Can you find area of the Green shaded Rectangle? | (Quarter Circle) |
HTML-код
- Опубликовано: 7 сен 2024
- Learn how to find the area of the Green shaded Rectangle. Important Geometry and Algebra skills are also explained: Pythagorean theorem; area of the triangle 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 area of t...
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 area of the Green shaded Rectangle? | (Quarter Circle) | #math #maths | #geometry
Olympiad Mathematical Question! | Learn Tips how to solve Olympiad Question without hassle and anxiety!
#FindGreenRectangleArea #QuarterCircle #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.
Fabulous sir🙋🏻♂️thank you for this great video🙏🏼👍🏼
My pleasure😀
Thanks for the feedback ❤️
1.46
Let the A and B be the length of the two legs of the triangle
Then Area = 5= 1/2 A*B
5=1/2 AB
10 =AB
and A= 10/B (solving for A) Equation F
Since the hypotenuse = 5 units, then
A^2 + B^2 = 5^2
A^2 + B^2 = 25 Equation G
(10/B)^2 + B^2 = 25 substituting Equation F into Equation B
100/B^2 + B^2 = 25
100 + B^4 = 25B^2 (multiply both sides by B^2)
Let B^2 = N (Introduce a new variable N)
then
100 + N^2 = 25 N
N^2 -25 + 100 = 0
(N-20)(N-5) factor
N=20 , and N= 5
Hence, B = sqrt 20 and B= sqrt 5
or B =4.4721359 and B= 2.236067977
4.4721359 is the longer leg of the triangle, and 2.236067977 is its shorter leg
Hence, the difference between the longer leg and 5 is the WIDTH of the small GREEN RECTANGLE, or 5 - 4.472136 = 0.527864
Hence, the difference between the shorter leg and 5 is the LENGHT of the small GREEN RECTANGLE or 5 - 2.2360678 =2.7639320
Hence, 0.527864 times 2.7639320 is the Area of the GREEN RECTANGLE.
0.527864 * 2.7639320 = 1.458 or 1.46 Answer
If you call length of green rectangle x then ed=5-x and since area of triangle eod is 5 then od=10/(5-x).Using pythag. You get an equation that gives 2 values for x,one being valid ie x=5- sqr5. da= 5-2sqroot 5.So green area= (5-sqr5)(5-2sqr5)=35-15sqr5.
Excellent!
Thanks for sharing ❤️
Instead of labelling the triangle sides as a & b, I labelled the rectangle sides a & b. So my triangle sides were 5-a & 5-b.
Theoretically, this should still work, but the math was getting nowhere.
great video!
Great job!
Thanks for the feedback ❤️
I used eqns 1 & 2 to find Base & Height:
H=10/B
B²+(10/B)²=(5)²
B²+(100/B²)=25
B²+(100/B²)-25=0
B⁴-25B²+100=0
(B²-5)(B²-20)=0
B=sqrt(5); 2sqrt(5)
Area=
(5-sqrt(5))•(5-2sqrt(5))
=25-5sqrt(5)-10sqrt(5)+10
=35-15sqrt(5)
Excellent!
Thanks for sharing ❤️
Let the unlabeled vertex of the green rectangle on BC be M and the unlabeled vertex on AB be N. Let ∠EOD = θ. As OE is a radius of quarter circle O, OE = OC = 5. This means that OD = OEcos(θ) = 5cos(θ) and DE = OEsin(θ) = 5sin(θ).
Triangle ∆ODE:
Aᴛ = OE(OD)sin(θ)/2
5 = 5(5cos(θ)sin(θ)/2
5sin(θ)cos(θ) = 2
2sin(θ)cos(θ) = 4/5
sin(2θ) = 4/5
sin(2θ) = O/H ==> O = 4, H = 5
cos(2θ) = A/H = (√H²-O²)/H
cos(2θ) = (√5²-4²)/5 = (√25-16)/5 = √9/5 = 3/5
sin(2θ/2) = √((1-cos(2θ))/2)
sin(θ) = √((1-3/5)/2) = √((2/5)/2) = 1/√5
DE = 5sin(θ) = 5(1/√5) = √5
OD² + DE² = OE²
OD² + (√5)² = 5²
OD² = 25 - 5 = 20
OD = √20 = 2√5
As OABC is a square and BMEN is a rectangle, then NB = ME = MD-DE = 5-√5 and BM = EN = OA-OD = 5-2√5.
Rectangle BMEN:
Aᵣ = wh = (5-√5)(5-2√5)
Aᵣ = 25 - 10√5 - 5√5 + 10
Aᵣ = 35 - 15√5 ≈ 1.459 cm²
Excellent!
Thanks for sharing ❤️
Nice! Appreciate how the algebraic identity was used to solve the problem.
Glad you liked it!
Thanks for the feedback ❤️
That’s very nice and understandable
Thanks Sir for this explain
With my respects
❤❤❤❤❤
Good luck
Let width=b ; Length=a
Area of rectangle=ab
Connect E to F
OD=EF=5-a
DE=OF= 5-b
Area of triangle=1/2((5-a)(5-b)=5
25-5a-5b+ab=10
ab-5(a+b)+15=0
ab=5a+5b-15 (1)
In ∆OEF
(5-a)^2+(5-b)^2=5^2
a^2+b^2-10(a+b)+50=25
(a+b)^2-2ab-10(a+b)+25=0
Let ab=x ; a+b=y
So (1) x=5y-15
And (2) y^2-2x-10y+25=0
so y^2-2(5y-15)-10y+25=0
So y=10+3√5 >5 reject ; y=10-3√5
x=50-15√5-15=35-15√5
ab=35-15√5
a+b=10-3√5
b=10-3√5-a
a(10-3√5-a)=35-15√5
a=5-2√5 ; a=5-√5
b=10-3√5-5+2√5=5-√5
b=5-2√5
So Green rectangle area=(5-√5)(5-2√5)=35-15√5 cm^2=1.46 cm^2❤❤❤
Excellent!
Thanks for sharing ❤️
Sistema ecuaciones: b*h=10 ; b²+h²=5²→ b=√5 ; h=2√5→ Área verde = (5-√5)*(5-2√5)=35-15√5 =1,45898....
Gracias y saludos.
Excellent!
Thanks for sharing ❤️
1.46
It would have been much easier if the area of the triangle was 6 (instead of 5) as we would avoid having to deal with decimals as
it would be a 3-4-5 right triangle, and the area of the triangle (at the top of the circle ) would be 2 * 1 =2
since 5-4=1 and 5-3 = 2. You still would deal with the same concept and must understand it, to solve the problem, but avoid dealing with radical or decimals
No other idea. It's very simple.
We can add that a =sqrt(5) and b = 2.sqrt(5), but it is of no real need for the final result.
Thanks for the feedback ❤️
Agreen=xy..x+b=5,y+h=5,bh=10...A=(5-b)(5-h)=25-5(b+h)+10=35-5(b+h)...5(b+h)=35-A...(^2)..25(b^2+h^2+2bh)=1225-70A+A^2...25(25+20)=1225-70A+A^2..A^2-70A+100=0...A=35-√1125=35-15√5
Excellent!
Thanks for sharing ❤️
Solution:
a = horizontal side of the right-angled blue triangle,
b = vertical side of the right-angled blue triangle.
Because of the area of the right-angled blue triangle, the following must apply:
(1) a*b/2 = 5 |*2/a ⟹ (1a) b = 10/a |in (2) ⟹
In addition, the Pythagorean theorem applies to the right-angled blue triangle:
(2) a²+b² = 5²
(2a) a²+100/a² = 25 |with a² = x ⟹
(2b) x+100/x = 25 |*x ⟹
(2c) x²+100 = 25*x |-25x ⟹
(2d) x²-25x+100 = 0 |p-q formula ⟹
(2e) x1/2 = 25/2±√(25²/4-100) = 25/2±1/2*√(625-400) = 25/2±1/2*√225 = 25/2±1/2*15 ⟹ (2f) x1 = 25/2+1/2*15 = 20 and (2g) x2 = 25/2-1/2*15 = 5 ⟹ 1. Case: a1² = x1 = 20 |√() ⟹ a1 = √20 = 2*√5 |in (1a) ⟹ (1b) b1 = 10/(2*√5) = √5 2nd case: a2² = x2 = 5 |√() ⟹ a2 = √5 |in (1a) ⟹ (1c) b2 = 10/√5 = 2*5/√5 = 2*√5
1st case:
Area of the green rectangle = (5-2*√5)*(5-√5) = 25-5*√5-10*√5+10 = 35-15*√5 ≈ 1.4590
2nd case:
Area of the green rectangle = (5-√5)*(5-2*√5) = 25-10*√5-5*√5+10 = 35-15*√5 ≈ 1.4590
a^2+b^2=25, a×b=10, to find (5-a)(5-b)=25-5(a+b)+10=35-5(a+b), (a+b)^2=25+20=45, thus the answer is 35-5sqrt(45)=5(7-3sqrt(5)).😊
Excellent!
Thanks for sharing ❤️
Great question as always! 👍👍
Glad you think so!
Thanks for the feedback ❤️
Great explanation.thank you sir.
Glad you liked it
You are very welcome!
Thanks for the feedback ❤️
ab/2 = 5 → ab = 10 → b = 10/a → a^2 + (a/10)^2 = 25 → k = a^2 = (1/2)(25 ± 15) → a = √5 → b = 2√5 →
green rectangle area = (5 - a)(5 - b) = 5(7 - 3√5)
Excellent!
Thanks for sharing ❤️
I decided to take Euclids theory since it was a no brainer that the hight is 2. I determined that the hight divided c in 4 and 1. Now I could calculate both sides of the triangle to be sqr of 5 and sqr of 20. So 5 - root 5 times 5 minus root 20 (2 times root 5) is the area were looking for and results in 35 minus 15 times root5😊
xy = 10
x^2 + y^2 = 5^2 = x^2 + (10/x)^2 = 25
x^4 - 25x^2 + 100 = 0
(x^2 - 5)(x^2 - 20) = 0
(x, y) = (✓5, 2✓5) or (2✓5, ✓5)
green area = (5 - ✓5)(5 - 2✓5) = 35 - 15✓5
If you set the area of the right triagle as (ODE)= 6 cm^2 then the anser is more simple (2 cm^2)
OE=5 5=(√5)² 1 : 2 : √5 OD=x DE=2x
x*2x/2=5 2x²=10 x=√5
Green Rectangle area = (5-√5)*(5-2√5) = (35 - 15√5)cm²
Excellent!
Thanks for sharing ❤️
S=5(7-3√5)≈1,4
A≈1.45898 !!! 😬
Excellent!
Thanks for sharing ❤️
3² + 4² = 5² The sides of the rectangle are 1 and 2 with a perimeter of 6 because 4 plus 1 equals 5 and 3 plus 2 equals 5 so (5-4)(5-3) is the area 1 times 2
Thanks for the feedback ❤️
Let's find the area:
.
..
...
....
.....
The triangle ODE is a right triangle, so we can apply the Pythagorean theorem:
OD² + DE² = OE²
DE² = OE² − OD²
DE = √(OE² − OD²)
The area of this triangle can be calculated as follows:
A(ODE) = (1/2)*OD*DE
2*A(ODE) = OD*DE
2*A(ODE) = OD*√(OE² − OD²)
4*A²(ODE) = OD²*(OE² − OD²)
4*(5cm²)² = OD²*[(5cm)² − OD²]
100cm⁴ = (25cm²)*OD² − OD⁴
OD⁴ − (25cm²)*OD² + 100cm⁴ = 0
OD² = (25/2)cm² ± √[(25/2)²cm⁴ − 100cm⁴]
OD² = (25/2)cm² ± √[(625/4)cm⁴ − (400/4)cm⁴]
OD² = (25/2)cm² ± √[(225/4)cm⁴]
OD² = (25/2)cm² ± (15/2)cm²
First solution:
OD² = (25/2)cm² + (15/2)cm² = (40/2)cm² = 20cm²
⇒ OD = (2√5)cm
⇒ DE = √[(5cm)² − 20cm²] = √(25cm² − 20cm²) = (√5)cm
Second solution:
OD² = (25/2)cm² − (15/2)cm² = (10/2)cm² = 5cm²
⇒ OD = (√5)cm
⇒ DE = √[(5cm)² − 5cm²] = √(25cm² − 5cm²) = (2√5)cm
Now we are able to calculate the area of the green rectangle. For both solutions we obtain:
A(green rectangle)
= (OA − OD)*(AB − DE)
= [5cm − (2√5)cm]*[5cm − (√5)cm]
= (25 − 5√5 − 10√5 + 10)cm²
= (35 − 15√5)cm²
≈ 1.459cm²
Best regards from Germany
Excellent!
Thanks for sharing ❤️
STEP-BY-STEP RESOLUTION PROPOSAL :
01) OD = X cm
02) DE = Y cm
03) OD * DE = X * Y = 10; and
04) X^2 + Y^2 = R^2 ; X^2 + Y^2 = 25
05) Solving a System of Two Nonlinear Equations with Two Unknowns :
06) X * Y = 10 and X^2 + Y^2 = 25
07) Positive Solutions :
08) X = sqrt(5) ; X ~ 2,24 cm
09) Y = 2 * sqrt(5) ~ 4,45 cm
10) Sides of Green Rectangle:
11) Length (L) = (5 - sqrt(5)) cm
12) Height (h) = (5 - 2*sqrt(5)) cm
13) Green Triangle Area = (L * h) sq cm
14) Green Triangle Area = [(5 - sqrt(5)) * (5 - 2*sqrt(5))] sq cm
15) Green Triangle Area = (35 - 15*sqrt(5)) sq cm
Thus,
OUR BEST ANSWER :
Green Rectangle Area is equal to (35 - 15*sqrt(5)) Square Centimeters or approx. equal to 1,46 Square Centimeters.
Greetings from The Great Land of Islam!! A Land of Wisdom, Peace and Knowledge.
Excellent!
Thanks for sharing ❤️🙏
The triangle is 3 x 4 x 5. 5 - 3 =2; 5 - 4 =1; area = 2 x 1 = 2
Thanks for the feedback ❤️
Ans; 15(3 - _/5) square Unit
My way of solution ▶
A(ΔODE)= 5 cm²
let OD= a
DE= b
⇒
ab/2= 5
ab= 10
r = 5 cm
EO= 5 cm which is equal to c
(a+b)²= a²+2ab+b²
a²+b²= c²
c= 5
c²= 25
⇒
(a+b)²= 25+2*10
(a+b)²= 45
a+b= √45
a+b= 3√5
a*b= 10
b= 10/a
⇒
a+10/a= 3√5
a²+10= 3√5a
a² - 3√5a+10=0
Δ= 45-4*1*10
Δ= 5
a₁= (3√5+√5)/2
a₁= 2√5
b₁= 10/2√5
b₁= √5
a₂= (3√5-√5)/2
a₂= √5
b₂= 10/√5
b₂= 2√5
⇒
it is irrelevant if we take a as a₁ or a₂:
a= 2√5
b= √5
Arectangle= (r-a)*(r-b)
= (5-2√5)*(5-√5)
= 25-5√5-10√5+2*√5*√5
= 25+10-15√5
= 35 -15√5
Arectangle= 5(7-3√5) cm²
Arectangle ≈ 1,46 cm²
I used the same method 😂
Thanks for the feedback ❤️