Can you find area of the Blue shaded rectangle? | (Circle) |
HTML-код
- Опубликовано: 12 сен 2024
- Learn how to find the area of the Blue shaded rectangle. Important Geometry and Algebra skills are also explained: Pythagorean theorem; Circle Theorem; 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 Blue shaded rectangle? | (Circle) | #math #maths
Olympiad Mathematical Question! | Learn Tips how to solve Olympiad Question without hassle and anxiety!
#BlueRectangleArea #Rectangle #Circle #Radius #GeometryMath #PythagoreanTheorem #ThalesTheorem
#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.
Wonderful math problem and clear-cut explanation. Many thanks
Glad it was helpful!
You are very welcome!
Thanks ❤️
Just extend AB such that X is a point on the circumference. XB is the diameter or 2r. BC = PC as they are tangents from the same external point and are also equal to r(OB). Then, since triangle EAB ~ triangle XEB( AA similarity theorem, angle XEB is angle subtended by a semicircle = 90 = angle EAB and angle EBA = angle EBX). Therefore EB / XB = AB / EB. 13/2r = AB/13. AB*r = 84.5 squnits which is the area of the rectangle.
Excellent!
Thanks ❤️
That's maybe the nicest way to make it. Prefer similar triangles over Pythagoraan if possible.
I solved the same as you but I posted a little bit later because of drinking my morning coffee😀
We note that E can be anywhere on AD, but we assume can't be the same at A or D. There are two limiting cases, AE is infinitesimal or DE is infinitesimal. Let's consider AE being infinitesimal. Then OA = OB = OP = 13/2 = 6.5 BC = OP = 6.5. Area ABCD = (6.5)(13) = 84.5. Let's consider DE being infinitesimal. BP = BE = 13. ABCD is very close to a square with diagonal 13, so side lengths 13/(√2), area = (13/(√2))(13/(√2)) = (13)²/2 = 84.5. Because the problem statement implies that the area is the same regardless of where point E is on AD, we can extend either of these two special cases to the general case.
Thanks ❤️
I "cheated" by using the knowledge that there is a solution even though the only number we are given is 13, which means the solution must therefore be independent of the length AB. So if we make A coincident with O we are left with finding the area of the square OPCB which has diameter 13. Pythagoras tells us the side length of such a square is 13/sqrt(2) so the area is 169/2 = 84.5.
Thanks ❤️
It's time for some geometry:
.
..
...
....
.....
Due to the law of cosines with R being the radius of the circle we can conclude:
BE² = OB² + OE² − 2*OB*OE*cos(∠BOE) = R² + R² − 2*R*R*cos(∠BOE) = 2*R²*[1 − cos(∠BOE)]
BE²/(2*R²) = 1 − cos(∠BOE)
⇒ cos(∠BOE) = 1 − BE²/(2*R²)
From the right triangle OAE we know:
OA/OE = OA/R = cos(∠AOE) = cos(180° − ∠BOE) = −cos(∠BOE) = BE²/(2*R²) − 1
⇒ OA = BE²/(2*R) − R
⇒ AB = OA + OB = [BE²/(2*R) − R] + R = BE²/(2*R)
Now we can calculate the area of the rectangle:
A(ABCD) = AB*AD = [BE²/(2*R)]*R = BE²/2 = 169/2
Best regards from Germany
Thanks ❤️
Sir I have one idea 💡. we will do a live stream on your birthday in that stream we will solve as many problems as your age number will be. 😊😅❤❤
I'll think about it!
Thanks dear ❤️
Shouldn’t we???
Area of the blue rectangle=169/2 square unit=84.5 square units.❤❤❤ Thanks
Thanks ❤️
this is an interesting problem, because the area does not depend on the value of l2 (see "l2=" in line 20):
10 print "premath-can you find area of the blue shaded rectangle"
20 dim x(3),y(3):l1=13:l2=6.2:sw=l1/(l1+l2):r=sw:goto 50
30 xe=r-sqr(abs(l1^2-(l2+r)^2)):dgu1=(xe-r)^2/l1^2:dgu2=l2^2/l1^2:dgu3=r^2/l1^2
40 dg=dgu1+dgu2-dgu3:return
50 gosub 30
60 dg1=dg:r1=r:r=r+sw:if r>10*l1 then print "nicht gefunden":end
70 r2=r:gosub 30:if dg1*dg>0 then 60
80 r=(r1+r2)/2:gosub 30:if dg1*dg>0 then r1=r else r2=r
90 if abs(dg)>1E-10 then 80
100 print r:ages=r*(l2+r):print "die gesuchte flaeche=";ages
110 x(0)=0:y(0)=0:x(1)=r:y(1)=0:x(2)=x(1):y(2)=r+l2:x(3)=0:y(3)=y(2)
120 xm=r:ym=r:masx=1200/2/r:masy=900/(l2+r):if masx
run in bbc basic sdl and hit ctrl tab to copy from the results window.
run in b
Extend length of rectangle inside circle (AB) up to intercept the circle; call that point P. The length of that extension AP = 2r - y
Extend top of rectangle (AD) to the right to intercept the circle; call that point Q. The length of that extension AQ = AE = √(13² - y²).
Using the intersecting chords theorem on chords BP and EQ intersecting at A:
(AQ)(AE) = (AP)(AB)
[ √(169 - y²) ]² = (2r - y) y
169 - y² = 2ry - y²
169 = 2ry
A = ry = 169/2 = 84.5. Done!!
Thanks ❤️
Wonderful watching the answer appear at the end.
Glad to hear that!
Thanks ❤️
I did it this way:
Extend line AB to meet the circle at the top of the diagram at point F and let length AB = a and AF = b.
Now, looking at △BEF and △ABE, let ∠AEF = α and ∠AFE = β. As ∠EAF = 90°, ∴ α + β = 90°.
However, ∠BEF = 90°(Thales' Theorem) and as ∠AEF = α, ∴ ∠AEB = β, as α + β = 90°.
∴ ∠ABE = α and △BEF ~ △ABE (Angle-Angle-Angle).
∴ AB/BE = BE/BF, and AB x BF = BE².
As AB = a and AF = b, ∴ BF (diameter of circle) = a + b.
AB x BF = a(a + b) and BE² = 13 x 13 = 169, so a(a + b) = 169.
As BF is the diameter of the circle with length a + b, therefore the radius, being half the diameter of the circle is equal to length (a + b)/2.
Now, length BC is equal to the length of the radius, and ∴ BC = (a + b)/2.
The area of the rectangle is given by AB x BC, which, in this case, is equal to [a(a + b)]/2.
As a(a + b) = 169, ∴ [a(a + b)]/2 = 169/2.
∴ Area [ABCD] = 169/2 units².
Thanks ❤️
Fabulous problem ❤, premath never Dissapoints us
Glad to hear that!
Thanks ❤️
Extend AB intersecting the circle at point F. We have: the 2 triangle BAE and BEF are similar so BE/BA=BF/BE----> 13/BA=BF/13---> BA.BF= sq 13
because BA = length of the rectangle and BF= 2R= 2 width of the rectangle
----> 2R. BA= 169 ----> Area =169/2= 84,5 sq units
Thanks ❤️
What a beautiful problem and a clear resolving ❤
Glad to hear that!
Thanks ❤️
+13)^2=169° (169°-360°)=√191 √1√9^√1 ° √1^1. √3^3 1 √ 7:21 1^3 1 31 (x+1x-3)
Let's use an adapted orthonormal. B(0;0) C(R;0) O(0;R) A(c,0) D(R,c) where R is the radius of the circle and c the hight BA of the rectangle.
The equation of the circle is x^2 = (y-R)^2 = R^2, or x^2 + y^2 - 2.R.y = 0 when developped. The equation of (AD) is y = c.
At the intersection of the circle with (AD) we have x^2 + c^2 - 2.R.c = 0, so x^2 = 2.R.c - c^2 and point E(sqrt(2.R.c - c^2); c).
Then we have BE = (2.R.c - c^2) + (c^2) = 2.R.c. Knowing that BE =13, we now have 2.R.c = 13^2 = 169, and R.c = 169/2 which is the area of the rectangle ABCD.
Thanks ❤️
At the beginning of line 4, please read BE^2 = ... instead of BE = ...
Prolongada BA, corta a la circunferencia enl F
Área ABCD =r*AB
Razón de semejanza entre los triángulos FEB y AEB, s=13/2r → AB=EB*s =13*13/2r =169/2r
ABCD =r*169/2r =169/2 =84.5
Un saludo cordial.
Thanks ❤️
Draw a radius that is perpendicular to EB and intersects it at point Q, it will cut it in two equal halves of length 13/2=6.5 , EQ and QB. Now OQB is similar to EAB, so y/13=6.5/r, therefore yr=13*6.5=84.5 units square.
Thanks ❤️
Not having constraints abot point E we can overlap it on point A and then diameter=13
Area = 13*13/2= 169/2
Thanks ❤️
+13)^2=169° (169°-360°)=√191 √1√9^√1 ° √1^1√3^3 1 √ 1^3 1 31 (x+1x-3)
Thank you!
You are very welcome!
Thanks ❤️
Neat. It's worth noting that the right triangle EAB with hypotenuse 13 is a Pythagorean triple, so the other two sides have to be 5 and 12, but this method gets around even having to solve for that.
Thanks ❤️
EAB can be a (5, 12, 13) a Pythagorean triangle, but it can also be any right triangle whose hypotenuse is 13.
For example, it can be a 30° - 60° - 90° triangle where AE = BE/2 or a 45³ - 45³ - 90° triangle where AE = BE/√2.
13 as Diameter of a Circle there can be many pairs of Rectangle sides ( Angle on perifary of Circle with Diameter as Hypotenuse.
Very difficult, no idea after looking at the figure for minutes😢 A special chord of length 13, 5, 12, 13, r=7?
Thanks ❤️
And how could wie find the Radius?
The Area of the Blue Rectangle is equal to:
11,628 li un * 7,268 li un = 84,5123 sq un
P.S. - I we'll be back later to explain my rational!!!
This is tougher than I expected.
Yes!
Thanks ❤️
🙂
Thanks ❤️
This appears to be a minimal information problem where we can still get the area without solving the sides. I think the area 84.5.
Excellent!
Thanks ❤️
If x is top side of rectangle then how can x is one side of triangle
Thanks ❤️
Try again, let the rectangle be r×h, then 13/h=(2r/13), 2rh=169, rh=84.5😂, I get it, but no information about r and H at all😮, we can just determine the area of this special rectangle. 😮
Thanks ❤️
Hindi me bol le English me to tere se bola jaa nahi raha hai 😏
Very difficult!!!
АО = х. ОН ⟂ ВЕ. ▲ОНВ ~ ▲АВЕ. (r + x)/13 = 6,5/r. r(r + x) = 13 × 6,5. S(ABCD) = r(r + x) = 84,5.
Thanks ❤️