All 3 orange triangles are similar. At 7:40, the area of the largest orange triangle has been found to be 60. For ΔJKP, the ratio of side PK to the corresponding side of the largest triangle, side HQ, is 8/12 = 2/3. So, the bases and heights are in the same ratio and the area is reduced by the square of that ratio: (2/3)² = (4/9) (base and height both reduced by 2/3, so product bh in A = (1/2)bh is reduced by factor (2/3)(2/3) = (2/3)²). Its area is (4/9)(60). Applying the same reasoning to ΔLMN, its sides are reduced by a factor of (4/12) = 1/3 and area by factor (1/3)² = 1/9, making its area (1/9)(60). Add the 3 areas: 60 + (4/9)(60) + (1/9)(60) = (14/9)(60) = (840)/9 = 280/3, as PreMath also found.
@ 5:32 the Cuckoo Clock bird is Cuckoo for Cocoa Puffs because it knows @ 9:25 you are going too use Triangles with same height have areas proportional too their bases. 🙂
I learned a lot from you so I was able to find quick ways and tricks to do this problem! big triangle: 12 × 10 × ½ = 60 medium triangle: 8 × [10 × ⅔] × ½ = 80/3 small triangle : 4 × [10 × ⅓] × ½ = 20/3 Total area = 60 + 80/3 + 20/3 = 180/3 + 80/3 + 20/3 = 280/3 = 93,333❤ Always and always support you from afar teacherr🎉
△HQA ~ △PKJ ~ △NML Since the ratio of the areas of similar shapes is proportional to the square, area △PKJ = (8/12)^2 * △QHA = 4/9 * 60 = 80/3, area △NML = (4/12)^2 * △QHA = 1/9 * 60 = 20/3 therefore △HQA + △PKJ + △NML = 60 + 80/3 + 20/3 = 280/3
Another method to find the area of the larger red triangle for example (ordinary methods are simpler, but it is just to remind this one): In an adapted orthonormal we have A(0,0) H(4,10) and Q(10,16), so VectorAH(4,10) and VectorAQ(10,16) We use a third dimension: VectorAH(4;10;0) and VectorAQ(10;16;0) We calculate the vectorial product of these two vectors, it is W(0;0;120) The area of the triangle is half of the norm of VectorW, so it is 60. (We also can use a determinant 3x3, the area of the triangle is half of the abs of the determinant which 1st column is (0;0;1), second column is (4;10;1) and third column is (10;16;1) I repeat that these methods are not interesting here, but they could be useful another time.
Let's use homotheties to change a little... The homothety of center A and of ratio 1/4 transforms [B,C] into [E,H], so EH = (1/4). BC = 4, and the area of triangle AEH is (1/2). EH.EA = (1/2).(4).(10) = 20.7 So the larger red triangle area is the area of triangle AEQ minus the area of triangle AEH, then it is (1/2).(16).(10) - 20 = 80 - 20 = 60. The homothety of center C and ratio 2/3 transforms the larger red triangle into the medium red triangle, so the area of the medium red triangle is the area of the larger red triangle multiplied by (2/3)^2, so the area of the medium red triangle is 60. (4/9) = 80/3 The homothety of center C and ratio 1/3 transforms the larger red triangle into the smaller red triangle, so the area of the smaller red triangle is the area of the larger red triangle multiplied by (1/3)^2, so the area of the smaller red triangle is 60. (1/9) = 20/3 Finally the area of the three red triangles is 60 + 80/3 + 20/3 = 280/3.
another solution... the tree triangles, Big(B), Medium(M), Small(S) are all simillar triangles ... by calculating the first we can calculate the others using symetry by symetry we get (xb *xh)/2 = x²A , A=Area of any triangle, x=scaling factor B = 60 M = 60 / (12/8)² = 60 * (8²/12²) = 60 * (64 / 144) S = 60 / (12/4)² = 60 * (4²/12²) = 60 * (16 / 144) Total area = 60 + ( 60 * (64 / 144) ) + ( 60 * (16 / 144) ) factoring out 60 give us 60( 144/144 + 64/144 + 16/144 -> Area = 60( (144 + 64 +16) / 144 ) = 60(224/144) [ not simplified ] Note: 60(144/144) = 60 * 1
11:22 Area(EKP)=10*8/2 (height*base)/2, but at 13:37 Area(FMN)=4*10/2 (base*height)/2. But as 10*8 and 8*10 are the same it doesn’t make difference in the result. At 9:03 PJ/EJ=2/1, but at 9:10 PJ=2k and EJ=k (not 2 and 1). But also didn’t seem affect the results.
By similar triangles EH=4, AEH=10*4/2=20 Also AEQ=10*16/2=80 AHQ=80 - 20=60 Triangles LMN, JKP and AHQ are all similar and in the side ratios 1:2:3 Hence Answer = 60*(9/9+4/9+1/9)=60*14/9 =280/3
1/ Let the areas of the small , medium and large orange triangles be S, M and L. Notice that all the orange triangles are similar. NM/QH=CN/CQ=10/30=1/3-----> S/L= sq[1/3] =1/9 Similarly, KP/QH=20/30=2/3----> M/L= 4/9 2/ EH/16=10/40 ---> EH=4 so area of the triangle AHE= 20; therefore L= 60;=, M= 80/3 and S= 20/3 The sum of the orange areas = 20/3 +80/3 + 60=93.33 sq units
The biggest triangle is 80 - 20 = 60 square units. The next smaller one is 4/9 * 60 = 26 2/3 square units. The smallest is 1/9 * 60 = 6 2/3 square units. The total area is 93 1/3 square units, which is 280/3 square units.
Let's do some math: . .. ... .... ..... Not only the orange triangles are similar, but also the triangles CMN, CKP and CHQ. So we can conclude: KP/MN = CP/CN = 2 ⇒ A(JKP) = 2²*A(LMN) HQ/MN = CQ/CN = 3 ⇒ A(AHQ) = 3²*A(LMN) Additionally the triangles CMN and AEH are congruent. The area of these triangles turns out to be: A(CMN) = A(AEH) = (1/2)*AE*EH = (1/2)*10*(16/4) = 20 A(AHQ) = A(AEQ) − A(AEH) = (1/2)*AE*EQ − A(AEH) = (1/2)*10*16 − 20 = 80 − 20 = 60 Now we can calculate the area of the orange triangles: A(orange) = A(LMN) + A(JKP) + A(AHQ) = A(AHQ)*(1/9 + 4/9 + 1) = 60*(14/9) = 280/3 Best regards from Germany
In the big red shaded triangle, you found d the area by taking the test area of the triangle with a height of 16 and base of 10 to get 80. Then you took the area smaller port of that triangle that wasn't shaded, 20 to get the area of the big red triangle of 60. However, if you rotated that triangle 90 degrees just to figure the area, you could have a base of 12 and a height of 10 that would give you 60. It seems to me you could use promotions to get the corresponding dimensions of the other shades triangles to get their areas, too.
Area large orange triangle Large, right angle triangle - small right angle triangle A = 1/2 x 16 x 10 - 1/2 x 4 x 10 A = 60 Ratio of area is not 8/12, but (8/12)^2 = 64/144 Area of medium triangle A = 64/144 x 60 A = 64 x5 /12 A = 80/3 Area of small triangle A = 16/144×60 A = 16 x 5/12 A = 20/3 Total orange area A = 60+80/3+20/3 A = 93.333 square units
Approximate Numeric Calculus Resolution!! 1) Triangle [AHQ] = (10*16)/2 - (4*10)/2 = 160/2 - 40/2 = 80 - 20 = 60 sq un The Area of the Orange Triangles are in a Geometrical Progression: U(n) = NM ; PK ; QH U(n) = 4 ; 8 ; 12 a(1) = 4 a(2) = 4 * 2 = 8 a(3) = 4 * 3 = 12 a(2)/a(1) = 2 (wich means that r^2 = 4) a(3)/a(2) = 1,5 (wich means that r^2 = 2,25 ) As we know that the Area of triangle [AHQ] = 60, the Area of triangle [JKP] must be equal to: [JKP] * 2,25 = 60 [JKP] = 26,7 sq un And triangle LMN must be equal to: [LMN] * 4 = 26,7 [LMN] = 6,675 sq un The sum all the Orange Areas is: 60 + 26,7 + 6,675 = 93,375 sq un Answer: The Area of all 3 Orange Areas is equal to approx. 93,375 sq un.
I calculated the distance of HA, KJ, and ML. Because these are scalene triangles, I used the are formula for s scalene triangle: A= Sq. Root S(S-A)(S-B(S-C). My final answer = 94.15. Am I right or wrong to use the scalene formula for area? Thanks.
By observation, PF bisects AC at K, so as NG bisects CK at M and QE bisects KA at H, and NG, PF, and QE are all parallel, QH = (3/4)QE = 12, PK = PF/2 = 8 = (2/3)QH, and NM = NG/4 = 4 = (1/3)QH. By observation, ∠LNM, ∠JPK, and ∠AQH are all congruent, ∠NML, ∠PKJ, and ∠QHA are all congruent, and ∠MLN, ∠KJP, and ∠HAQ are all congruent, so ∆LNM, ∆JPK, and ∆AQH are all similar. Triangle ∆AQH: A = bh/2 = (12)10/2 = 60 Triangle ∆JPK: A = bh/2 = 8(20/3)/2 = 80/3 Triangle ∆LNM: A = bh/2 = 4(10/3)/2 = 20/3 Total orange area: A = 60 + 80/3 + 20/3 = 280/3 ≈ 93.33
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All 3 orange triangles are similar. At 7:40, the area of the largest orange triangle has been found to be 60. For ΔJKP, the ratio of side PK to the corresponding side of the largest triangle, side HQ, is 8/12 = 2/3. So, the bases and heights are in the same ratio and the area is reduced by the square of that ratio: (2/3)² = (4/9) (base and height both reduced by 2/3, so product bh in A = (1/2)bh is reduced by factor (2/3)(2/3) = (2/3)²). Its area is (4/9)(60). Applying the same reasoning to ΔLMN, its sides are reduced by a factor of (4/12) = 1/3 and area by factor (1/3)² = 1/9, making its area (1/9)(60). Add the 3 areas: 60 + (4/9)(60) + (1/9)(60) = (14/9)(60) = (840)/9 = 280/3, as PreMath also found.
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@ 5:32 the Cuckoo Clock bird is Cuckoo for Cocoa Puffs because it knows @ 9:25 you are going too use Triangles with same height have areas proportional too their bases. 🙂
Excellent!
Thanks ❤️
I learned a lot from you so I was able to find quick ways and tricks to do this problem!
big triangle: 12 × 10 × ½ = 60
medium triangle: 8 × [10 × ⅔] × ½ = 80/3
small triangle : 4 × [10 × ⅓] × ½ = 20/3
Total area = 60 + 80/3 + 20/3
= 180/3 + 80/3 + 20/3
= 280/3
= 93,333❤
Always and always support you from afar teacherr🎉
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Thanks dear ❤️ You are the best 😀
△HQA ~ △PKJ ~ △NML
Since the ratio of the areas of similar shapes is proportional to the square, area △PKJ = (8/12)^2 * △QHA = 4/9 * 60 = 80/3, area △NML = (4/12)^2 * △QHA = 1/9 * 60 = 20/3
therefore △HQA + △PKJ + △NML = 60 + 80/3 + 20/3 = 280/3
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Another method to find the area of the larger red triangle for example (ordinary methods are simpler, but it is just to remind this one):
In an adapted orthonormal we have A(0,0) H(4,10) and Q(10,16), so VectorAH(4,10) and VectorAQ(10,16)
We use a third dimension: VectorAH(4;10;0) and VectorAQ(10;16;0) We calculate the vectorial product of these two vectors, it is W(0;0;120)
The area of the triangle is half of the norm of VectorW, so it is 60.
(We also can use a determinant 3x3, the area of the triangle is half of the abs of the determinant which 1st column is (0;0;1), second column is (4;10;1) and third column is (10;16;1)
I repeat that these methods are not interesting here, but they could be useful another time.
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Beautiful work 👍
Thank you! Cheers!❤️
❤❤ interesting as always
Let's use homotheties to change a little...
The homothety of center A and of ratio 1/4 transforms [B,C] into [E,H], so EH = (1/4). BC = 4, and the area of triangle AEH is (1/2). EH.EA = (1/2).(4).(10) = 20.7
So the larger red triangle area is the area of triangle AEQ minus the area of triangle AEH, then it is (1/2).(16).(10) - 20 = 80 - 20 = 60.
The homothety of center C and ratio 2/3 transforms the larger red triangle into the medium red triangle, so the area of the medium red triangle is the area of the larger red triangle multiplied by (2/3)^2, so the area of the medium red triangle is 60. (4/9) = 80/3
The homothety of center C and ratio 1/3 transforms the larger red triangle into the smaller red triangle, so the area of the smaller red triangle is the area of the larger red triangle multiplied by (1/3)^2, so the area of the smaller red triangle is 60. (1/9) = 20/3
Finally the area of the three red triangles is 60 + 80/3 + 20/3 = 280/3.
Please read 20 and not 20/7 at line 3.
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another solution...
the tree triangles, Big(B), Medium(M), Small(S) are all simillar triangles ...
by calculating the first we can calculate the others using symetry
by symetry we get (xb *xh)/2 = x²A , A=Area of any triangle, x=scaling factor
B = 60
M = 60 / (12/8)² = 60 * (8²/12²) = 60 * (64 / 144)
S = 60 / (12/4)² = 60 * (4²/12²) = 60 * (16 / 144)
Total area = 60 + ( 60 * (64 / 144) ) + ( 60 * (16 / 144) )
factoring out 60 give us 60( 144/144 + 64/144 + 16/144 ->
Area = 60( (144 + 64 +16) / 144 ) = 60(224/144) [ not simplified ]
Note: 60(144/144) = 60 * 1
11:22 Area(EKP)=10*8/2 (height*base)/2, but at 13:37 Area(FMN)=4*10/2 (base*height)/2. But as 10*8 and 8*10 are the same it doesn’t make difference in the result.
At 9:03 PJ/EJ=2/1, but at 9:10 PJ=2k and EJ=k (not 2 and 1).
But also didn’t seem affect the results.
Many methods to get the correct result.
Thanks ❤️
QHA=12＊10/＊1/2=60
PKJ=8＊20/3＊1/2=80/3
NML=4＊10/3＊1/2=20/3
area of Orange triangles :
60+80/3+20/3=280/3
Thanks ❤️
By similar triangles
EH=4, AEH=10*4/2=20
Also
AEQ=10*16/2=80
AHQ=80 - 20=60
Triangles LMN, JKP and AHQ are all similar and in the side ratios 1:2:3
Hence
Answer = 60*(9/9+4/9+1/9)=60*14/9
=280/3
Thanks ❤️
(1 + (8/12)^2 + (4/12)^2)60 =(1 + 4/9 + 1/9)60 = 14/9 *60= 280/3(square unit)
1/ Let the areas of the small , medium and large orange triangles be S, M and L. Notice that all the orange triangles are similar.
NM/QH=CN/CQ=10/30=1/3-----> S/L= sq[1/3] =1/9
Similarly, KP/QH=20/30=2/3----> M/L= 4/9
2/ EH/16=10/40 ---> EH=4 so area of the triangle AHE= 20; therefore L= 60;=, M= 80/3 and S= 20/3
The sum of the orange areas = 20/3 +80/3 + 60=93.33 sq units
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The biggest triangle is 80 - 20 = 60 square units.
The next smaller one is 4/9 * 60 = 26 2/3 square units.
The smallest is 1/9 * 60 = 6 2/3 square units.
The total area is 93 1/3 square units, which is 280/3 square units.
Thanks ❤️
Let's do some math:
.
..
...
....
.....
Not only the orange triangles are similar, but also the triangles CMN, CKP and CHQ. So we can conclude:
KP/MN = CP/CN = 2 ⇒ A(JKP) = 2²*A(LMN)
HQ/MN = CQ/CN = 3 ⇒ A(AHQ) = 3²*A(LMN)
Additionally the triangles CMN and AEH are congruent. The area of these triangles turns out to be:
A(CMN) = A(AEH) = (1/2)*AE*EH = (1/2)*10*(16/4) = 20
A(AHQ) = A(AEQ) − A(AEH) = (1/2)*AE*EQ − A(AEH) = (1/2)*10*16 − 20 = 80 − 20 = 60
Now we can calculate the area of the orange triangles:
A(orange) = A(LMN) + A(JKP) + A(AHQ) = A(AHQ)*(1/9 + 4/9 + 1) = 60*(14/9) = 280/3
Best regards from Germany
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AHQ = 60 60 + 60(8/12)^2 + 60(4/12)^2 = 60(14/9)
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In the big red shaded triangle, you found d the area by taking the test area of the triangle with a height of 16 and base of 10 to get 80. Then you took the area smaller port of that triangle that wasn't shaded, 20 to get the area of the big red triangle of 60. However, if you rotated that triangle 90 degrees just to figure the area, you could have a base of 12 and a height of 10 that would give you 60. It seems to me you could use promotions to get the corresponding dimensions of the other shades triangles to get their areas, too.
Thanks ❤️
Area large orange triangle
Large, right angle triangle - small right angle triangle
A = 1/2 x 16 x 10 - 1/2 x 4 x 10
A = 60
Ratio of area is not 8/12,
but (8/12)^2 = 64/144
Area of medium triangle
A = 64/144 x 60
A = 64 x5 /12
A = 80/3
Area of small triangle
A = 16/144×60
A = 16 x 5/12
A = 20/3
Total orange area
A = 60+80/3+20/3
A = 93.333 square units
Thanks ❤️
Approximate Numeric Calculus Resolution!!
1) Triangle [AHQ] = (10*16)/2 - (4*10)/2 = 160/2 - 40/2 = 80 - 20 = 60 sq un
The Area of the Orange Triangles are in a Geometrical Progression:
U(n) = NM ; PK ; QH
U(n) = 4 ; 8 ; 12
a(1) = 4
a(2) = 4 * 2 = 8
a(3) = 4 * 3 = 12
a(2)/a(1) = 2 (wich means that r^2 = 4)
a(3)/a(2) = 1,5 (wich means that r^2 = 2,25 )
As we know that the Area of triangle [AHQ] = 60, the Area of triangle [JKP] must be equal to:
[JKP] * 2,25 = 60
[JKP] = 26,7 sq un
And triangle LMN must be equal to:
[LMN] * 4 = 26,7
[LMN] = 6,675 sq un
The sum all the Orange Areas is:
60 + 26,7 + 6,675 = 93,375 sq un
Answer:
The Area of all 3 Orange Areas is equal to approx. 93,375 sq un.
I calculated the distance of HA, KJ, and ML. Because these are scalene triangles, I used the are formula for s scalene triangle: A= Sq. Root S(S-A)(S-B(S-C). My final answer = 94.15. Am I right or wrong to use the scalene formula for area? Thanks.
Beautiful work❤❤❤❤
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By observation, PF bisects AC at K, so as NG bisects CK at M and QE bisects KA at H, and NG, PF, and QE are all parallel, QH = (3/4)QE = 12, PK = PF/2 = 8 = (2/3)QH, and NM = NG/4 = 4 = (1/3)QH. By observation, ∠LNM, ∠JPK, and ∠AQH are all congruent, ∠NML, ∠PKJ, and ∠QHA are all congruent, and ∠MLN, ∠KJP, and ∠HAQ are all congruent, so ∆LNM, ∆JPK, and ∆AQH are all similar.
Triangle ∆AQH:
A = bh/2 = (12)10/2 = 60
Triangle ∆JPK:
A = bh/2 = 8(20/3)/2 = 80/3
Triangle ∆LNM:
A = bh/2 = 4(10/3)/2 = 20/3
Total orange area:
A = 60 + 80/3 + 20/3 = 280/3 ≈ 93.33
Thanks ❤️
▲MNL ~ ▲PKJ (K = 1/2). ▲MNL ~ ▲AHQ (K = 1/3). S(MNL) = x.
S(orange) = x + 4x + 9x = 14x. S(AHQ) = 3/8S(AEQD) = 60.
S(AHQ) = 9/14S(orange). S(orange) = 14/9S(AHQ) = 14/9 × 60 = 280/3.
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(16)^2=256 ,(10)^4=10.000 (10.000+256)=10 256 (10 256-360°)= √9
896 √7^√14 3^√32 √7^1√^2^√7 3^√2^√16 √1^√1√1^√1 3^√1√4^√4 3√2^√2√2^2 3^√1^√1√1^2 3^2 (ABCD+2ABCD-3)
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Hello, pro. I'm here to learn this concept.
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60 + 40 + 20 = 120 sq. units.
hi, could you explain me why have you used to solve the medium triangle JKP the height FE (10) because 1/2 8 X 10 solve the area of triangle EFK
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@@PreMath what is the answer of my question
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I solved this question in 6 mins and iam grade 10
For which grade this problems was created
جميل جدا
Très belle
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Just saying that the triangle scale for side similarity but not for area. Silly Me.
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Mais um belíssimo problema de geometria. Obrigado.
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5th view
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1st comment
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