Let x = side of square. Once we determine the height of ABC to be 12, then we know that CP = 12-x. ABC is similar to CDF. So there is the ratio 14/x = 12/(12-x). Solve for x and get 6.46. So x squared is 41.75.
Alternative to Heron's formula: As done later in the video, drop a perpendicular from C to AB and label the intersection as point E. Let CE have length h and AE have length b, then BE has length 14 - b. For ΔACE, b² + h² = (15)² = 225. For ΔBCE, h² + (14 - b)² = (13)² = 169. Expanding, h² + 196 - 28b + b² = 169, Replace b² + h² with 225: 225 + 196 - 28b = 169. Solve for b and find b = 9. Substitute b = 9 into b² + h² = 225 and find h = 12 or CE = 12. Proceed from 5:05 in the video. While k is a perfectly valid designation for a variable, many of us, especially electrical engineers, associate it with 1000, so 12k looks like 12,000. Using a different letter would avoid this confusion. However, we really don't need variable k. Let the sides of the square have length s. ΔDFC and ΔABC are similar. Corresponding sides are in the same ratio and also in the same ratio as corresponding heights. Sides DF and AB are corresponding and heights CP and CE also corresponding, so DF/AB = CP/CE, s/14 = (12 - s)/12, 12s = (14)(12 - s), 12s = 168 - 14s, 26s = 168, s = 84/13. The square's area = s² = (84/13)² = (7056)/(169), as PreMath also found.
Use Heron's formula to solve for area of triangle ABC. Determine height of triangle ABC, which equals 12. Show by angle-angle similarity that triangle CDF is similar to triangle ABC. Make a proportion of height over base: 12/14 = (12-x)/x, which when solved for x is 84/13. x is the side length of the square, so x^2 = 84^2/13^2 = 7056/169
Equation for AC y=12x/9, convert to x1=9y/12, equation for BC y=(-12x+168)/5 convert to x2=-5y/12 + 14. Height of square = y, width of square =x2-x1, set equal, solve for y. Area =y²
If you let side of square = x then using similar triangles CPF and CEB knowing that CP=12-x you get PF= (60-5x)/12. Then using similar triangles CDP and CAE you get DP=(36-3x)/4. Adding DP and PF and putting equal to x you get x=84/13 etc.
We already know that 13-14-15 is a special triangle that can be divided in two Pitágorean triplet right triangles 5-12-13 and 9-12-15 Therefore, height of triangle is h=12 cm Similarity of triangles: s / (h-s) = b / h h.s = b (h-s) 12.s = 14 (12-s) = 168 - 14s 26 s = 168 --> s = 6,4615cm s² = 41,75 cm² ( Solved √ )
At 4:41, Let the purple square be DFGH, All lengths are measured in units. DF=FG=GH=GD=x with EB=y and AE= 14-y. CB=13,BA=14 and AC=15. By Pythagoras, CE^2=EB^2+CE^2 CE^2=CB^2-EB^2 CE^2= 13^2-y^2=169-y^2...........................(1) AC^2=AE^2+CE^2 CE^2=AC^2-AE^2 =15^2-(14-y)^2=225-196+28y-y^2=29+28y-y^2......................(2) From (1) and (2), 169-y^2=29+28y-y^2 28y=140 y=5 units. CE^2=CB^2-y^2 =13^2-5^2=(13+5)(13-5)=18*8=144 CE=12 units,PE=x units and CP=12-x units. Consider similar triangles CDF and CAB, CP/DF=CE/AB (12-x)/x= 12/14 =6/7 7(12-x)=6x 84-7x=6x 13x=84 x= 84/13 units Area of pink square=x^2= (84/13)^2= 7056/169 =41.75 square units approximately. There are many ways to skin this particular cat professor. Thanks for the puzzle.
We use an orthonormal center A and first axis (AB). The equation of the circle center A and radius 15 is x^2 + y^2 -225 = 0 and the equation of the circle center B and radius 13 is (x -14)^2 + y^2 -169 = 0 or x^2 + y^2 -28.x + 27 = 0. At the intersection we have -28.x +27 = -225 and that gives x = 9, then y^2 = 225 -9^2 = 144, and y = 12. So we now have point C: C(9; 12). The equation of (AC) is (x).(12) - (y).(9) = 0 or y = (4/3).x. Then VectorBC(-5; 12) and the equation of (BC) is (x -14).(12) - (y).(-5) = 0 or 12.x +5.y -168 = 0 Point D(a; (4/3).a). Point F is on (BC) and has the same ordinate (4/3).a, so its abscissa x verifies 12.x = -5.(4/3).a +168, so x = (-5/9).a +14 and then we have F((-5/9).a +14; (4/3).a). Now DF = (-5/9).a +14 -a = (-14/9).a + 14, and this distance is also the ordinate of D: (4/3).a So we have (-14/9).a + 14 = (4/3).a and that gives a = 63/13, and (4/3).a = 84/13 which is the side length of the square. Finally the area of the square is (84/13)^2 = 7056/169.
S=(13+14+15)/2=16 Area of the triangle ABC=√21(21-13)(21-14)(21-15)=84 Draw CE is attitude 1/2(CE)(14)=84 CE=168/14=12 Let DF=x ∆CDF~∆CAB x/14=(12-x)/12 So x=84/13 So Purple Square Area=(84/13)^2=7056/169 square units=41.75 square units.❤❤❤
Square's sides are a. Triangle's area is sqrt((21)(6)(7)(8)) = sqrt(7056) = 84. Triangle height is 12 due to 168/14. Therefore two right triangles of 5, 12, 13 and 9, 12, 15. 5/12=(5-(5/14)a)/a 5a=12(5-(5/14)a) 5a=(60-(30/7)a a=(12-(30/35)a a+(6/7)a=12 7a+6a=84 13a=84 a=84/13 a^2=7056/169
Solution: First calculate the area using Heron's Formula. Knowing the area and the value of the base, calculate the total height. Apply the Pythagorean Theorem, calculate the two measurements of the base which are 9 and 5, from left to right. After that, by similarity of triangles (proportion) calculate the constants of proportion of the triangles. It will be 12/5 on one side and 12/9 on the other side of the base. We arrive at the value of 14k for the side of the square. Knowing that the total height is 26k (14k + 12k) and that the height is 12, I calculate the value of k = 6/13. Finally I calculate the value of the side of the square (14k = 14 × 6/13 = 84/13) and then the Area of the Square = (84/13)² = 41.751 units of area.
41.75 Drop a perpendicular line from the vertex to form two right triangle 9-12-15 (3-4-5 triplet scaled up by 2) and a 5-12- 13 (Pythagorean triplets) Let's label the length of the square 'x' Notice that the top triangle is similar to the largest triangle in terms of x, its height is 12/14 x, since in the 13-14-15 largest triangle , the height is 12 Hence, the height of the triangle in terms of x = x + 12/14x = 26/14 x Hence, 26/ 14 x= 12 x = 14/26 * 12 x = 6.4615 4 x^2 = 41.75147 Answer
A method without Heron's formula 1) Let area of square is a ^2 2) sum of area of the two triangles beside the square on base of14 units =1/2*(14-a) *a sq units =(7a - a^2/2) sq units 3) I) Now Pythagorean triplet says when we draw height on 14 unit base One rt angle will be (9,12,15) ref [3*(3,4,5)] II) Then the we see 13^2 =12^2 +5^2 Hence the other rt triangle will be (5,12,13) III) Then height of the big triangle on base of 14 units will be 12 units if) Height of triangle above the square is (12-a) units v) Area of the triangle above the square =1/2*a*(12-a)sq units = (6a -a^2/2)sq units 4)Then a^2 + 7a - a^2/2 +6a -a^2/2=1/2*12*14 a=84/13 units 5) a^2=(84/13)^2=41.75 sq units (approx)
RUSSIA! Провёл из угла АВС биссектрису к стороне АС, которая яволяется и высотой BD.Площадь треугольника АВС нашёл по формуле Герона Александрийского, она равна 84; далее нашёл высоту треугольника по основанию 14, h=12; после этого из подобия треугольника с основанием х и треугольника с основанием 14, составил пропорцию, 12/14=(12-а)/а, откуда 26а=168; а=6,461538461...; площадь пурпурного квадрата равна 41,7514792898... округлять не буду.
Let's find the area: . .. ... .... ..... First of all we calculate the area of the triangle ABC according to the formula of Heron: s = (AB + AC + BC)/2 = (14 + 15 + 13)/2 = 42/2 = 21 A = √[s*(s − AB)*(s − AC)*(s − BC)] = √[21*(21 − 13)*(21 − 14)*(21 − 15)] = √(3*7*2³*7*2*3) = 3*7*2² = 84 May E and G be the lower left and the lower right corner of the square, respectively. Now we add H on AB such that CH is the height of the triangle ABC according to its base AB: A = (1/2)*AB*h(AB) = (1/2)*AB*CH ⇒ CH = 2*A/AB = 2*84/14 = 12 The triangles ACH and BCH are right triangles, so we can apply the Pythagorean theorem: AC² = AH² + CH² ⇒ AH² = AC² − CH² = 15² − 12² = 225 − 144 = 81 ⇒ AH = √81 = 9 BC² = BH² + CH² ⇒ BH² = BC² − CH² = 13² − 12² = 169 − 144 = 25 ⇒ BH = √25 = 5 May s be the side length of the square. Since the triangles BFG and BCH are similar and the triangles ADE and ACH are similar as well, we can conclude: BG/BH = FG/CH BG/5 = s/12 ⇒ BG = 5*s/12 AE/AH = DE/CH AE/9 = s/12 ⇒ AE = 9*s/12 AB = AE + EG + BG 14 = 9*s/12 + s + 5*s/12 = s*(9/12 + 12/12 + 5/12) = 26*s/12 = 13*s/6 ⇒ s = 6*14/13 = 84/13 Now we are able to calculate the area of the purple square: A(purple) = s² = (84/13)² = 7056/169 ≈ 41.75 Best regards from Germany
*Solution:* The triangle ABC ~ DCF. Now, Consider DF=x. Hence, CD/15 = x/14 → *CD =15x/14.* Hence, AD= AC - CD→ *AD =15 - 15x/14* Law of cosines in ∆ABC with respect to angle A: 13²=15² + 14² - 2×15×14 cos A 169= 421 - 420 cos A Cos A = 252/420 = 3/5. Hence, sin² A= 1 - cos² A sin² A = 1 - 9/25= 16/25 sin A = 4/5.On the other hand, sin A = x/AD →AD = 5x/4 and how AD=15 - 15x/14, we have: 15 - 15x/14 = 5x/4 ÷(5) 3 - 3x/14 = x/4 x/4 + 3x/14 = 3 ×(28) 7x + 6x = 84 → x= 84/13, Therefore, the area of the square is: *(84/13)² square units* or *≈41.75 square units*
I think that this problem is an application of the « Pythagorean triangles », i. e. the right triangles whose sides are all integers. If we note (a, b, c) the triplet of the sides in decreasing order, the (5, 4, 3) triangle is very well known (and of course its multiples : (10, 8, 6) and (15, 12, 9) and etc. An other well-known Pythagorean triangle is (13, 12, 5) and its multiples… And an interesting but easy theorem says that there are infinitely many Pythagorean triangles, and that any integer belongs to at least one of them. Back to the problem, considering the (15, 12, 9) and (13, 12, 5) triangles, we can « glue » them side by side using the side 12, and rebuild the (15, 14, 13) triangle (which is NOT a Pythagorean triangle). Knowing the vertical height 12 in the (15, 14, 13) triangle, we easily obtain the result x = 84/13. Thank you for your videos ! 😊
1/ By Heron theorem, area of the triangle ABC= 84 --> CE = h= 12 2/ Label a as side of the square and h’ =CP We have: h’= 12-a Area of the triangle ABC= Area of the triangle DCF+ Area of the trapezoid ADFB -> (12-a).a/2 + (a+14) x a/2=84 -> a/2.(12-a+a+14)=84 -> a=84/13 Area of the sqare = sq (84/13) = 41.75 sq units😅😅😅
My way of solution ▶ For the given triangle ΔABC, the values are: [AB]= 14 [BC]= 13 [CA]= 15 By drawing the height of the triangle, we have two triangles, the triangle on the left ΔADC and the trinagle on the right ΔDBC. D ∈ [AB] and [AD] ⊥ [DC] For the triangle ΔADC, we can write: [AD]= x [DC]= h [CA]= 15 By applying the Pythagorean theorem, we get: [AD]²+[DC]²= [CA]² x²+h²= 15² x²+h²= 225...........................Eq-1 For the triangle ΔDBC, we can write: [DB]= 14-x [CD]= h [BC]= 13 By applying the Pythagorean theorem, we get: [DB]²+[CD]²= [BC]² (14-x)²+h²= 13² 196-28x+x²+h²= 169............Eq-2 By subtracting equation two from equation one, we get: x²+h²= 225 -196+28x-x²-h²= -169 ⇒ 28x-196= 56 28x= 252 x= 9 ⇒ [AD]= 9 h= 12 [CD]= 12 [DB]= 5 b) Let's find the value of ∠CAD tan(α)= [DC]/[AD] tan(α)= 12/9 tan(α)= 4/3 In the same way the value of ∠CAD tan(β)= [CD]/[DB] tan(β)= 12/5 c) Let's consider the small triangle on the left: ΔAEF E ∈ [AD] F ∈ [AC] in this triangle: ∠FAE = α tan(α)= a/[AE ] 4/3= a/[AE ] ⇒ [AE]= 3a/4 d) Let's consider the small triangle on the right: ΔGBH G ∈ [DB] H ∈ [BC] in this triangle: ∠GBH = β tan(β)= a/[GB] 12/5= a/[GB ] ⇒ [GB]= 5a/12 d) We know that: [AE]+[EG]+[GB]= 14 [AE]= 3a/4 [GB]= 5a/12 [EG]= a ⇒ 3a/4 + a+ 5a/12 = 14 9a+12a+5a= 14*12 26a= 168 a= 84/13 length units ⇒ Apurple= a² Apurple= (84/13)² Apurple= 7056/169 square units
je n'ai pas trouvé la solution mais j'ai fait une découverte. Cos(A)=(15²+14²-13²)/(2x14x15)=(225+14+13)/14x30=252/420=3/5 Sin²(A)=1- 3²/5²=(25-9)/25=6/25,sin(A)= 4/5 (triangle rectangle(3,4,5)semblable au triangle rectangle(9,12,15)) sin(A)/13=sin(B)/15=sin(C)/14, Sin(B)=12/13 et sin(C)=56/65 cos²(B)=1-(12/13)²=(13²-12²)/13²=(13+12)/13²=25/13²=5²/13², cos(B)=5/13 (triangle rectangle 5,12,13) cos²(C)=1-(56/65)²=(65²-56²)/65²=(9x121)/65²=(3²x11²)/65²=33²/65²,cos(C)= 33/65 encore 1 triangle rectangle avec uniquement des carrés parfaits: 33², 56² et 65² On a le sinus de l'angle en A donc je peux trouver la hauteur = 15 sin(A)=5x4/5=12 La surface du triangle =15x14 sin(A)/2=(5x14x4/5)/2=84 Le triangle CDF est semblable au triangle ABC
Last part is unnecessary. Could have been solved easily by adding areas of three triangle and square and equating it to the area of the given triangle.
I love it❤
Glad to hear that!
Thanks for the feedback ❤️
Let x = side of square. Once we determine the height of ABC to be 12, then we know that CP = 12-x. ABC is similar to CDF. So there is the ratio 14/x = 12/(12-x). Solve for x and get 6.46. So x squared is 41.75.
Alternative to Heron's formula: As done later in the video, drop a perpendicular from C to AB and label the intersection as point E. Let CE have length h and AE have length b, then BE has length 14 - b. For ΔACE, b² + h² = (15)² = 225. For ΔBCE, h² + (14 - b)² = (13)² = 169. Expanding, h² + 196 - 28b + b² = 169, Replace b² + h² with 225: 225 + 196 - 28b = 169. Solve for b and find b = 9. Substitute b = 9 into b² + h² = 225 and find h = 12 or CE = 12. Proceed from 5:05 in the video.
While k is a perfectly valid designation for a variable, many of us, especially electrical engineers, associate it with 1000, so 12k looks like 12,000. Using a different letter would avoid this confusion.
However, we really don't need variable k. Let the sides of the square have length s. ΔDFC and ΔABC are similar. Corresponding sides are in the same ratio and also in the same ratio as corresponding heights. Sides DF and AB are corresponding and heights CP and CE also corresponding, so DF/AB = CP/CE, s/14 = (12 - s)/12, 12s = (14)(12 - s), 12s = 168 - 14s, 26s = 168, s = 84/13. The square's area = s² = (84/13)² = (7056)/(169), as PreMath also found.
Thanks for the feedback ❤️
Oh ! The famous Brahmagupta's triangle ! Nice video !
Use Heron's formula to solve for area of triangle ABC. Determine height of triangle ABC, which equals 12. Show by angle-angle similarity that triangle CDF is similar to triangle ABC. Make a proportion of height over base: 12/14 = (12-x)/x, which when solved for x is 84/13. x is the side length of the square, so x^2 = 84^2/13^2 = 7056/169
Equation for AC y=12x/9, convert to x1=9y/12, equation for BC y=(-12x+168)/5 convert to x2=-5y/12 + 14. Height of square = y, width of square =x2-x1, set equal, solve for y. Area =y²
If you let side of square = x then using similar triangles CPF and CEB knowing that CP=12-x you get PF= (60-5x)/12. Then using similar triangles CDP and CAE you get DP=(36-3x)/4. Adding DP and PF and putting equal to x you get x=84/13 etc.
Thank you!
You're welcome! Glad you found it helpful. ❤️
Thanks for the feedback ❤️
We already know that 13-14-15 is a special triangle that can be divided in two Pitágorean triplet right triangles 5-12-13 and 9-12-15
Therefore, height of triangle is h=12 cm
Similarity of triangles:
s / (h-s) = b / h
h.s = b (h-s)
12.s = 14 (12-s) = 168 - 14s
26 s = 168 --> s = 6,4615cm
s² = 41,75 cm² ( Solved √ )
The rt 🔺 s are (5, 12,13) and (9,12,15)
@@PrithwirajSen-nj6qq
Exactly
@@marioalb9726 yes you have corrected .Thanks. Request to see my sol.
@@PrithwirajSen-nj6qq
I wrote the following just to understand your solution:
A₁= s²
A₂= ½bh = ½(14-s)s = 7s-½s²
A₃= ½bh = ½s(12-s) = 6s-½s²
A₁+A₂+A₃= ½bh= ½14*12= 84
s² + (7s-½s²) + (6s-½s²) = 84
s² + 13s - s² = 84
13s = 84 ---> s=6,4615cm
s² = 41,75 cm² ( Solved √ )
!!! Very nice solution !!!
@@marioalb9726 O K.
Thanks
At 4:41,
Let the purple square be DFGH,
All lengths are measured in units.
DF=FG=GH=GD=x with EB=y and AE= 14-y. CB=13,BA=14 and AC=15.
By Pythagoras,
CE^2=EB^2+CE^2
CE^2=CB^2-EB^2
CE^2= 13^2-y^2=169-y^2...........................(1)
AC^2=AE^2+CE^2
CE^2=AC^2-AE^2 =15^2-(14-y)^2=225-196+28y-y^2=29+28y-y^2......................(2)
From (1) and (2),
169-y^2=29+28y-y^2
28y=140
y=5 units.
CE^2=CB^2-y^2
=13^2-5^2=(13+5)(13-5)=18*8=144
CE=12 units,PE=x units and CP=12-x units.
Consider similar triangles CDF and CAB,
CP/DF=CE/AB
(12-x)/x= 12/14 =6/7
7(12-x)=6x
84-7x=6x
13x=84 x= 84/13 units
Area of pink square=x^2= (84/13)^2= 7056/169 =41.75 square units approximately.
There are many ways to skin this particular cat professor.
Thanks for the puzzle.
We use an orthonormal center A and first axis (AB). The equation of the circle center A and radius 15 is x^2 + y^2 -225 = 0 and the equation of the circle center B and radius 13 is (x -14)^2 + y^2 -169 = 0 or x^2 + y^2 -28.x + 27 = 0. At the intersection we have -28.x +27 = -225 and that gives x = 9, then y^2 = 225 -9^2 = 144, and y = 12. So we now have point C: C(9; 12).
The equation of (AC) is (x).(12) - (y).(9) = 0 or y = (4/3).x. Then VectorBC(-5; 12) and the equation of (BC) is (x -14).(12) - (y).(-5) = 0 or 12.x +5.y -168 = 0
Point D(a; (4/3).a). Point F is on (BC) and has the same ordinate (4/3).a, so its abscissa x verifies 12.x = -5.(4/3).a +168, so x = (-5/9).a +14
and then we have F((-5/9).a +14; (4/3).a). Now DF = (-5/9).a +14 -a = (-14/9).a + 14, and this distance is also the ordinate of D: (4/3).a
So we have (-14/9).a + 14 = (4/3).a and that gives a = 63/13, and (4/3).a = 84/13 which is the side length of the square.
Finally the area of the square is (84/13)^2 = 7056/169.
S=(13+14+15)/2=16
Area of the triangle ABC=√21(21-13)(21-14)(21-15)=84
Draw CE is attitude
1/2(CE)(14)=84
CE=168/14=12
Let DF=x
∆CDF~∆CAB
x/14=(12-x)/12
So x=84/13
So Purple Square Area=(84/13)^2=7056/169 square units=41.75 square units.❤❤❤
Good Morning Teacher
Forte Abraço do Brasil
Deus lhe Abençoe
Obrigada querido❤️
Amor e orações dos EUA! 😀
@@PreMath Muitíssimo obrigado
Deus abençoe o Sr pelas vossas aulas
Square's sides are a.
Triangle's area is sqrt((21)(6)(7)(8)) = sqrt(7056) = 84.
Triangle height is 12 due to 168/14.
Therefore two right triangles of 5, 12, 13 and 9, 12, 15.
5/12=(5-(5/14)a)/a
5a=12(5-(5/14)a)
5a=(60-(30/7)a
a=(12-(30/35)a
a+(6/7)a=12
7a+6a=84
13a=84
a=84/13
a^2=7056/169
Solution: First calculate the area using Heron's Formula. Knowing the area and the value of the base, calculate the total height. Apply the Pythagorean Theorem, calculate the two measurements of the base which are 9 and 5, from left to right. After that, by similarity of triangles (proportion) calculate the constants of proportion of the triangles. It will be 12/5 on one side and 12/9 on the other side of the base. We arrive at the value of 14k for the side of the square. Knowing that the total height is 26k (14k + 12k) and that the height is 12, I calculate the value of k = 6/13. Finally I calculate the value of the side of the square (14k = 14 × 6/13 = 84/13) and then the Area of the Square = (84/13)² = 41.751 units of area.
41.75
Drop a perpendicular line from the vertex to form two right triangle
9-12-15 (3-4-5 triplet scaled up by 2) and a 5-12- 13 (Pythagorean triplets)
Let's label the length of the square 'x'
Notice that the top triangle is similar to the largest triangle in terms of
x, its height is 12/14 x, since in the 13-14-15 largest triangle , the height is 12
Hence, the height of the triangle in terms of x = x + 12/14x = 26/14 x
Hence, 26/ 14 x= 12
x = 14/26 * 12
x = 6.4615 4
x^2 = 41.75147 Answer
Excellent!
Thanks for sharing ❤️
CE = 12, DF = Side length
AB/CE = DF/(12-DF)
14/12 = DF/(12-DF)
DF = 168/26
Side length = DF
A method without Heron's formula
1) Let area of square is a ^2
2) sum of area of the two triangles beside the square on base of14 units =1/2*(14-a) *a sq units
=(7a - a^2/2) sq units
3)
I) Now Pythagorean triplet
says when we draw height on 14 unit base
One rt angle will be (9,12,15) ref [3*(3,4,5)]
II) Then the we see 13^2 =12^2 +5^2
Hence the other rt triangle will be (5,12,13)
III) Then height of the big triangle on base of 14 units will be 12 units
if) Height of triangle above the square is (12-a) units
v) Area of the triangle above the square =1/2*a*(12-a)sq units
= (6a -a^2/2)sq units
4)Then
a^2 + 7a - a^2/2 +6a -a^2/2=1/2*12*14
a=84/13 units
5) a^2=(84/13)^2=41.75 sq units (approx)
RUSSIA! Провёл из угла АВС биссектрису к стороне АС, которая яволяется и высотой BD.Площадь треугольника АВС нашёл по формуле Герона Александрийского, она равна 84; далее нашёл высоту треугольника по основанию 14, h=12; после этого из подобия треугольника с основанием х и треугольника с основанием 14, составил пропорцию, 12/14=(12-а)/а, откуда 26а=168; а=6,461538461...; площадь пурпурного квадрата равна 41,7514792898... округлять не буду.
Wait...I just took AST POSN exam and it's the same question 💀👍ty for the solution
Let's find the area:
.
..
...
....
.....
First of all we calculate the area of the triangle ABC according to the formula of Heron:
s = (AB + AC + BC)/2 = (14 + 15 + 13)/2 = 42/2 = 21
A = √[s*(s − AB)*(s − AC)*(s − BC)] = √[21*(21 − 13)*(21 − 14)*(21 − 15)] = √(3*7*2³*7*2*3) = 3*7*2² = 84
May E and G be the lower left and the lower right corner of the square, respectively. Now we add H on AB such that CH is the height of the triangle ABC according to its base AB:
A = (1/2)*AB*h(AB) = (1/2)*AB*CH ⇒ CH = 2*A/AB = 2*84/14 = 12
The triangles ACH and BCH are right triangles, so we can apply the Pythagorean theorem:
AC² = AH² + CH² ⇒ AH² = AC² − CH² = 15² − 12² = 225 − 144 = 81 ⇒ AH = √81 = 9
BC² = BH² + CH² ⇒ BH² = BC² − CH² = 13² − 12² = 169 − 144 = 25 ⇒ BH = √25 = 5
May s be the side length of the square. Since the triangles BFG and BCH are similar and the triangles ADE and ACH are similar as well, we can conclude:
BG/BH = FG/CH
BG/5 = s/12
⇒ BG = 5*s/12
AE/AH = DE/CH
AE/9 = s/12
⇒ AE = 9*s/12
AB = AE + EG + BG
14 = 9*s/12 + s + 5*s/12 = s*(9/12 + 12/12 + 5/12) = 26*s/12 = 13*s/6
⇒ s = 6*14/13 = 84/13
Now we are able to calculate the area of the purple square:
A(purple) = s² = (84/13)² = 7056/169 ≈ 41.75
Best regards from Germany
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شكرا لكم على المجهودات
يمكن استعمال
cosB =[14^2 +13^2 -15^2]/2×13×14
=5/13
sinB=12/13
CF=13×x/14
FB=13-CF
sinB =x/(13-CF)
12/13=x/(13-13x/14)
x=84/13
Muito boa a questão. Eu a resolvi por outro método, mas gostei do seu. Parabéns pela escolha. Brasil - Outubro de 2024.
Excelente!
Fico feliz em ouvir isso!
Obrigado pelo feedback ❤️
Amor e orações dos EUA! 😀
41.75
Excellent!
Thanks for sharing ❤️
All you really need to solve this is finding the area and height of the big triangle. Nothing else, really.
*Solution:*
The triangle ABC ~ DCF. Now, Consider DF=x. Hence,
CD/15 = x/14 → *CD =15x/14.* Hence,
AD= AC - CD→ *AD =15 - 15x/14*
Law of cosines in ∆ABC with respect to angle A:
13²=15² + 14² - 2×15×14 cos A
169= 421 - 420 cos A
Cos A = 252/420 = 3/5. Hence,
sin² A= 1 - cos² A
sin² A = 1 - 9/25= 16/25
sin A = 4/5.On the other hand,
sin A = x/AD →AD = 5x/4 and how
AD=15 - 15x/14, we have:
15 - 15x/14 = 5x/4 ÷(5)
3 - 3x/14 = x/4
x/4 + 3x/14 = 3 ×(28)
7x + 6x = 84 → x= 84/13, Therefore, the area of the square is:
*(84/13)² square units*
or
*≈41.75 square units*
I think that this problem is an application of the « Pythagorean triangles », i. e. the right triangles whose sides are all integers. If we note (a, b, c) the triplet of the sides in decreasing order, the (5, 4, 3) triangle is very well known (and of course its multiples : (10, 8, 6) and (15, 12, 9) and etc.
An other well-known Pythagorean triangle is (13, 12, 5) and its multiples…
And an interesting but easy theorem says that there are infinitely many Pythagorean triangles, and that any integer belongs to at least one of them.
Back to the problem, considering the (15, 12, 9) and (13, 12, 5) triangles, we can « glue » them side by side using the side 12, and rebuild the (15, 14, 13) triangle (which is NOT a Pythagorean triangle).
Knowing the vertical height 12 in the (15, 14, 13) triangle, we easily obtain the result x = 84/13.
Thank you for your videos ! 😊
The area is 7059/169 units square. Looks like this is a problem that I need to practice over and over again!!!
1/ By Heron theorem, area of the triangle ABC= 84
--> CE = h= 12
2/ Label a as side of the square and h’ =CP
We have: h’= 12-a
Area of the triangle ABC= Area of the triangle DCF+ Area of the trapezoid ADFB
-> (12-a).a/2 + (a+14) x a/2=84
-> a/2.(12-a+a+14)=84
-> a=84/13
Area of the sqare = sq (84/13) = 41.75 sq units😅😅😅
Veni,vidi,vici ...spliced it all together to solve thanks too Heron's Formula and Similar Triangles. 🙂
My way of solution ▶
For the given triangle ΔABC, the values are:
[AB]= 14
[BC]= 13
[CA]= 15
By drawing the height of the triangle, we have two triangles, the triangle on the left ΔADC and the trinagle on the right ΔDBC.
D ∈ [AB] and [AD] ⊥ [DC]
For the triangle ΔADC, we can write:
[AD]= x
[DC]= h
[CA]= 15
By applying the Pythagorean theorem, we get:
[AD]²+[DC]²= [CA]²
x²+h²= 15²
x²+h²= 225...........................Eq-1
For the triangle ΔDBC, we can write:
[DB]= 14-x
[CD]= h
[BC]= 13
By applying the Pythagorean theorem, we get:
[DB]²+[CD]²= [BC]²
(14-x)²+h²= 13²
196-28x+x²+h²= 169............Eq-2
By subtracting equation two from equation one, we get:
x²+h²= 225
-196+28x-x²-h²= -169
⇒
28x-196= 56
28x= 252
x= 9
⇒
[AD]= 9
h= 12
[CD]= 12
[DB]= 5
b) Let's find the value of ∠CAD
tan(α)= [DC]/[AD]
tan(α)= 12/9
tan(α)= 4/3
In the same way the value of ∠CAD
tan(β)= [CD]/[DB]
tan(β)= 12/5
c) Let's consider the small triangle on the left: ΔAEF
E ∈ [AD]
F ∈ [AC]
in this triangle: ∠FAE = α
tan(α)= a/[AE ]
4/3= a/[AE ]
⇒
[AE]= 3a/4
d) Let's consider the small triangle on the right: ΔGBH
G ∈ [DB]
H ∈ [BC]
in this triangle: ∠GBH = β
tan(β)= a/[GB]
12/5= a/[GB ]
⇒
[GB]= 5a/12
d) We know that:
[AE]+[EG]+[GB]= 14
[AE]= 3a/4
[GB]= 5a/12
[EG]= a
⇒
3a/4 + a+ 5a/12 = 14
9a+12a+5a= 14*12
26a= 168
a= 84/13 length units
⇒
Apurple= a²
Apurple= (84/13)²
Apurple= 7056/169 square units
CAB=α..cosα/2=√(21*8/15*14)=√(4/5)..CBA=β..cosβ/2=√(21*6/13*14)=√(9/13)..14=l/tgα+l+l/tgβ=l(ctgα+1+ctgβ)=l(3/4+1+5/12)=l*26/12..l=12*14/26=84/13.mah..!!?
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Thanks for sharing ❤️
je n'ai pas trouvé la solution mais j'ai fait une découverte.
Cos(A)=(15²+14²-13²)/(2x14x15)=(225+14+13)/14x30=252/420=3/5
Sin²(A)=1- 3²/5²=(25-9)/25=6/25,sin(A)= 4/5 (triangle rectangle(3,4,5)semblable au triangle rectangle(9,12,15))
sin(A)/13=sin(B)/15=sin(C)/14, Sin(B)=12/13 et sin(C)=56/65
cos²(B)=1-(12/13)²=(13²-12²)/13²=(13+12)/13²=25/13²=5²/13², cos(B)=5/13 (triangle rectangle 5,12,13)
cos²(C)=1-(56/65)²=(65²-56²)/65²=(9x121)/65²=(3²x11²)/65²=33²/65²,cos(C)= 33/65 encore 1 triangle rectangle avec uniquement des carrés parfaits: 33², 56² et 65²
On a le sinus de l'angle en A donc je peux trouver la hauteur = 15 sin(A)=5x4/5=12
La surface du triangle =15x14 sin(A)/2=(5x14x4/5)/2=84
Le triangle CDF est semblable au triangle ABC
Hii sir😊😊
😂😂😂
I tapped on translate on hi sir with the 2 smiling emojies, and I got this is a secret
Hello dear❤️
Thanks ❤️
Last part is unnecessary. Could have been solved easily by adding areas of three triangle and square and equating it to the area of the given triangle.
61
35 square unit ?
12&5&9 is rong 5.133333&8.86666666&11.943571 your az nswar & math is ronge