Find area of the Yellow triangle in the Square | (Important math skills explained) | #math #maths

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Find area of the Yellow triangle in the Square | (Important math skills explained) | #math #maths

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[ashishx]

Yes I also solved it,
I am from India 🇮🇳

[kennethstevenson]

Very effective method of simplified solution by factoring out a common factor to simplify the problem then bringing the factor back to find the final solution. Great solution of a problem with three variables.

[mohabatkhanmalak]

I will have to go back in the clip several times till I master it. Thanks for posting proffessor!

[Bingurmom]

Sir, I love your maths problems.I watch it everyday🎉

[alster]

Easy, although in the middle part, I used FOIL method, basic substitution and quadratic factoring


Fast forwarding to 5:36, here is my my technique


(a-4/a)(a-8/a)= 6


By using FOIL, we get

a² -8-4+32/a²= 6
a²-12+32/a²= 6
a²+32/a-18= 0



Then by substitution, let a²= x


(x+32/x-18= 0)(x)

x²-18x+32= 0

(x-16)(x-2)= 0

x= 16, 2



Since 16 is a valid solution and a perfect square

a²= 16 (this is the area of the square)

and the rest of the video follows

[tombufford]

At a quick glance, each 2 outer triangles have common base b1=b2=, s, b3= s-h1. s =square side length, s3 if h1, h2, h3 are the triangles respective heights, 0.5s*h1=34, 0.5s*h2=68 and 0.5(s-h1)*h3=51. h2+h3 =s. substituting using these equations. 0.5(s-72/s)*h3=51 (s^2-72)*h3-102*s=0, h3+h2=s. h2*s=136, h3*s=s^2-136 . (s^2*h3-72*h3-102*s=0, h3=s-h2, h2=136/s. Then This has become complicated. I think the factoring by 17 simplifies this greatly, 34, 68 and 51 factored to 2, 4 and 3 at the

[felistrix]

uppercase letters for formulas
lowercase letters for variables

A: 68 = ab
B: 136 = ac
C: 102 = (a-b)(a-c) = a²-ab-ac+bc
D: 153+x = a²

A+B+C-D: 153-x = bc
A*B/D: (68*136)/(153+x) = bc

set last both formulas equal and solve for x
x² = 153²-68*136
x = 119

[bigm]

Thanks, Professor, for another wonderful solution!❤

[SanjaySinghaniaIN]

Instead of 3 variable this problem can be easily reduced to 2 variables. Take any side with triangle vertex not on the corner of square and demarcate the segments a and b; then this gives the area of square as (a+b)^2 and sides of triangles as(taking t = 17): 2t area triangle = a+b, b
4t area triangle = a+b, 2b (by equating this area to the double of 2t area)
3t area triangle = a-b, a
Evaluating a^2 + b^2 as adding up areas of triangle 2t and 3t = 10t.
Again, by subtracting the area of 2t from the area of 3t, we get the common quadratic in ab (replacing a^2-b^2 by (a-b)(a+b) by (6t/a)(4t/b)) solving which gives -t+7t for 2ab. Taking non-negative value for 2ab as 6t, we've (a+b)^2 = 10t + 6t = 16t
Now subtracting areas of triangles from area of square yields the answer 16t-9t = 7t, where t = 17.

[johnbrennan]

I used the given areas to solve. I ended up with x^4 - 204x^2+ 9248= 0. This factorised as ( x^2- 272)( x^2- 34)= 0. That gives x^2 = 272 which is the area of the . Subtract 153 to get 119 as area of yellow figure.

[chelliahRaveedrarajah]

I love, understand and learn this wonderful geometry question 🎉🎉🎉. Thank you sir for solve my doubt.❤❤❤

[cyruschang]

If the length of each side of the square is x,
the short side of the pink triangle is 68/x
the short side of the green triangle is 136/x
area of the blue triangle = [x - (68/x)] [x - (136/x)] / 2 = 51
x^2 - 204 + (68•136/x^2) = 102
x^2 - 306 + (68•136/x^2) = 0
x^4 - 306x^2 + 68 (136) = 0
x^4 - 306x^2 + 34 (272) = 0
(x^2 - 34)(x^2 - 272) = 0
x^2 = 34 or 272
x = ✓272 or ✓34 (not possible)
area of the yellow triangle = area of the square - (34 + 51 + 68) cm^2 = (✓272)(✓272) cm^2 - (34 + 51 + 68) cm^2 = (272 - 153) cm^2 = 119 cm^2

[giuseppemalaguti]

Con semplici calcoli risulta l'area del quadrato 272...Ay=272-153=119

[santiagoarosam]

Una línea vertical por el vértice superior intermedio y una horizontal po el vértice intermedio izquierdo, divide el cuadrado en cuatro celdas de áreas (2×51); (2×68 -a); (2×34 -a) y (a)》102/(68-a) = (136-a)/a》a^2 -306a +9248=0》a=34 》Las cuatro celdas suman: 102+102+34+34=272》 Área amarilla =272-51-68-34=119
Gracias y saludos.

[wackojacko]

@ 7:41 one needs a strong cup of coffee too start tweaking this biquadratic or "quartic" equation. Math sure is fun! 🙂

[yalchingedikgedik]

Thats very nice and enjoyable algebra exercises.
Thanks Sir
For your efforts

[Ctrl_Alt_Sup]

I started writing 68=34x2 and 51=68-(34/2). I found some interesting relationships between the surfaces and the sides of the triangles but I got lost looking for the side of the square... it was better to work with 2x17, 3x57, 4x17 and you named the sides wisely. Thanks !

[じーちゃんねる-vn]

Think x1/√17 (1/2)(a-8/a)(a-4/a)=3 a^2=16 yellow area=16-2-3-4=7 real yellow area=7x17=119

[Copernicusfreud]

Yay! I solved the equation. I let the sides of the square = x. Using x to define the areas of the triangles gives (1/2)*(x)*(x/68) = 34 and (1/2)*(x)*(x/136)=68 and (1/2)*(x - 68/x)*(x-136/x)=51. Using (1/2)*(x - 68/x)*(x-136/x)=51, I ended up with a quadratic equation of x^4 - 306*x^2 + 9248 = 0. I let u = x^2, which is the area of the square. u^2 - 306*u + 9248 = 0. Using the quadratic formula to solve for u, u = x^2 = 272 and u = x^2 = 34. x^2 = 34 is impossible and rejected because the areas of the 3 given triangles are 34 or greater, so x^2 = 272. 272-34-51-68 = 119. The answer is 119 cm^2.

[vacuumcarexpo]

Even if the given square were a rectangle, not a square, we could find the area of the yellow triangle though there are 4 variables but only 3 equations.

Why?