Can you find area of the six Yellow shaded Squares? | (Rectangle) | #math #maths | #geometry

Today I will teach you tips and tricks to solve the given olympiad math question in a simple and easy way. Learn how to prepare for Math Olympiad fast!

Need help with solving this Math Olympiad Question? You're in the right place!

I have over 20 years of experience teaching Mathematics at American schools, colleges, and universities. Learn more about me at

Can you find area of the six Yellow shaded Squares? | (Rectangle) | #math #maths | #geometry

Olympiad Mathematical Question! | Learn Tips how to solve Olympiad Question without hassle and anxiety!

#FindYellowArea #Squares #Rectangle #SimilarTriangles #GeometryMath #PythagoreanTheorem
#MathOlympiad #IntersectingChordsTheorem #RightTriangle #RightTriangles
#OlympiadMathematicalQuestion #HowToSolveOlympiadQuestion #MathOlympiadQuestion #MathOlympiadQuestions #OlympiadQuestion #Olympiad #AlgebraReview #Algebra #Mathematics #Math #Maths #MathOlympiad #HarvardAdmissionQuestion
#MathOlympiadPreparation #LearntipstosolveOlympiadMathQuestionfast #OlympiadMathematicsCompetition #MathOlympics #CollegeEntranceExam
#blackpenredpen #MathOlympiadTraining #Olympiad Question #GeometrySkills #GeometryFormulas #Angles #Height #ComplementaryAngles
#MathematicalOlympiad #OlympiadMathematics #CompetitiveExams #CompetitiveExam

How to solve Olympiad Mathematical Question
How to prepare for Math Olympiad
How to Solve Olympiad Question
How to Solve international math olympiad questions
international math olympiad questions and solutions
international math olympiad questions and answers
olympiad mathematics competition
blackpenredpen
math olympics
olympiad exam
olympiad exam sample papers
math olympiad sample questions
math olympiada
British Math Olympiad
olympics math
olympics mathematics
olympics math activities
olympics math competition
Math Olympiad Training
How to win the International Math Olympiad | Po-Shen Loh and Lex Fridman
Po-Shen Loh and Lex Fridman
Number Theory
There is a ridiculously easy way to solve this Olympiad qualifier problem
This U.S. Olympiad Coach Has a Unique Approach to Math
The Map of Mathematics
mathcounts
math at work
Pre Math
Olympiad Mathematics
Two Methods to Solve System of Exponential of Equations
Olympiad Question
Find Area of the Shaded Triangle in a Rectangle
Geometry
Geometry math
Geometry skills
Right triangles
imo
Competitive Exams
Competitive Exam
Calculate the length AB
Pythagorean Theorem
Right triangles
Intersecting Chords Theorem
coolmath
my maths
mathpapa
mymaths
cymath
sumdog
multiplication
ixl math
deltamath
reflex math
math genie
math way
math for fun

Subscribe Now as the ultimate shots of Math doses are on their way to fill your minds with the knowledge and wisdom once again.

[Birol]

My way of solution ▶
this yellow shape is touching two points on the right and left of the given rectangle;
d= 13
this length is equal to the diagonal of an rectangle build by these yellow blocks:
x= 2a
y= 3a
according to the Pythagorean theorem we can write:
(2a)²+(3a)²= d²
4a²+9a²= 13²
13a²= 13²
a²= 13

Ayellow= 6a²
Ayellow= 6*13
Ayellow= 78 square units

[unknownidentity]

Let's find the area:
.
..
...
....


Let A, B, C and D be the corners of the rectangle with the following coordinates:

A: ( 0 ; 0 )
B: ( 13 ; 0 )
C: ( 13 ; 11 )
D: ( 0 ; 11 )

Let E, F, G and H be the points of intersection on AB, BC, CD and DA, respectively. Then we obtain the following coordinates:

E: ( xE ; 0 )
F: ( 13 ; yF )
G: ( xG ; 11 )
H: ( 0 ; yH )

Now we need the following two vectors which are perpendicular to each other:

r = ( (xG − xF)/2 ; (yG − yF)/2 ) = ( (xG − 13)/2 ; (11 − yF)/2 )
s = ( (yF − yG)/2 ; (xG − xF)/2 ) = ( (yF − 11)/2 ; (xG − 13)/2 )

With these vectors we obtain:

v(G) = v(F) + 2*r
v(E) = v(F) + 1*r + 3*s
v(H) = v(F) + 3*r + 2*s

0 = yE = yF + 1*[(11 − yF)/2] + 3*[(xG − 13)/2]
0 = xH = xF + 3*[(xG − 13)/2] + 2*[(yF − 11)/2]

0 = yF + 11/2 − yF/2 + 3*xG/2 − 39/2
0 = 13 + 3*xG/2 − 39/2 + yF − 11

0 = yF/2 + 3*xG/2 − 14
0 = yF + 3*xG/2 − 35/2

0 = yF/2 − 7/2
⇒ yF = 7

0 = yF + 3*xG/2 − 35/2
0 = 7 + 3*xG/2 − 35/2
0 = 3*xG/2 − 21/2
⇒ xG = 7

Let's check these results:

r = ( (xG − 13)/2 ; (11 − yF)/2 ) = ( (7 − 13)/2 ; (11 − 7)/2 ) = ( −3 ; +2 )
s = ( (yF − 11)/2 ; (xG − 13)/2 ) = ( (7 − 11)/2 ; (7 − 13)/2 ) = ( −2 ; −3 )

xE = xF + 1*rx + 3*sx = 13 + 1*(−3) + 3*(−2) = 4
yE = yF + 1*ry + 3*sy = 7 + 1*(+2) + 3*(−3) = 0 ✓

xH = xF + 3*rx + 2*sx = 13 + 3*(−3) + 2*(−2) = 0 ✓
yH = yF + 3*ry + 2*sy = 7 + 3*(+2) + 2*(−3) = 7

The length of FG corresponds to twice the side length s of one yellow square:

(2*s)² = (xG − xF)² + (yG − yF)² = (7 − 13)² + (11 − 7)² = (−6)² + 4² = 36 + 16 = 52

Now we are able to calculate the area of the yellow region:

A(yellow) = 6*s² = (3/2)*(4*s²) = (3/2)*(2*s)² = (3/2)*52 = 78

Best regards from Germany

[MrPaulc]

Rectangle is 143 un^2
Call the squares' sides a.
Drop a perpendicular from the top point where a square's vertex meet the horizontal.
Drop another line from that point to the base where a square's opposite diagonal touches the base. These are two sides of a right triangle. The perpendicular is 11 (given), the diagonal is sqrt(10a^2. The short distance along the base is x.)

[stephenbrand]

What an excellent problem and solution. Thank you.

[marcgriselhubert]

Point of tangency of yellow area with feft side of the rectangle is A, with lower dide is B, with right side is C and with upper side is D.
c is the side length of each yellow square.
We use an orthonormal center A and first axis the lower side of the yellow square which contains A
We have A(0; 0) B(2.c; -c) C(3.c; 2.c) D(c; 2.c)
The equation of the left side of the rectangle is y = m.x
The distance from C to this line is
abs(3.m.c - m.c)/sqrt(m^2 +1) = 13 (1)
The equation of the lower side of the recangle is y + c = (-1/m).(x - 2.c) as it is perpendicular to the precedent, or -x/m -y +2.c/m -c = 0
The distance from D to this line is:
abs(-c/m -2.c +2.c/m -c)/ sqrt1 + 1/m^2) = abs(-3.c.m + c)/sqrt(m^2 +1) when simplified and that is 11.(2)
We compare (1) and (2) and can say that:
11;abs(3.m - 2) = 13. abs(-3.m + 1)
so: 33.m -22 = -39.m + 13, giving m = 35/6
or 33.m -22 = 39.m -13, giving m = -3/2
m is negative of the, so m = -3/2
Now with (1) we get
c;abs(-9/2 -2) = 13.sqrt(9/4 +1),
giving c = sqrt(13).
Then the area of each square is 13, the total yellow area is the 6.13 = 78.

[zdrastvutye]

if a square has the dimension or sidelength of l5,
its rotation or location can be described using the distances
l3 and l4 with:
l5=sqr(l3^2+l4^2)
with l2=13 and l1=11, this leads to an equation system with 2 unknown dimensions:
2*l3+3*l4=l2
l3+ 3*l4=l1 so l4=3 and l3=2 so l5=sqr(13)
and there is another solution without counting x and y components:
10CLS:PRINT "premath-can you find area of the yellow shaded squares"
20DIM x(1, 5, 3), y(1, 5, 3), u(3), v(3):l1=11:l2=13:WINDOW OPEN:WINDOW FULL
30sw=.1:la=1:x(0, 0, 0)=la:y(0, 0, 0)=0:x(0, 0, 1)=2*la:y(0, 0, 1)=0
40x(0, 0, 2)=2*la:y(0, 0, 2)=la:x(0, 0, 3)=la:y(0, 0, 3)=la
50REM ***for a=1 TO 2:FOR b=0 TO 3:y(0, 2*a, b)=y(0, 0, b)+a*la
60REM ***x(0, 2*a, b)=x(0, 0, b):NEXT b:NEXT a
70FOR a=0 TO 3:x(0, 1, a)=x(0, 0, a)-la:y(0, 1, a)=y(0, 0, a)+la:NEXT a
80FOR a=1 TO 2:FOR b=0 TO 3:x(0, 1+a, b)=x(0, 1, b)+a*la:y(0, 1+a, b)=y(0, 1, b)
90NEXT b:NEXT a:FOR a=1 TO 2:FOR b=0 TO 3:x(0, 3+a, b)=x(0, 1+a, b)
100y(0, 3+a, b)=y(0, 1+a, b)+la:NEXT b:NEXT
110w=0:GOTO 260
120FOR au=0 TO 5:FOR bu=0 TO 3:x(1, au, bu)=x(0, au, bu):y(1, au, bu)=y(0, au, bu)
130NEXT bu:NEXT
140FOR au=0 TO 5:FOR bu=0 TO 3:xu=x(1, au, bu):yu=y(1, au, bu)


170x(1, au, bu)=xv:y(1, au, bu)=yv:NEXT bu:NEXT au
180FOR au=0 TO 5:FOR bu=0 TO 3
190IF xmin>x(1, au, bu) THEN xmin=x(1, au, bu)
200IF xmax<x(1, au, bu) THEN xmax=x(1, au, bu)
210IF ymin>y(1, au, bu) THEN ymin=y(1, au, bu)
220IF ymax<y(1, au, bu) THEN ymax=y(1, au, bu)
230NEXT bu:NEXT au:lx=xmax-xmin:ly=ymax-ymin

250RETURN
260GOSUB 120
270dg1=dg:w1=w:w=w+sw:IF w>2*PI THEN STOP
280w2=w:GOSUB 120:IF dg1*dg>0 THEN 270
290w=(w1+w2)/2:GOSUB 120:IF dg1*dg>0 THEN w1=w ELSE w2=w
300IF ABS(dg)>1e-5 THEN 290
310PRINT w, la, lx, ly:pa=6*la^2/lx/ly:ag=l1*l2*pa
320PRINT "die";:PRINT COLOUR(6)"gelbe";:PRINT "flaeche=";ag
330PRINT "das sind";pa*100;"%der gesamtflaeche"


360IF masx<masy THEN mass=masx ELSE mass=masy
370GOTO 390

390col=1:x=u(0):y=v(0):GOSUB 380:xba=xbu:yba=ybu:FOR a=1 TO 4
400ia=a:IF ia=4 THEN ia=0
410x=u(ia):y=v(ia):GOSUB 380:xbn=xbu:ybn=ybu:GOTO 430
420LINE xba;yba, xbn;ybn COLOUR col:xba=xbn:yba=ybn:RETURN
430GOSUB 420:NEXT a:col=6:FOR a=0 TO 5:x=x(1, a, 0):y=y(1, a, 0)
440GOSUB 380:xba=xbu:yba=ybu:FOR b=1 TO 4:ib=b:IF ib=4 THEN ib=0
450x=x(1, a, ib):y=y(1, a, ib):GOSUB 380:xbn=xbu:ybn=ybu
460GOSUB 420:NEXT b:NEXT a
this will vary the rotation of the geometric formation or ensemble and...
install dosbox and amstrad basic from reenigne to run this

[jimlocke]

Doing it the hard, but maybe a little more straightforward, way. Let's designate the corners of the rectangle as A (upper left), B (upper right), C(lower right) and D(lower left). Let's designate the points where the yellow squares touch the sides of the rectangle as E (on AB), F (on BC), G on CD and H (on AD). Construct line segments EH, GH and FG. Let the squares have side length a. Note that EH and GH are hypotenuses of right triangles with sides of length a and 2a, therefore have length a√5. We can prove that ΔAEH and DHG are similar, and, furthermore, they are congruent by angle - side - angle. Let AH have length m, so corresponding side DG has length m. Then, DH has length 11 - m and corresponding side AE has length 11 - m. Since AB has length 13, this leaves BE with a length of m + 2. We also have ΔBEF with hypotenuse 2a and ΔCFG with hypotenuse a√(10), neither similar to the first 2 and not similar to each other.

Apply the Pythagorean theorem to ΔAEH, with sides of length m and 11 - m. The hypotenuse squared is m² + (11 - m)² = m² + 121 - 22m + m² = 2m² - 22m + 121 but also equals 5a². So, a² = (2m² - 22m + 121)/5. Note that ΔBEF has hypotenuse EF = 2a and side BE length (m + 2). (EF)² is 4a² and (BE)² = m² + 4m + 4. Substitute (2m² - 22m + 121)/5 for a², so EF squared = 4(2m² - 22m + 121)/5 = (8m² - 88m + 484)/5. The square of BF is hypotenuse minus BE = (8m² - 88m + 484)/5 - (m² + 4m + 4) = (3m² - 108m + 464)/5.

Consider ΔCFG. Its hypotenuse = a√(10) and side CG has length 13 - m. Side BF = √((3m² - 108m + 464)/5), so side CF = 11 - √((3m² - 108m + 464)/5). We apply the Pythagorean theorem. The square of the hypotenuse is 10a². Replace a² with (2m² - 22m + 121)/5 and we get 4m² - 44m + 242. The square of CG is 169 - 26m + m². (CF)² = (FG)² - (CG)² = 4m² - 44m + 242 - (169 - 26m + m²) = 3m² - 18m + 73. (CF)² = (11 - √((3m² - 108m + 464)/5))² = 121 - 22√((3m² - 108m + 464)/5) + (3m² - 108m + 464)/5. So, we have the task of solving the equation 121 - 22√((3m² - 108m + 464)/5) + (3m² - 108m + 464)/5 = 3m² - 18m + 73. The standard approach would be to move everything except the radical to one side and square both sides, leaving us with a 4th order equation to solve. If someone wants to try it, please do so! Left to my own devices, I would replace m by x, let y = the difference of the two sides of the equation and plot (x, y) for a range of x from 0 to 5.5. Then, I'd do a half interval search for y = 0. This could be done with a computer program, but x = 4 might pop out early if searching by hand with the aid of a calculator.

We can check for errors in the equations using PreMath's solution. He got a = √(13), so a² = 13. Then, a² = (2m² - 22m + 121)/5 and 13 = (2m² - 22m + 121)/5, 65 = 2m² - 22m + 121, m² - 11m + 28 = 0, which has roots 4 and 7. m = 7 doesn't make sense, since we can do some rough calculations and decide m can't be more than half the length of AD, so m = 4. So, (3m² - 108m + 464)/5 becomes 16 and its square root is 4. 121 - 22√((3m² - 108m + 464)/5) + (3m² - 108m + 464)/5, the left side, becomes 121 - 22(4) + 16 = 49. The right side 3m² - 18m + 73 = (3)(16) - (18)(4) + 73 = 49. So, m = 4 is a solution and a = √(13). Each yellow square has area 13 and there are 6 squares, total area 78.

At first glance, this solution method seems more straightforward than PreMath's, but it isn't until you generate equations for ΔCFG that you find that you have a 4th order equation to solve!

[jmcxzcs]

Draw a line to get a square(11, 11). then we can get a triangle( a, 2, ?). By using the similarity, we can find a right triangle(3a, 6, 9). therefore a²=13 and the answer is 78. Thanks.

[yalchingedikgedik]

Very nice
Thanks Sir
With my respects ❤❤❤❤

[wcdaniel]

Why was ‘x’ on the left-hand side, bottom left corner not included in the equation, 2x+3y=13? The ‘x’ on the left-hand side, bottom left corner was included in the equation, x+3y=11.

[LucasBritoBJJ]

Is it possible to prove that the points the squares touches the “11 units” lines are at the same height?

This would give us a parallel “13 unit” line

I drew this line connecting this two points, than (3a)²+(2a)²=13²

Which give us a²=13 as well

[giuseppemalaguti]

Le equazioni usate sono, (l lato del i calcoli risulta a=6, l^2=13.. Ayellow=6*13=78...PS, a=2y nella figura

[santiagoarosam]

Los cuadrados amarillos tienen dimensiones a*a→ Una recta vertical que pase por el vértice inferior del cuadrado extremo derecho, dista 11 ud del lado azul vertical izquierdo y 13-11=2 ud del vertical derecho→ otra vertical que pase por el vértice inmediato inferior dista 13-2-2=9 ud del lado azul vertical izquierdo→ 9 ud es la proyección horizontal de la recta amarilla de longitud 3a cuyo extremo es el punto de contacto con el azul vertical izquierdo→ Con los datos anteriores podemos descomponer el cuadrado amarillo del extremo derecho en cuatro triángulos rectángulos congruentes de lados [(2)/(9/3=3)/(a)] y un cuadrado central de lados (3-2=1)→ Área de cada cuadrado amarillo =4(2*3/2)+(1*1) =13 ud²→ Área de la zona amarilla =6*13=78 ud².
Gracias y un saludo cordial.

[gaskellr]

At 5.29, please explain how you come up with labelling the 3x and 3y please, as you do not explain this.

[jarikosonen]

This complementary angle theory looks important, if one just find the right auxiliary lines.

[calvinmasters]

I presumed that the left-hand-most and right-hand-most yellow corners formed a horizontal line.
Then (2a)^2 + (3a)^6 = 169
13a^2 = 169
a = √ 13
area of yellow square = 13

[michaeldoerr]

The answer is 78 units square. Golly looks like most of the within area shape problems involve a great deal of AA similarity!!! I mean it was why you made use of complementary angles was it not???. I hope that I am improving my reasoning. And maybe make a playlist of area problems that DO NOT involve subtraction of other areas. And I better remain caught up in order to BE a mathematician!!!

[LuisdeBritoCamacho]

STEP-BY-STEP RESOLUTION PROPOSAL :


01) Total Area = 143 sq un, so the Area of the 6 Squares < 143 sq un

02) Drawing a Parallel Line to Length with 13 lin un passing the 2 Points that touch the 2 Widths with 11 lin un, we get a Hypotenuse with Length 13 lin un.

03) Let the Side of each Yellow Square be "a".

04) Now we have a Right Triangle with Sides (2a ; 3a ; 13)

05) (2a)^2 + (3a)^2 = 13^2 ; 4a^2 + 9a^2 = 169 ; 13a^2 = 169 ; a^2 = 169 / 13 ; a^2 = 13 ; a = sqrt(13).

06) Side of Yellow Square = sqrt(13) lin un

07) Area of Yellow Square = 13 sq un

08) Six Yellow Squares Area = (6 * 13) = 78 sq un

Therefore,


OUR ANSWER :

The 6 Yellow Squares Area must be equal to 78 Square Units.