Calculate area of the Yellow shaded Triangle | Blue area is 8 | Important Geometry skills explained

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

Calculate area of the Yellow shaded Triangle | Blue area is 8 | Important Geometry skills explained

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

#FindAreaOfYellowTriangle #AreaOfTriangle #GeometryMath #MathOlympiad #RightTriangle
#OlympiadMathematicalQuestion #HowToSolveOlympiadQuestion #MathOlympiadQuestion #MathOlympiadQuestions #OlympiadQuestion #Olympiad #AlgebraReview #Algebra #Mathematics #Math #Maths #MathOlympiad #HarvardAdmissionQuestion
#MathOlympiadPreparation #LearntipstosolveOlympiadMathQuestionfast #OlympiadMathematicsCompetition #MathOlympics #CollegeEntranceExam
#blackpenredpen #MathOlympiadTraining #Olympiad Question #GeometrySkills #GeometryFormulas #InscribedCircle #PythagoreanTheorem #SimilarTriangles #CircleTheorem #Tangents
#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
Olympiad Question
Geometry
Geometry math
Geometry skills
Right triangles
imo
Competitive Exams
Competitive Exam
Inscribed square
Pythagorean Theorem
Similar triangles
Right Triangle
Square area formula
Triangle area
Area of Triangle formula

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

[bigm]

Thanks for the recent collection of fun problems!❤❤🥂😀😊

[montynorth]

The 2 yellow triangles in the top half of the rectangle are 8 each = 16.
The yellow triangle in the lower part has the same area since both the base length formed by the 2 upper triangles and their heights of 1 square length are the same.
Thus 16 x 2 = 32.

[bluejeep]

Thinking outside the box, a simple approach is to extend the square grid one layer below and then extend (duplicate and reverse) the yellow triangle to create a rectangle comprising double the size of the yellow triangle. That rectangle contains 8 small triangles of 8cm2, for a total of 64cm2. The yellow triangle is half that, ie. 32cm2.

[sngmn]

Pick's formula: A = I + B/2 - 1
I - inner nodes count
B - boundary nodes count
A - area of our shape

I = 1
B = 4
A = 1 + 4/2 - 1 = 2 s. u.
1 s. u. = 8 + 8 = 16 cm^2
A = 2 * 16 = 32 cm^2

[theoyanto]

Very good example, as it turned out it was mental arithmetic, love it

[nineko]

Nice one, but you did it in an extremely roundabout way, it's much simpler to realise that there are four yellow triangles with the same base and the same height as the blue one, so it's really just 4 • 8 = 32 :)

[jonathanjose]

So the side of each square is 4, I completed the rectangle, area of the rectangle minus the unshaded triangles, and got 32 Square units as well

[misterenter-izrz]

Clearly the side of square is 4, the answer is the difference of done.😍

[kenb]

Consider the area of the bottom two yellow triangles. Moving the bottom right vertex horizontally won't change the this area since it's base/height remain unchanged. Now move the bottom right vertex so it mirrors the above two triangles, and you got 4 identical triangle, each with an area of 8 cm^2, total 32 cm^2.

[murdock]

Nice and awesome, many thanks!
a^2 = 16 → 6a^2 - (a^2/2)(1 + 4 + 3) = 32

[TurquoizeGoldscraper]

I took a point D at the mid-point of B and C and connected A to D. This gives me two triangles ABD and ACD, each have a base of 8 and a height of 4. Both triangles are 1/2 * 8 * 4 = 16, so that adds up to 32.

[spiderjump]

Side of each of the small squares= square root of (2x8)=4
Triangle can divided into 2 triangles with base of 8 and height of 4
So area = 2(1/2x8x4)
=32

[quentin]

Hi,
It was also possible without Pythagorean theorem and without square roots:
Top area : 8 + 8 = 16 cm²
For the bottom area, we act as if half of the right square were colored in yellow, and then remove it.
Bottom area :
((((3*4)*4)/2)-8)
= 24 - 8 cm²
= 16 cm²
Top area + Bottom area = 16 + 16 = 32 cm²
Greetings from France

[pwmiles]

Draw in the bottom left hand square making 6 squares total
Area of 6 squares = 2x8x6 = 96 cm^2
Areas not occupied by yellow triangle = 16/2 + 4*16/2 + 3*16/2 = 64 cm^2
Area of yellow triangle = 96 - 64 = 32 cm^2

[devondevon]

32
Since the yellow is a right triangle (from two 45-degree angles) and since the diagonal of each square is
sqrt 32 ( the square root of 4^2 + 4^2), the height of the yellow triangle is 2 sqrt 32, and the base
is sqrt 32.
Hence the area is sqrt 32*sqrt 32*2 / 2 = 32 Answer

[SladeMacGregor]

This problem was fun! Thanks Professor!

[philipkudrna]

5 sec: half a Square is 8 cm^2. The two top left parts are 8+8. if you take the small triangle in the bottom left and use it to complete the lower right half-triangle, you get another 8. what remain is the center bottom square. The yellow area by symmetry is also half of the square, ie also 8. this means the area is 4 times 8 or 32 cm^2!

[EnnioPiovesan]

If AB is extended beyond A until it meets the straight line passing through C at P, a triangle CBP is obtained, which is half of a square, formed by 2 full squares and 4 half squares, for a total area of ​​4 squares, i.e. 64 cm². From this we must subtract the area of ​​the triangle CAP which is (called l the side of each square) (1/2) (4·l) l = 2 l² = 32 cm².
The searched area is then 32 cm².

[balki]

Great video. Pick’s theorem is a great way to solve it fast

[GetMatheFit]

Blue triangle: A(Blue)=base*hight/2 = 8
Top triangle: = 16
Bottom triangle: = 16
Yellow triangle: = 4*A(Blue) = 4*8= 32
LG Gerald