Calculate Area of the Blue shaded Circle | Learn these simple Geometry Tools fast | Math Olympiad

Need help with finding the area of a Semicircle? 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 Blue shaded Circle | Learn these simple Geometry Tools fast | Math Olympiad

#AreaOfCircle #FindAreaOfBlueShadedCircle #GeometryMath #OlympiadPreparation #Area
#CollegeEntranceExam #OlympiadMathematics #PythagoreanTheorem
#FindAreaOftheshadedSemicircle #HowtofindtheAreaof Semicircle #PythagoreanTheorem #ASA #AreaOfCircle #AreaofTheCircle #RightTriangle #TrigonometricRatios #Radius #Square#IsoscelesTriangle #IsoscelesTriangles #Rectangle #Circle #Radius
#blackpenredpen #OlympiadMathematics

How to Find the Area of a Circle
How to Find the Area of a Semicircle
How to Find the Area of a SemiCircle Step by Step
How to solve the hard geometry problems
How to find area of shaded region
blackpenredpen
Olympiad Mathematics
pre math
Po Shen Loh
Find area of the Blue shaded Circle
imo
Competitive exams
college entrance exam
Pythagorean Theorem

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

[-wx--]

Horizontal chord from A and vertical diameter gives (by symmetry and intersecting chords theorem) 3·(2r−3) = (r−6)², which resolves to the same quadratic equation. Root r = 3 corresponds to degenerate case when red rectangle is a half of the square.

[MWorl]

There’s a lot of people making math videos on YouTube and after some time of watching different people, I think you might be the best.

[natewright]

It’s interesting how two different thought processes lead to the same equation.

For this problem I plotted the circle in the x-y plane so the bottom left of the square is the origin.
Then by symmetry and the given rectangle, we know one of the points of the circle is (6, 3)
We also see the circle is shifted to the right of the origin by r and also up by r.

So the equation for the circle would be (x-r)^2 + (y-r)^2 = r^2
Using the point (6, 3) leads to the same quadratic equation for r which gives r=3 and r=15. Rejecting 3 for the same reason and leading to 225pi for the area.

[claudiogrossi]

Un problema apparentemente irrisolvibile brillantemente risolto. Complimenti!

[chwnam]

This one is easy. I solved! Thank you!

[murdock]

Nice and awesome, many thanks, Sir!
sin⁡(φ) = (a - 6)/a
cos⁡(φ) = (a - 3)/a
sin^2(φ) + cos^2(φ) = 1 → (a - 6)^2 + (a - 3)^2 = a^2 → a = 15 → πa^2 = 225π
btw: sin⁡(φ) = 3/5 → ∆ = pyth. triple (9-12-15)

[devapriyaguharoy]

Very inspiring ! Would like to have more with brain teasing stuffs.

[n.]

Think you! It's the beauty and simple problem!

[alexundre]

Boa Tarde
Obrigado pelos Ensinamentos
Deus Lhe Abençoe

[HappyFamilyOnline]

Great explanation👍
Thanks for sharing😊😊

[zsoltszigeti]

It's a bit easier with coordinategeometry: we are looking for a circle with one point at 6;3. So the circle is (x-r)^2+(y-r)^2=r^2, and x=6, y=3. We have to find r: (6-r)^2+(3-r)^2=r^2 => r^2-18r+45=0. We have two solutions: 3, 15; 3 rejected, the radius is 15 => area is 225pi.

[rishudubey]

Thankyou so much sir 🙏for your hardwork 🙏

[philrobson]

I took out my dividers and found that 2 units of 6 plus 1 unit of 3 made the radius of 15. The rest was easier.

[spnrx]

I used a CAD program to solve this. Maybe it wasn't as precise but it was mighty close, within a rounding error.

[KenFullman]

I always like to look at the alternative value given by the quadratic equation, rather than just discount it out of hand. The outer square has a side length of 2xR so if we look at that value of 3=R (that you discounted as not possible) we would have a square with side lengths of 6. So if you were to draw the square with side lengths of 6 you'd find that it fits over the brown rectangle perfectly, with the rectangle occupying exactly the top half of the square. If we were to draw an inscribed circle in that square, the bottom corner of the brown rectangle would, indeed, lie on the circle. So this is what that alternative value is telling us.

So it's not that the value of 3 is wrong, it's just telling you an alternative solution. (albeit not fulfilling all of the parameters of the original problem)

[sandanadurair]

Applying intersecting chords rule we get
3*(2r-3) = (r-6)*(r-6)

Simplifying this
r^2-15r+45=0
r=3 or 15

Area 225pi

[vidyadharjoshi]

Draw horizontal from BC to left parallel to top side. Draw a perpendicular AD from A to BC.
Let R be the radius. AC sq = ADsq + CDSq. Rsq = (R - 3)sq + (R - 6)sq.
Solve for R. R = 9 +/- 6. Reject 3. R = 15. Area = pi*225 = 706.85.

[theoyanto]

Splendiferous, fantasmagorical, even though I make many errors, these problems are such great fun 😊👍🏻

[Taigan_HSE]

I took a slightly different route to get the same result. Using the formula for a circle (x-a)^2+(y-b)^2=r^2. Then plugging in the three points: (6, -3), (r, 0), (0, -r). Then realizing that |a|=|b|=|r|. Lets us solve for r=15.

[Navin_bawaskar_]

Very easy question it's also in our maths module