Area ABC = S + 46 (14 + 32)
Area AED/Area ABC = n²/m²
S/(S + 46) = (28/b)²/(64/b)²
S/(S + 46) = (28/b/64/b)²
S/(S + 46) = (28/64)²
S/(S + 46) = (7/16)²
S/(S + 46) = 49/256
256S = 49S + 2254
207S = 2254
S = 2254/207 (÷23)
S = 98/9 cm² ✅
S ~= 10,889 cm² ✅
Please elaborate about the hight of triangle CDE
Extend ED up to F, where CF is parallel to EB. As CF and EB are parallel and FE and BC are parallel, and ∠FEB = ∠EBC = 90°, then CFEB is a rectangle. As EC is the diagonal of CFEB, then the area of CFEB is twice the area of ∆EBC, or 2(32) = 64 cm².
As the area of ∆CDE is 14 and the area of ∆EBC is 32, the area of triangle ∆CFD = 64-(32+14) = 64-46 = 18.
As ∠CFD = ∠AED = 90° and ∠FDC and ∠EDA are veryical angles and thus congruent, then ∆CFD and ∆AED are similar triangles. As ∆CFD is 18 cm² and ∆CDE is 14 cm², then the ratio of their bases FD to DE is 18/14 = 9/7, as the triangles have the same height.
As ∆CFD and ∆AED are similar, the ratio of all their sides are the same, so the ratio of their areas equals the square of the ratio of their sides.
A = (7/9)²18
A = 49(18)/81 = 98/9 = 10.8̅ cm²
❤❤
RUSSIA! Рассмотрим трапецию BCDE, её площадь равна (ВС+DЕ)*BE/2=46. BC=BE=8, DE=a; (8+а)*8/2=46; а=7/2;
АЕ=х; исходя из подобия треугольников AED и ABC, составляем пропорцию 7/2х=8/(8+х), откуда находим х=56/9; теперь находим площадь розового треугольника, А=56*7/9*2*2=98/9.
Very nice
Bom dia Mestre
Essa eu acertei, fiquei feliz porque estou aprendendo Geometria com o Sr
Grato
Excelente!
Fico feliz em ouvir isso!
Obrigado pelo feedback ❤️
@@PreMath Muito obrigado pela instrução
O Sr é um Homem de Bom ❤️
xy = 28
xz = 64
y/z = 28/64 = 7/16
pink triangle is (7/16)^2 = 49/256 of the large triangle
(14 + 32) cm^2 is 1 - (7/16)^2 = 207/256 of the large triangle
pink triangle = 46 cm^2 (49/207) = (2254/207) cm^2 = (10 + 184/207) cm^2
DE*EB/2=14 CB*EB/2=32 DE=28/EB CB=64/EB DE : CB = 7 : 16
⊿DEA∞⊿CBA ⊿DEA=7*7*s=49s ⊿CBA=16*16*s=256s
DECB= 256s-49s=207s=46(cm²) s=2/9
⊿DEA=Purple area=49s=49*2/9=98/9(cm²)
BC =16x
DE = 7x
AE =7 p
AB= 16p
BE =( 16-7)p=9p
Area of 🔺 CBE =
1/2*BC*BE=1/2 *16x *9p
= 32 sq units
xp =32/72=4/9
Area of 🔺 ADE = 1/2*7x *7p =49xp/2 =49*4/9*2=98/9 sq units
Nice work, please how did you arrive at values(show formulas), BC? DE? AB? What does x and p mean in your solution?
2.area of EDC = 28 = ED.EB and 2.area of EBC = BC.EB, so by division: ED/BC = 28/64 = 7/16
Triangles ADE and ACB are similar (same angles), so AE/AB = ED/BC = 7/16. Let's note AE = 7.k and AB = 16.k. Then by difference EB = 9.k
2.area of AED = AE.ED = 7.k.ED and 2.area of EDC = ED.EB = ED.9.k. So we have area of AED/ area of EDC = 7/9, and as area of EDC = 14, then we have that area of AED = (7/9).(14) = 98/9.
6:00, why h square over k square? thanks.
Because the ratio of the areas of two similar triangles equals the square of the similarity ratio.
Please can you prove this theorem you just stated here? I was about asking the same question that he asked. Let's not assert without proofs, please.
Thanks for stating the theorem. You are absolutely right, I have seen the proof of the theorem. Thank you so much for letting me know that
It’s tricky but fun😅
1/ Label the area of the purple area as A
We have
1/2 DExEB= 14 (1)
1/2 BCxEB=32. (2)
-> (1)/(2)-> DE/BC=14/32=7/16
The purple triangle and ABC triangle are similar
so, DE/BC=AE/AB=7/16
-> A/(A+46) = sq(7/16)=49/256
-> A=46x49/207= 10.89 sq cm😅😅😅
x/14=(x+14)/32 => x =98/9
Answer 10.88888 round to 10.89
Different approach
Let's label the length of the blue A and the height C
Let's label the base of the yellow B and the height C (the same height as the blue)
then BC = 28
and AC = 64 Equation 1
B/A = 7/16 (divide BC by AC)
Hence, B = 7/16 A (multiply both sides by A). This is the base of the purple
Let's label the base of the purple P
Since the purple is similar to the large triangle, then
(P+C)/A = P/7/16 A
(P+ C) = P/7/16 (multiply both sides by A)
7/16 P + 7/16 C = P (cross multiply)
7/16 P + 7/16 C= 16/16 P ( 16/16P = P)
7/16 C = 9/16 P
7/16 * 16/9 * C = P
7/9 C = P This is the height of the purple
Hence, the base and height of the purple in terms of A and C
are 7/9 C and 7 /16 A
Hence, the area of the purple in terms of A and C is
7/9 C * 7/16 A * 1/2 = 49/144 * 1/2 * AC = 49/ 288 AC
But recall AC =64 (see equation 1)
Hence, the area of the purple is 49/288 * 64 = 3136/288
3136/288 = 10.888888889
Can you teach..... hacsigun into the circle minus or circle into the hacsigun
First comment from Yorkshire UK👍
Triangle ABC is not defined uniquely.
Let EB = BC = 8.
Comparing blue & yellow triangle areas.
8 / 32 = DE / 14. ( same heights EB).
DE = 3.5.
Similar triangles ABC & AED.
8 / (AE + 8) = 3.5 / AE.
Cross multiplying.
8AE = 3.5AE + 28.
4.5 AE = 28.
AE = 28 / 4.5.
Area of purple triangle.
1/2 x (28 / 4.5) x 3.5.
14 x 3.5 / 4.5.
10.89
Professor,I don't understand ∆CDE height because k is not perpendicular.
10 100%raite full area 8×14÷2=56 )(3.4285714286×6÷2=10
Beginning @ 8:30 , For some odd reason the dimensions of stuff change for me standing on the South Pole or the Equator. Just can't trust myself. Sometimes I'm exact but not so precise. Maybe if I did all my measurements on the 45th Parallel North or South everything would be okay. 🙂
Aunque más frío, yo me siento más aplomado en el círculo polar ártico.
😊😊😊
98/9
RESOLUTION PROPOSAL :
01) DE = b(ase)
02) BC = B(ase)
03) BE = h(eigth)
04) Trapezoid [BCDE] Area = (B + b) * (h/2) = 46 ; h*B + h*b = 92
05) b * h = 28 sq cm
06) B * h = 64 sq cm
07) h = 28/b and h = 64/B
08) 28/b = 64/B ; 28 * B = 64 * b ; 7B = 16b ; b = 7B/16
09) There are many infinite Solutions for this Equation : b = 7B/16 ; within the Variable Domain D = ]1 ; 14[
10) Prime Factors of 28 = {2 ; 2 ; 7}
11) Prime Factors of 64 = {2 ; 2 ; 2 ; 2 ; 2 ; 2}
12) Then I saw that : 8 * 3,5 = 28. What's the right Solution?
13) Could it be (Possible Solution) BE = BC = 8 cm? And DE = 3,5 cm? I am not sure!!
14) Let's try.
15) 8/(8 + X) = 3,5/X ; 8X = 3,5 * (8 + X) ; 8X = 28 + 3,5X ; 4,5X = 28 ; X = 56/9
16) 2 * Purple Area = 35/10 * 36/9 = 1.260 / 90 = 126 / 9 = 14
17) PA = 14 / 2 ; PA = 7 Square Centimeters.
My way of solution ▶
A(ΔEBC)= 32 cm²
A(ΔECD)= 14 cm²
[EB]= a
[BC]= b
[DE]= c
[AE]= d
A(ΔEBC)= 32 cm²
[EB]*[BC]/2
a*b/2= 32
ab= 64
A(ΔECD)= 14 cm²
[DE]*[EB]/2
c*a/2= 14
ca= 28
Let's divide ab to da :
ab/ca= 64/28
b/c= 16/7
b) ΔAED ~ ΔABC
[AE]/[AB]= [DE]/[BC]
d/(d+a)= c/b
c/b= 7/16
⇒
d/(d+a)= 7/16
16d= 7d+7a
9d= 7a
d= 7a/9
c) [AB]= a+d
d= 7a/9
[AB]= 7a/9 + a
[AB]= 16a/9
ab= 64
(16a/9)*b= s
⇒
s= 64*(16/9)ab/ab
s= 1024/9
A(ΔABC)= 1024/9/2
A(ΔABC)= 1024/18 cm²
d) Apurple= A(ΔABC) - A(ΔEBC) - A(ΔECD)
A(ΔEBC)= 32 cm²
A(ΔECD)= 14 cm²
⇒
Apurple= 1024/18 - 32 - 14
Apurple= 98/9 cm² ✅
10.88888
Do you to mee in the area math wanli 20%
Pagrium