As usual, a somewhat alternative approach: The big triangle is integer Pythagorean, 5-12-13. Because everything is perpendicular, all three smaller triangles are similar to each other and to the big one by angle-angle-angle. Using your notation, consider triangles ADE and EBF and compare each to the big triangle ABC: From triangle ADE, L/x = 13/12; cross-multiply and 13x = 12L; so x = 12L/13. From triangle EBF, (12 -- x)/L = 13/5; cross-multiply and we get 60 -- 5x = 13L; so x = (60 -- 13L)/5. We now have two independent expressions for x in terms of L. equating them, we get: x = 12L/13 = (60 -- 13L)/5; cross-multiply to get 60L = (13)(60 -- 13L) = 780 -- 169L. Solve for L: 60L + 169L = 229L = 780; so L = 780/229, just as you determined; and the area of the square is (780/229)^2. We could use any two of the smaller triangles and it would work out the same way. (I almost said "similarly" but wouldn't want to make a bad pun.) Cheers. 🤠
Length BC = the sum of lengths CP, PF and FB. By Pythagorous BC = 13. Using similar triangles CP = 5L/12 and FB = 12L/5. Therefore 5L/12 + L + 12L/5 = 13. Using a common denominator of 60 on the LHS expression, (60L+25L+144L)/60 =13 which simplifies to L = 780/229.
Perhaps you have figured out these kind of problems "in your head?" I decided to focus on the [5] side of the triangle. first though, calculating the hypotenuse of [13] was helpful. I used 𝒔 as the side of the yellow square, "𝒔" for "Square" [1.1] △CPD … DP is on the '12' leg. [1.2] △ADE … DE is on the '13' leg. Put 'em together: [2.1] 13(𝒔/12) ⊕ 5(𝒔/13) = 5 The rationale is "divide by the normalized side-length, then multiply by the side-length that we need in proportion". Find a common denominator … [3.1] (13 / 13)13(𝒔/12) + (12 / 12)5(𝒔/13) = ((12 × 13) / (12 × 13)) × 5 Then eliminate the denominator [4.1] 169𝒔 + 60𝒔 = 780 [4.2] 229𝒔 = 780 [4.3] 𝒔 = 780 ÷ 229 and the yellow square [5.1] 𝒔² = 3.406² = 11.602 Ta, da!
C² = 5²+12² C = 13 cm L / cos α + L sin α = 5 13 L /12 + 5 L 13 = 5 L (13/12 + 5/13) = 5 1,4679 L = 5 L = 3,406 cm Area = L² Area = 11.6 cm² ( Solved √ )
Once known that ABC is the Pythagorean triplet (5,12,13) we can easily show that all the right triangles inside ABC are similar to ABC, so I wrote this system: 12x+13z=12 13y+5x=5 5y+12y+12z=13 (x for DAE, y for DCP and z for EFB) and found y = 845/2977 = 65/229 = 0,2838... so side of square = 12y = 12*0,28= 3,406... area = 11,60...
If you start with 13x = 12y = 5z (i.e. the side of the square), 5x + 13y = 5 or 12x + 13z = 12 is enough to solve for x, y or z. For example: 13x = 12y ⇒ y = 13x / 12 so 5x + 13y = 5 becomes 5x + 13 (13x / 12) = 5 and therefore x = 5 / (5 + 13² / 12)
The numbers made the final calculations a little unwieldy to perform with pen and paper alone (at least for me) so I stopped after finding L = 780/229 and convincing myself that this fraction could not be simplified. An excellent problem apart from that. Thanks.
Darn, almost had it by calculating in my head. Made one error though. 😞 I only used one variable, the one for L. AE = 12/13*L and EB = 13/5*L So L = 12 / (12/13 + 13/5) = 12 * 5 * 13 / ( 12*5 + 13*13) here I made the error and used 12*13=156 instead of 13*13, so 216 instead of 229. Too bad... So L^2 = (780 / 229)^2 CD + DA = 5 would have been a similar route. 13/12*L + 5/13*L = 5 with again L = 5*12*13/(13*13+5*12).
Solution: BC = √(5²+12²) = 13. There are 4 similar right triangles, ABC, DPC, AED and EBF. The side x of the square is in triangle AED hypotenuse, in triangle DPC long leg and in triangle EBF short leg. AD = a. Similarity triangle DPC to triangle ABC: (1) x/(5-a) = 12/13 Similarity triangle AED to triangle ABC: (2) a/x = 5/13 |*x ⟹ (2a) a = 5/13*x |in (1) ⟹ (1a) x/(5-5/13*x) = 12/13 |*(5-5/13*x) ⟹ (1b) x = 12/13*(5-5/13*x) = 60/13-60/169*x |+60/169*x ⟹ (1c) x+60/169*x = 229/169*x = 60/13 |*169/229 ⟹ (1d) x = 60/13*169/229 = 60*13/229 = 780/229 ⟹ area of the yellow square = x² = (780/229)² ≈ 11,6016
Answer 11.6016 or 11.6 I did something different to see if it works First, all four triangles are similar since at least two angles of each are congruent Let the side of the square = 780 b. I used 780 b (13*12*5= 780) in order to avoid decimals and to make it easier to calculate the other sides since these triangles are similar Hence area of the square = 780b * 780 b Hence CP = 325b (5/12 of 780) since CP is the '5' of triangle CPB CD =845 b (13/12 of 780) Recall I let the length of the square = 780 b DA= 300 b (5/13 of 780) AE = 720 b (12/13 *780 ) EB= 2028 b (13/5 * 780 ) as EF is the '5' here and BE is the '13' FB =1872 b (12/5 * 780) EF again is the '5' here but is the "12" Hence AC=CD+ DA = 1145b AB= AE + EB = 2748 b So the height and base of the square are 1145 b and 2748 b Hence the area of the triangle in terms of b =3,146,460/2 = 1,573,230 b^2 And the area of the square in terms of b = 608,400 b^2 (780b * 780b) Hence the square as a percentage of the triangle = 608, 400/1,573, 230 = 0.38672 or 38.672% Since the area of the triangle = 30 [ 5 * 12 * 1/2], then The area of the square is 0.38672 * 30 = 11.6016 Answer The square is approximately 38.67% of the main triangle.
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As usual, a somewhat alternative approach:
The big triangle is integer Pythagorean, 5-12-13. Because everything is perpendicular, all three smaller triangles are similar to each other and to the big one by angle-angle-angle. Using your notation, consider triangles ADE and EBF and compare each to the big triangle ABC:
From triangle ADE, L/x = 13/12; cross-multiply and 13x = 12L; so x = 12L/13.
From triangle EBF, (12 -- x)/L = 13/5; cross-multiply and we get 60 -- 5x = 13L; so x = (60 -- 13L)/5.
We now have two independent expressions for x in terms of L. equating them, we get:
x = 12L/13 = (60 -- 13L)/5; cross-multiply to get 60L = (13)(60 -- 13L) = 780 -- 169L. Solve for L:
60L + 169L = 229L = 780;
so L = 780/229, just as you determined; and the area of the square is (780/229)^2.
We could use any two of the smaller triangles and it would work out the same way. (I almost said "similarly" but wouldn't want to make a bad pun.)
Cheers. 🤠
Thanks for video.Good luck sir!!!!!!!!!!!!!
Nice solution:)
Length BC = the sum of lengths CP, PF and FB. By Pythagorous BC = 13. Using similar triangles CP = 5L/12 and FB = 12L/5. Therefore 5L/12 + L + 12L/5 = 13. Using a common denominator of 60 on the LHS expression, (60L+25L+144L)/60 =13 which simplifies to L = 780/229.
Perhaps you have figured out these kind of problems "in your head?" I decided to focus on the [5] side of the triangle. first though, calculating the hypotenuse of [13] was helpful. I used 𝒔 as the side of the yellow square, "𝒔" for "Square"
[1.1] △CPD … DP is on the '12' leg.
[1.2] △ADE … DE is on the '13' leg. Put 'em together:
[2.1] 13(𝒔/12) ⊕ 5(𝒔/13) = 5
The rationale is "divide by the normalized side-length, then multiply by the side-length that we need in proportion".
Find a common denominator …
[3.1] (13 / 13)13(𝒔/12) + (12 / 12)5(𝒔/13) = ((12 × 13) / (12 × 13)) × 5
Then eliminate the denominator
[4.1] 169𝒔 + 60𝒔 = 780
[4.2] 229𝒔 = 780
[4.3] 𝒔 = 780 ÷ 229
and the yellow square
[5.1] 𝒔² = 3.406² = 11.602
Ta, da!
Generalized: s = abc / (ab + c²) where s = DE (side of the square), a = CA, b = AB and c = BC
5x/12+x+12x/5=13
(144x+25x+60x)/60=13
229x=780
Area=11.6 sq units
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Very interesting. The various equations to get the sides are really a version of trigonometry.
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C² = 5²+12²
C = 13 cm
L / cos α + L sin α = 5
13 L /12 + 5 L 13 = 5
L (13/12 + 5/13) = 5
1,4679 L = 5
L = 3,406 cm
Area = L²
Area = 11.6 cm² ( Solved √ )
Once known that ABC is the Pythagorean triplet (5,12,13) we can easily show that all the right triangles inside ABC are similar to ABC, so I wrote this system:
12x+13z=12
13y+5x=5
5y+12y+12z=13
(x for DAE, y for DCP and z for EFB)
and found y = 845/2977 = 65/229 = 0,2838... so
side of square = 12y = 12*0,28= 3,406...
area = 11,60...
If you start with 13x = 12y = 5z (i.e. the side of the square), 5x + 13y = 5 or 12x + 13z = 12 is enough to solve for x, y or z.
For example: 13x = 12y ⇒ y = 13x / 12 so 5x + 13y = 5 becomes 5x + 13 (13x / 12) = 5 and therefore x = 5 / (5 + 13² / 12)
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Yay! I solved the problem.
The numbers made the final calculations a little unwieldy to perform with pen and paper alone (at least for me) so I stopped after finding L = 780/229 and convincing myself that this fraction could not be simplified. An excellent problem apart from that. Thanks.
Darn, almost had it by calculating in my head. Made one error though. 😞
I only used one variable, the one for L.
AE = 12/13*L and EB = 13/5*L
So L = 12 / (12/13 + 13/5) = 12 * 5 * 13 / ( 12*5 + 13*13)
here I made the error and used 12*13=156 instead of 13*13, so 216 instead of 229. Too bad...
So L^2 = (780 / 229)^2
CD + DA = 5 would have been a similar route. 13/12*L + 5/13*L = 5 with again L = 5*12*13/(13*13+5*12).
Pythagorean theorem (right triangle ABC):
BC² = AB² + AC² = 12² + 5² = 144 + 25 = 169
⇒ BC = 13
Similar triangles:
ADE: AD:AE:DE = 5:12:13
CDP: CP:DP:CD = 5:12:13
AD/DE = 5/13
DP/CD = 12/13
DE = DP
⇒ (AD/DE)*(DP/CD) = AD/CD = (5/13)*(12/13) = 60/169
AD + CD = AC = 5
AD = (60/169)CD
(60/169)CD + CD = 5
(229/169)CD = 5
CD = 5*169/229
DP = (12/13)*CD = (12/13)*5*169/229 = 780/229
A(square) = DP² = (780/229)²
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Ab + dc
Solution:
BC = √(5²+12²) = 13.
There are 4 similar right triangles, ABC, DPC, AED and EBF. The side x of the square is in triangle AED hypotenuse, in triangle DPC long leg and in triangle EBF short leg.
AD = a.
Similarity triangle DPC to triangle ABC:
(1) x/(5-a) = 12/13
Similarity triangle AED to triangle ABC:
(2) a/x = 5/13 |*x ⟹
(2a) a = 5/13*x |in (1) ⟹
(1a) x/(5-5/13*x) = 12/13 |*(5-5/13*x) ⟹
(1b) x = 12/13*(5-5/13*x) = 60/13-60/169*x |+60/169*x ⟹
(1c) x+60/169*x = 229/169*x = 60/13 |*169/229 ⟹
(1d) x = 60/13*169/229 = 60*13/229 = 780/229 ⟹
area of the yellow square = x² = (780/229)² ≈ 11,6016
Answer 11.6016 or 11.6
I did something different to see if it works
First, all four triangles are similar since at least two angles of each are congruent
Let the side of the square = 780 b. I used 780 b (13*12*5= 780) in order to avoid decimals and to make
it easier to calculate the other sides since these triangles are similar
Hence area of the square = 780b * 780 b
Hence CP = 325b (5/12 of 780) since CP is the '5' of triangle CPB
CD =845 b (13/12 of 780) Recall I let the length of the square = 780 b
DA= 300 b (5/13 of 780)
AE = 720 b (12/13 *780 )
EB= 2028 b (13/5 * 780 ) as EF is the '5' here and BE is the '13'
FB =1872 b (12/5 * 780) EF again is the '5' here but is the "12"
Hence AC=CD+ DA = 1145b
AB= AE + EB = 2748 b
So the height and base of the square are 1145 b and 2748 b
Hence the area of the triangle in terms of b =3,146,460/2 = 1,573,230 b^2
And the area of the square in terms of b = 608,400 b^2 (780b * 780b)
Hence the square as a percentage of the triangle = 608, 400/1,573, 230 = 0.38672 or 38.672%
Since the area of the triangle = 30 [ 5 * 12 * 1/2], then
The area of the square is 0.38672 * 30 = 11.6016 Answer
The square is approximately 38.67% of the main triangle.
11.6
11.56 I say