Two Methods! | Calculate area of the Yellow shaded rectangle | Important Geometry skills explained
Тәжірибелік нұсқаулар және стиль
Learn how to find the area of the Yellow shaded rectangle. Important Geometry and algebra skills are also explained: 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
• Two Methods! | Calcula...
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
Two Methods! | Calculate area of the Yellow shaded rectangle | Important Geometry skills explained
Olympiad Mathematical Question! | Learn Tips how to solve Olympiad Question without hassle and anxiety!
#FindAreaOfYellowRectangle #AreaOfRectangle #GeometryMath #MathOlympiad #Square #Rectangle
#PreMath #PreMath.com #MathOlympics #HowToThinkOutsideTheBox #ThinkOutsideTheBox #HowToThinkOutsideTheBox? #FillInTheBoxes #GeometryMath #Geometry #RightTriangles #AreaOfRectangle
#OlympiadMathematicalQuestion #HowToSolveOlympiadQuestion #MathOlympiadQuestion #MathOlympiadQuestions #OlympiadQuestion #Olympiad #AlgebraReview #Algebra #Mathematics #Math #Maths #MathOlympiad #HarvardAdmissionQuestion
#MathOlympiadPreparation #LearntipstosolveOlympiadMathQuestionfast #OlympiadMathematicsCompetition #MathOlympics #CollegeEntranceExam
#blackpenredpen #MathOlympiadTraining #Olympiad Question #GeometrySkills #GeometryFormulas #AreaOfTriangles #AreaOfRectangle #Length #Width
#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
Square
imo
Competitive Exams
Competitive Exam
Area of a blue square
premath
premaths
Area
Area of a rectangle
Subscribe Now as the ultimate shots of Math doses are on their way to fill your minds with the knowledge and wisdom once again.
Пікірлер: 60
This channel deserves much more attention considering how much work goes into it. Cheers and thank you for sharing friend
@bigm383
10 ай бұрын
Absolutely!😀
@PreMath
10 ай бұрын
Much appreciated! So nice of you. Thank you! Cheers! 😀
Height of rectangles = y. Area of yellow rectangle = A. A + 80 = 32y. A + 176 = 40y. Dividing :- ( A + 176 ) / ( A + 80 ) = 40y / 32y = 5 / 4. Cross multiplying :- 5 ( A + 80 ) = 4 ( A + 176 ) 5A + 400 = 4A + 704. A = 304.
@PreMath
10 ай бұрын
Thank you! Cheers! 😀
@williamlyerly3114
10 ай бұрын
Think this soln is most obvious. No x or y Calc. Or if drop out A then calc y directly. Substitute to get A.
@jhill4874
10 ай бұрын
Exactly what I did. Cheers.
The second solution is out of this world...Amazing!
Love your outside the box method! Super!
@PreMath
10 ай бұрын
Glad you like it!
Thanks PreMath . We are enjoying and learning solve Math. Exercise .
@PreMath
9 ай бұрын
Glad to hear that
تمرين جيد جميل .رسم واضح مرتب. شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويحميكم ويرعاكم جميعا . تحياتنا لكم من غزة فلسطين.
@PreMath
10 ай бұрын
Thanks dear🙏
Easy and refreshing geometry problem
@PreMath
10 ай бұрын
Great! Thank you! Cheers! 😀
(176-80)/(40-32)=12=Altura → [32-(80/12)]x12=304 =Área amarilla Gracias y saludos.
Very nice solution!🥂👍❤️
@PreMath
10 ай бұрын
Awesome! So nice of you. Thank you! Cheers! 😀
Fantastic alternative method at the end.
@PreMath
10 ай бұрын
Glad you liked it Thank you! Cheers! 😀
Thanks for video.Good luck sir!!!!!!!!
@PreMath
10 ай бұрын
Thank you too
thankyou so much dear ❤
@PreMath
10 ай бұрын
You are so welcome! So nice of you, dear Thank you! Cheers! 😀
Thank you
@PreMath
9 ай бұрын
You're welcome
You are wonderful my friend❤
@PreMath
10 ай бұрын
You are the best, dear So nice of you. Thank you! Cheers! 😀
great again!
@PreMath
10 ай бұрын
So nice of you. Thank you! Cheers! 😀
Most of us came up with 1st Method.. But the 2nd one is mind blowing.
@PreMath
10 ай бұрын
Glad you liked it
I got to xy-40y=80 and xy-32y=176. A quick inspection told me that 8y=96 so y=12. Exceelent channel. The help is appreciated.
@PreMath
10 ай бұрын
So nice of you. Thank you! Cheers! 😀
Hi Premath I like your explanation second method is awesome I like your channel
@PreMath
10 ай бұрын
So nice of you. Thank you so much 🙂
I just mentally calculated that green had 8 cm longer base and 96 cmsq more area so ht is 12. Width of blue is 80/12 and that is 6 2/3, gr base is 14 2/3. Width of yellow is 40-14 2/3 or 25 1/3. 25 1/3 base times 12 ht is 304.
بفرض عرض المستطيل الكلي y 32+176/y=40+80/y 176/y_80/y=40_32 96/y=8 y=12 S+176=40×12 S=304
Very nice, sir. :::::::: Let the area of yellow rect be A 80+A = 32y.......(1) 176+A =40y......(2) Divide (2) by (1) and get for A.
@PreMath
10 ай бұрын
So nice of you. Thank you! Cheers! 😀
Nice to see different methods for solving these problems. Here's a third method: Let A = area of center (yellow) rectangle Combined area of yellow and blue rectangles = A + 80 (with base = 32) Combined area of yellow and green rectangles = A + 176 (with base = 40) Since these rectangles have same height, the ratio of their areas = ratio of their bases: (A + 80) / (A + 176) = 32 / 40 (A + 80) / (A + 176) = 4 / 5 5A + 400 = 4A + 704 5A - 4A = 704 - 400 A = 304
@PreMath
10 ай бұрын
Thanks for sharing!
I solved the problem as Method #1 in the video. I missed the second method. I did get the answer.
@PreMath
10 ай бұрын
Great! Thank you! Cheers! 😀
(32-x)y=80 (40-x)y=176 80/(32-x)=176/(40-x) 3200-80x=5632-176x 96x=2432 x=76/3 y=12 area of Yellow Rectangle : xy = 76/3*12 = 304
@PreMath
10 ай бұрын
Thank you! Cheers! 😀
area = A width = l so... 32 x l = A + 80 .............(1) 40 x l = A + 176 .............(2) (2) - (1) = 176 -80 = (40-32) x l l = 96/8 = 12 1/ A = (32 x 12) - 80 = 304 cm² or... 2/ A = (40 x 12) - 176 = 304 cm²
Let H be the height & X be the width of yellow rectangle. H*(32 - X) = 80 & H*(40 - X) = 176. H = 12. HX is the yellow area = 32*12 - 80 = 304.
@PreMath
10 ай бұрын
Thank you! Cheers! 😀
3 methode: 80/12=6 2/3. 176/12=14 2/3. 32*12=384. 384-80=304. 304/12=25 1/3. 25 1/3 + 14 2/3 = 40 so correct.
Let A be the area, A+80/A+176=32/40=4/5, so 5A+400=4A+704, then A=304.😃
@PreMath
10 ай бұрын
Thank you! Cheers! 😀
2nd method is quicker and easy
Ar=196
X=24cm
Traditional method of solving equated equations is really defining a past participle. The Woke version of solving we come out of the darkness of the box and wallah!...Boom! ,.... we see the light and readily come up with the answer. 🙂
@PreMath
10 ай бұрын
Thanks for sharing!
Green Area - Blue area = 176 - 80 = 96 cm² has a base of : b= 40 - 32 = 8 cm Its area is A = b.H. , then H = A/b = 92 / 8 H = 12 cm Yellow area: B = 32 - 80/12 (or 40-176/12 ) A = B. H = (32-80/12).12 A = 304 cm² ( Solved √ )
Blue rectangle Area = 80 = b.H b = 80 / H Green rectangle: Area = 176 = B.H B = 176 / H Total Base of larger rectangle: 40.+ (80/ H) = 32 + (176 / H) 40 - 32 = 176 / H - 80 / H 8 = 96 / H H = 96 / 8 H = 12 cm Bt = 40 + 80/12 = 46,66 cm Total Area : At = Bt H = 46,66 . 12 At = 560 cm² Yellow area: Aw = At - 80 - 176 Aw = 304 cm² ( Solved √ )