Can you find the area of the Blue Square? | (Rectangle) |

Подписаться 411 тыс.

Просмотров 14 тыс.

50% 1

Learn how to find the area of the Blue Square. Important Geometry and Algebra skills are also explained: Pythagorean theorem; area of the square formula; area of the rectangle formula. Step-by-step tutorial by PreMath.com
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!
Step-by-step tutorial by PreMath.com
• Can you find the area ...
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
/ premath
Can you find the area of the Blue Square? | (Rectangle) | #math #maths | #geometry
Olympiad Mathematical Question! | Learn Tips how to solve Olympiad Question without hassle and anxiety!
#FindBlueSquareArea #Square #Rectangle #GeometryMath #PythagoreanTheorem
#MathOlympiad #IntersectingChordsTheorem #RightTriangle #RightTriangles
#PreMath #PreMath.com #MathOlympics #HowToThinkOutsideTheBox #ThinkOutsideTheBox #HowToThinkOutsideTheBox? #FillInTheBoxes #GeometryMath #Geometry #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.

Опубликовано:

17 сен 2024

Ссылка:

Скачать:

Готовим ссылку...

Добавить в:

Мой плейлист

Посмотреть позже

Комментарии : 72

@GablesDude 7 дней назад

This was great! Thank you for your clear instructions!

@PreMath 6 дней назад

Glad to hear that! You are very welcome! Thanks for the feedback ❤️

@marioalb9726 7 дней назад

A₁ = 15 cm² = b.h = 5h h = 3 cm Pytagorean theorem: (s-3)²+(s-5)²=10² (s²-6s+9)+(s²-10s+25)=100 2s² - 16s + 34 = 100 s² - 8s - 33 = 0 s = 11 cm s² = 121 cm² ( Solved √ )

@PreMath 7 дней назад

Excellent! Thanks for sharing ❤️

@quigonkenny 7 дней назад

Green rectangle: Aɢ = hw 15 = h(5) h = 15/5 = 3 Draw CA, where C is the point on the top side of the blue square where BC and CA are perpendicular. If s is the side length of the blue square, then by observation, BC = s-5 and CA = s-3. Triangle ∆ABC: BC² + CA² = AB² (s-5)² + (s-3)² = 10³ s² - 10s + 25 + s² - 6s + 9 = 100 2s² -16s - 66 = 0 s² - 8s - 33 = 0 (s-11)(s+3) = 0 s = 11 | s = -3 ❌ s > 0 Blue square: Aʙ = s² = 11² = 121 cm²

@PreMath 7 дней назад

Excellent! Thanks for sharing ❤️

@alastairjames8723 6 дней назад

X=-3 represents a square whose top right corner is the bottom left corner of the green rectangle and whose bottom left corner is the point B. Don't reject it just because it's 'out of the box'

@devondevon4366 6 дней назад

121 The width of the rectangle = 3 (15/5) Draw a line from the edge of the rectangle to the square to form a right triangle . Let the length of the square = n, then the bases of the triangle are , n-5, and n-3. Hence, 10^2 = (n-5)^2 + (n-3)^2 =n^2 + 25-10n + n^2 +9-6n =2n^2 + 34- 16n 50 = n^2 + 17-8n 0 = n^2 - 33-8n 0 = (n-11) (n+3) 11=n 121= n^2

@PreMath 6 дней назад

Excellent! Thanks for sharing ❤️

@jamestalbott4499 7 дней назад

Thank you!

@PreMath 7 дней назад

You are very welcome! Thanks for the feedback ❤️

@sergioaiex3966 7 дней назад

Solution: A = Lenght × Width 15 = 5 × W W = 3 cm Applying Pythagorean Theorem, we'll have: (a - 5)² + (a - 3)² = 10² a² - 10a + 25 + a² - 6a + 9 = 100 2a² - 16a + 34 = 100 2a² - 16a - 66 = 0 (÷2) a² - 8a - 33 = 0 a = [8 ± √(64 + 132)]/2 a = [8 ± 14]/2 a = 11 A = a² A = (11)² A = 121 cm² =========

@PreMath 6 дней назад

Excellent! Thanks for sharing ❤️

@gelbkehlchen 7 дней назад

Solution: a = horizontal side of the green rectangle = 5, b = vertical side of the green rectangle = 15/5 = 3, x = side of the blue square. Pythagoras: (x-5)²+(x-3)² = 10² ⟹ x²-10x+25+x²-6x+9 = 100 |-100 ⟹ 2x²-16x-66 = 0 |/2 ⟹ x²-8x-33 = 0 |p-q-formula ⟹ x1/2 = 4±√(16+33) = 4±7 ⟹ x1 = 4+7 = 11 and x2 = 4-7 = -3 [not allowed in geometry] ⟹ Blue square area = 11² = 121[cm²]

@PreMath 6 дней назад

Excellent! Thanks for sharing ❤️

@unknownidentity2846 7 дней назад

Let's find the area: . .. ... .... ..... The height of the green rectangle can be calculated as follows: A = (base b)*(height h) ⇒ h = A/b = (15cm²)/(5cm) = 3cm Now let's add point C on the right side of the blue square such that ABC is a right triangle. In this case we can apply the Pythagorean theorem and with s being the side length of the square we obtain: AB² = AC² + BC² AB² = (s − b)² + (s − h)² (10cm)² = (s − 5cm)² + (s − 3cm)² 100cm² = s² − (10cm)*s + 25cm² + s² − (6cm)*s + 9cm² 0 = 2*s² − (16cm)*s − 66cm² 0 = s² − (8cm)*s − 33cm² s = 4cm ± √[(4cm)² + 33cm²] = 4cm ± √(16cm² + 33cm²) = 4cm ± √(49cm²) = 4cm ± 7cm Since the side length is a positive quantity, the only useful solution is: s = 4cm + 7cm = 11cm Now we are able to calculate the area of the blue square: A = s² = (11cm)² = 121cm² Best regards from Germany

@PreMath 7 дней назад

Excellent! Thanks for sharing ❤️

@Waldlaeufer70 6 дней назад

(a - 5)² + (a - 3)² = 10² a² - 10a + 25 + a² - 6a + 9 = 100 2a² - 16a + 34 = 100 a² - 8a + 17 = 50 a² - 8a = 33 a² - 8a + 16 = 33 + 16 = 49 (a - 4)² = 49 a - 4 = 7 (negative solution invalid) a = 11 A(square) = 11² = 121 [cm²]

@calvinmasters6159 7 дней назад

Thank you. More please.

@PreMath 7 дней назад

More to come! You are very welcome! Thanks for the feedback ❤️

@giuseppemalaguti435 7 дней назад

(l-3)^2+(l-5)^2=100...l=11

@PreMath 7 дней назад

Excellent! Thanks for sharing ❤️

@Birol731 7 дней назад

My way of solution ▶ Agreen= 15 cm² x= 5 cm ⇒ 15= 5y y= 3 cm [AB]= 10 cm The length [AB] is the hypothenuse of a right triangle: the length of the blue square a the base of this rectangular = a-5 the height of this triangular= a-3 By applyling the Pythagorean theorem we get: 10²= (a-3)²+(a-5)² 100= a²-6a+9+a²-10a+25 2a²-16a-66=0 a²-8a-33=0 Δ= 196 √Δ= 14 ⇒ a₁= (8+14)/2 a₁= 11 cm a₂= (8-14)/2 a₂= -3 < 0 ❌ ⇒ a= 11 cm Asquare= 11² Asquare= 121 cm²

@PreMath 7 дней назад

Excellent! Thanks for sharing ❤️

@pas6295 6 дней назад

First ly name the points of intersection as A. B. C. D. E And F. Extend EF to meet BC at G. Now the four sides of the square are AB. BC. CD. and. DA. Let that be =X cms. Take the green figure Area is 25 cm^2. One side is 3 cm. Hence the other side Namely ED and FH =3cm. Now the triangle BFG you have BG as =X-3and FG as =X-5. So in the triangle Rt angle FB^2= (x-3)^2+(x-5)^2. 100=2x^2-16x+44. So 2x^2-16x=56. By dividing by 2 .We have. X^2-8X-28=0. So in that Quadratic equation you get two values. Take the positive one as length is positive. That when multiples by X itself becomes the area.

@yalchingedikgedik8007 6 дней назад

Thanks Sir Thanks PreMath That’s very good and enjoyable Good luck with glades ❤❤❤❤❤❤❤

@prossvay8744 7 дней назад

Let x is the side of square So (x-3)^2+(x-5)^2=10^2 So x=11cm Area of the blue square=11^2=121 cm^2.❤❤❤

@PreMath 7 дней назад

Excellent! Thanks for sharing ❤️

@himo3485 7 дней назад

15/5=3 5+x=3+y y=x+2 x²+(x+2)²=10² 2x²+4x+4=100 2x²+4x-96=0 x²+2x-48=0 (x+8)(x-6)=0 x>0 , x=6 y=8 5+6=3+8=11 Blue Square area = 11*11 = 121cm²

@PreMath 7 дней назад

Excellent! Thanks for sharing ❤️

@satrajitghosh8162 6 дней назад

green rectangle is of 5 cm long and 3 cm high Side of blue square be a This implies ( a - 5)^2 + ( a - 3)^2 = 10 ^2 2 a ^2 - 16 a + 34 = 100 a^2 - 8 a + 64 = 130 a = 8 + √ (130) Hereby a^2 = 194 + 16 √ (130)

@cyruschang1904 5 дней назад

(x - 3)^2 + (x - 5)^2 = 10^2 2x^2 - 16x - 66 = 0 x^2 - 8x - 33 = 0 x = (4 + ✓49) cm = 11 cm area = (11 cm)^2 = 121 cm^2

@LuisdeBritoCamacho 7 дней назад

STEP-BY-STEP RESOLUTION PROPOSAL : 01) Lets close the Rigth Triangle creating Point C. 02) BC = X 03) AC = Y 04) AB = 10 05) 100 = X^2 + Y^2 06) Square Side = 5 + Y 07) Square Side = 3 + X 08) 5 + Y = 3 + X ; Y = X - 2 09) 100 = X^2 + (X - 2)^2 ; 100 = X^2 + X^2 - 4X + 4 ; 2X^2 - 4X + 4 = 100 ; 2X^2 - 4X - 96 = 0 ; X^2 - 2X - 48 = 0 10) Two Solutions : X = - 6 or X = 8 11) So, X = 8 and Y = 6 12) Square Side = 5 + 6 = 3 + 8 = 11 13) Area = 11^2 = 121 Therefore, OUR ANSWER : Blue Square Area equal 121 Square Centimeters.

@PreMath 7 дней назад

Excellent! Thanks for sharing ❤️

@sorourhashemi3249 7 дней назад

Easy

@PreMath 6 дней назад

Thanks for the feedback ❤️

@countysecession 7 дней назад

Knew the answer by looking. A right triangle with a hypotenuse of 10 gives it away.

@PreMath 7 дней назад

Excellent! Thanks for the feedback ❤️

@phungpham1725 7 дней назад

I agree with you! It is indeed a 6-8-10 triangle in this problem😊!

@batavuskoga 7 дней назад

I made one stupid mistake (x-3)²+(x-5)²=10² x²-6x+9=x²-10x+25=100 By adding I made the mistake 2x²-16x+36=100 instead of 2x²-16x+34=100 Buthey, this happens to all of us. We're all human beings. One of the best math channels on youtube, clear explanation. You can't say that of some other math channels

@PreMath 7 дней назад

Excellent! Glad to hear that! Thanks for the feedback ❤️🙏