Math Olympiad | Can you find area of the Red shaded square? | (Right triangles) | #math #maths

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Math Olympiad | Can you find area of the Red shaded square? | (Right triangles) | #math #maths

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[maxforsberg]

You can solve this very easily by calculating cos(beta)=2/sqrt(5) and sin(beta)=1/sqrt(5).

Then define the length CF as x and BF thus has the length 18-x.

Let s be the side length of the square.

We can then set up the following relations:

Sin(beta) = s/x = 1/sqrt(5)
Cos(beta) = (18-x)/s = 2/sqrt(5).

This very easy to solve for s.

[phungpham]

1/ Notice that all the right triangles are similar and the ratio of the short leg/long leg= 1/2
2/ Let a be the side of the square. we have AD= a/2 and similarly EF/EC=2---->EC= the hypotenuse AC= 7a/2
3/ AC= 9sqrt5 so Area of the square= sqa=33.1 sq units

[prossvay]

AC=√18^2+9^2=9√5
Right triangle ABC~CEF
Let x is side of square; CE=a
9/18=x/a
a/x=1/2
a=2x
Right triangle ABC~ADG
AD=9√5-(x+a)=9√5-3x
AD/DG=AB/BC
(9√5-3x)/x=9/18=1/2
18√5-6x=x
7x=18√5
So: x=18√5/7
Area of the red square units.
Thanks ❤❤❤

[andybriggs]

What a great little puzzle. I always look forward to my daily PreMath fix, and can get it right 2/3 of the time. Thanks for your hard work, it is much appreciated.

[misterenter-izrz]

Generally s, b, h, s=bh/(b+h), in this case, b=sqrt(5)9, h=18/sqrt(5), then

[tombufford]

At a quick glance, Four similar triangles. Then base AC = sqrt(81 + (81 * 4)) = 20.125. Square DEFG, side lengths, s . Then s/CE = AD/s = 9/18 . Then s = 2* AD and s = 0.5 * CE. Then the base = 0.5 *s + s + 2*s = 3.5 * s. Then s = 20.125/3.5 = 5.75 and the area of the Red square = 5.75^2 = 33.1 Area units.

[santiagoarosam]

Los triángulos ABC, ADG y FEC son semejantes y la razón entre catetos es 9/18=1/2》 Si c= lado del cuadrado 》 AC=AD+DE+EC =c/2+c+2c =7c/2 》9^2 +18^2=(7c/2)^2》 c^2=1620/49 =33.06
Gracias y saludos.

[karphuntr]

Let x equal the side of the square.
Length AC = sq root 405 from Pythagoras
From similar triangles AD = x/2 and EC = 2x.
AD + x + EC = sq root 405
Substitute in the values for AD and EC in terms of x gives and rearranged gives
7x = 2 x sq root 405
So x = 5.75
X^2 = 33.1

[MrPaulc]

I got 34.08 but did it in my head, so I'll write it down for more clarity.
In ascending order the similar triangles' sides are a, 2a, and a*sqrt(5).
Square's sides are x.
ADE is (1/2)x, x, and x*sqrt(5)
GBF is x/sqrt(5), 2x/sqrt(5), and x
So length AB can be represented as (x*sqrt(5)) + x/sqrt(5) = 9
5x^2 + x^2/5 ==81
AC is (1/2)x + x + 2x, so (7/2)x = 9*sqrt(5)
(7/18)x = sqrt(5)
7x = 18*sqrt(5)
49x^2 = 324*5=1620
x^2 = 1620/49=33.06
I just saw where I went wrong. I used the calculator for the very final calculation and keyed in 1670 instead of 1620. Ho hum.

[yalchingedikgedik]

Thanks Sir
Thanks PreMath
With my glades

[jsEMCsquared]

Here is how I would do it . I would make all the triangles into squares and multiply area. Then I would divide by two. Then make the entire triangle into a square and multiply and then divide by two then deduct the difference.

[himo]

9 : 18 = 1 : 2
AD=x DE=2x EC=4x
AC=√[9^2+18^2]=√405=9√5
side of the Red square :
9√5*2x/7x=18√5/7
area of the Red square :
18√5/7＊18√5/7=1620/49

[marcgriselhubert]

Let's use an adapted orthobormal B(0;0) C(18;0) A(0;9) F(a;0), and G(a/2;0) as triangles BGF and BAC are similar.
VectorGF(a;-a/2) and so GF =sqrt((a)^2 + (a/2)^2) = a.(sqrt(5)/2)
VectorAC(18;-9) or VectorU(-2;1) are directing (AC), so the equation of (AC) is: (x-18).(1) - (y).(-2) = 0 or x + 2y -18 = 0
The distance from F to (AC) is then abs(a -2.0 -18) / sqrt ((1)^2 + (2)^2) = (18 - a) /sqrt(5) (knowing that a < 18), so FE = (18 - a) /sqrt(5)
Now, as EFGD is a square, we have FE = GF, or (18 - a) /sqrt(5) = a. (sqrt(5)/2)., or 18 - a = a. (5/2), or 18 = a. (7/2), and then a = 36/7.
The area of the square EFGD is GF^2 = (a. (sqrt(5)/2))^2 = (5/4).(a^2) = (5/4). (1296/49) = 1620/49.

[misterenter-izrz]

s/(9sqrt(5))(18/sqrt(5))+s=18/sqrt(5), (2/5+1)s=18/sqrt(5), 7/5 s=18/sqrt(5), s=(18×5)/(7sqrt(5))=(18/7)sqrt(5), therefore the answer is (18/7)^2×5=1620/49.😊

[batavuskoga]

I also solved this, the same way, but with comparison of the triangles ABC and BGF
You can see BC=2*AB, so because of the angle-angle theorem, you can conclude BF=2*BG --> BF=2k, BG=1*k
Using the Pythagoras theorem : FG²=BG²+BF² --> FG=sqrt(5)*k
CF=18-2k
BE/CF=BG/FG --> sqrt(5)*k/(18-2k)=k/sqrt(5)*k -->
18k-2k²=5k² --> 18k=7k² --> k=18/7
FG=sqrt(5)*k=sqrt(5)*18/7
area
area square = app. 33.06 square units

[wackojacko]

Around the world with Alpha and Beta. @ 3:03 the assumption that Alpha and Beta are complementary and concluding AA Similarity Theorem...a philosophical problem that there's no such thing as exact distance, time, weight or temperature so we can't have an exact answer because we can't provide in this case exact angles. 🙂

[alster]

The triangle similarity part may be tricky at first, but the other solutions are straightfoward.

[devondevon]

33.061277
This my approach
The length of AC = 20.12461 Pythagorean

Let the square side = N, and then the area of the three triangles are as follows:

BFG is similar to ABC . Hence, its sides in terms of N are 9/20.12461 and

18/20.12461 N. Hence its area = N^2 ( I made a mistake here by
putting N earlier instead of N^2 )

Triangles ADG and ECF both have sides N and length AD and EC. But
AD and EC length is 20 .12461 - N
Hence, the area of both is N ( 20.12461 - N)/2 = ( 20.12461 N- N^2)/2

Since the area of the ABC = 81 (18* 9* 1/2), then
81 = the Sum of the three rectangles and the square
81= N + (20.12461 N - N^2)/2 + N^2

N + 20.12461N - N^2 + 2N^2 (multiply both sides by 2)

0 = N^2 + 20.12461N - 162 ( quadratic equation)

Using the quadratic calculator N= 5.749889364804

So, the length of the square is 5.749889364804.

Its area N^2 = 5.749889364804^2 = 33.061227

[LuisdeBritoCamacho]

Algebraic Geometry Resolution:
I know that the side of the Square (s) must be : 5 < s < 6 lu
Therefore the Area of the Square (A) must be : 25 < A < 36 su
The Equation that defines the given Triangle [A B C] is : y = -2x + 18
The General Equation of the Straight Line that Defines the Triangle [B F G] is : y = -2x + b
I have to find "b" in a way that the distance between these two Straight Lines; y = -2x + 18 and y = -2x + b; is equal.
Or in a way that the hypotenuse of the Triangle [B F G] is equal to the distance between these two Straight Lines.
The side of the Square must be approximately equal to 5, 8 lu; therefore the Area of the Square is ~ 33, 64 su.

[hcgreier]

I did this as follows:
Call a = side of square.
AC = √(18²+9²) = 9√5.
GB = x. Since x/BF = 9/18 = 1/2 → BF = 2x, therefore a = √[x²+(2x)²] = x√5
AG² = (AC-DC)² + a², but DC = 3a, so
(9 - x)² = (9√5 - 3a)² + a² with a = x√5 one gets
(9 - x)² = (9√5 - 3x√5)² + (x√5)², so x = 18/7
a = x√5 = 18√5/7 → A = (18√5/7)² ≈ 33.06 square units
And now I see I did it way too complicated 🤣