Can you find the red area? - Geometry puzzle

[@MathQueenSusanne]

Hey math friends! If you’re enjoying this video, could you double-check that you’ve liked it and subscribed to the channel? It’s a simple equation: your support + my passion = more great content! Thanks for helping me keep this going - you’re the best!

[@meisievannancy]

Such a sweetheart and good teacher too. So did what you asked.

[@kewitt1]

I used to do math for fun, 25+ years ago, I am loving your channel. I've forgotten so many of the things I never used but only 1 out of every 4-5 videos I can figure out in my head, Ones like this one I had no clue how to do it. But as you solved it, I started recalling the process.

[@feerrnn]

You talk so clearly and calmly. thanks for the video

[@jake4974]

it's true, right?

[@quercusquercus532]

Agreed. Great channel, even if it's aimed at a level that's not quite mine. Would definitely be all over it if Math Queen gave some deeper theoretical tutorials.

[@incarnateTheGreat]

Certainly better than your parents screaming at you when you can't understand it (and they feel like they can't explain it).

[@DavidTobolowsky]

I earned a BS degree in math fifty years ago before attending medical school and have long forgotten most everything mathematical, but your channel brings back fond memories. Your explanations are crystal clear and your demeanor charming, putting the fun into math, where it belongs.

[@quercusquercus532]

I'm a similar story a couple decades later. I usually click these in cases where there's basically one step I'm missing. In this case, the observation that the radius must be perpendicular to the triangle. But I end up sticking around for the whole thing, cuz of the soothing vibe.

[@OmerAhmed-d4s]

Your way of teaching math is incredible.

[@stevenkenny2213]

Super cool. Could also use the formula for 30-60-90 triangles to avoid use of Sin and a calculator I think.

[@LTVoyager]

That was my approach. Everyone should know the side lengths of 30-60-90 and 45-45-90 triangles.

[@Z-eng0]

Exactly, besides anyone who ever studied trigonometry knows the sine and cosine of at least all the special angles (0, 30, 45, 60, 90, 180). Still I much prefer the 30-60-90 triangle approach as it feels easier

[@wolf1066]

to be fair, I *_did_* need a calculator to work out 1.5 times the square root of 3 (and to work out the square root of 3, tbh)

[@LTVoyager]

@ The square root of both 2 and 3 are burned into my memory. 😂

[@domosautomotive1929]

You are right.....leave the trig out of the problem and just use properties of 30-60-90.

[@philabu7376]

sin(60) = √3/2, can easily be derived by calculating the height of an equilateral triangle with pythagoras' theorem.. so the "final" solution should be 27π/8 m²

[@henrikholst7490]

First make it work. Then make it pretty. 🎉

[@eekmag]

Plug that in your calculator... What is it, engineering chanel?

[@Splattle101]

Yep, and that allows you to arrive at a decimal answer with a single calculator operation instead of five. When you're doing chemistry or other long problems the number of operations escalates very quickly and the error bars get wider with every rounding. Reducing the number of operations = tighter result, so carrying exact values is sometimes worth it.

[@Patrik6920]

or u can expand on that .. sin(60) = √3/2 -> ( sin(60) )² = 3/4 = 0.75 -> A = 9π(3/4)/2 OR A = 3.375π .. or a head count 3² *.75 * π /2

[@donmoore7785]

Bizarre. She calls herself the Math Queen.

[@crosstraffic187]

Thank you for making an old man better at math.

[@jonraybould-videos]

Hi Susannah, I love your videos! I am a 65-year-old musician but I studied A-level maths (plus music and physics) in the UK 48 years ago! Not done any maths since! My teachers had a good way of getting us to remember SOH-CAH-TOA - "Selly Oak Hospital Can Amputate Hands, Toes, Or Arms"! (Selly Oak was my local hospital....!!). Really enjoying going back to my teenage maths experience. You have brought the joy back for me. Thank you.

[@FlyBird]

I keep forgetting that solving math equations has this magical quality of clearing my mind. Solving this alongside with you put me in a brief moment of peace, thank you!

[@wayneyadams]

We know that the sides of a 30-60-90 are in the ratios 1-2-sqrt(3). and that gives us a messy value of sqrt(3)/2 for sin(60) We are going to square the sin(60), which will give us a nice value of 3/4. We can substitute those values in our area equation to get 27pi/8.

[@minxythemerciless]

It's so good to have a clearly articulated solution that is correct. This is in contrast to the recent flood of worked examples (usually Indian) that find the solution in a really weird way and are often wrong.

[@bdphourde]

You are blessed as an excellent teacher! If I only had you in elementary school, I wouldn't have had so much trouble wth my addition tables. ;)

[@orenmane06]

Once i found your channel, I start liking and understanding math 😁🤗🌷

[@abdulqadersaidi7658]

Fantastic video and very useful mathematical information. Excellent. Thanks

[@professor_stevens6784]

Neat! My solution did not use trig. Set the length of the triangle's to be "s." The point where the line r touches a side at right angles divides that side into segments of length a and s - a. So r^2 = a^2 + s^/4 and r^2 + (s - a)^2 = 3s^2 / 4 (both by the Pythagorean Theorem, which you also use to get the height of the triangle). Solve those two equations for a and you get a = s / 4, which you substitute in and get r = s * sqrt(3) / 4. Then pi r^2 / 2 for the area. (For s = 6, r is about 2.5981.)

[@xanx1234]

I had forgotten that it was an equilateral triangle, once you started to explain that then sohcahtoa jumped into my mind and it became straight forward to solve. Thanks for that!

[@softshells]

What a beautiful problem-I enjoyed solving it. It’s been 30+ years since a classroom. I ended up buying a geometry 📐 box (since I was very organized student!) and then remembering some of my trig functions. I think key for me was recalling that where Radius line meets the tangent, angle is 90 degrees. Thanks 🙏

[@WilliamRoeder-bw7ed]

I enjoyed this problem. Normally I can see how to solve your questions almost immediately and it takes a few minutes to get the answer. This time it took a few minutes just to figure out how to solve it. Thank you for the mental exercise!

[@dangermouse8466]

Wow! I enjoyed getting to that solution. Thank you Susanne!

[@Mal1234567]

I found the red area immediately. It had a question mark on it.

[@Recipe-Review]

Just spent the last 15min wondering why I didn't get 10.6 on my calculator. Then I read a comment and set mine to degrees 👍 it's been a lot of years since I ventured into these sorts of math problems. But well worth the brain shake. Thanks.

[@tonyennis1787]

Nice. Didn't care for the calculator solution, but setting up the problem is more important.

[@henrikholst7490]

No need for it as sin 60° is a nice round number: ✓3 over 2. 🎉

[@ASINGH-li8eq]

done very good .good explaination.!

[@gottfriedschuss5999]

There are multiple ways to solve this, with and without trig. Regardless of the method, I think, pedagogically, it’s important to show the desired area is 27 pi / 8 first, which is indeed equal to 10.6029. Otherwise, you do a great job explaining, at an appropriate level, how to solve this.

[@surrealistidealist]

Where does 27 come from?

[@vespa2860]

@@surrealistidealist Sin 60 degrees = (sqrt 3)/2) So (3 X sqrt 3)/2)^2 = (9 X 3)/4=27/4 All multiplied by pi/2 = (pi X 27)/8 which is same as 27pi/8

[@nathanc6516]

I got that too.

[@commonmancrypto1648]

=(PI() * (SQRT(6^2 - 3^2) / 2)^2) / 2 = 10.6028 is the first solution I thought of.

[@ASINGH-li8eq]

This is the easiest way what the Madam explained

[@davida6299]

Love your channel! What software/program and equipment do you use for the videos? Thank you, keep the great content coming!

[@Joao_133]

love these videos ❤

[@meisievannancy]

I actually calculated it a bit differently. I actually used the Cos function of 60 to get the other side of the small triangle which gives 1/2 of 3 =3/2. So then using Pythagoras r = sqrt (3^2 - (3/2) ^2) = 3/2*sqrt3 So half area of circle =(pi*r^2)/2 = 27/8*pi

[@danielnevesdepaiva]

Excelent!!! This way was more simple!!!

[@rrodriguez9001]

The sin(30deg) is 0.5. Using that, the inner right triangle has sides of 3m (hypotenuse), 1.5m (short side) and 1.5 √5 (long side and radius of circle). This makes the area of the half circle equal to 3π/2√5...which is 10.537m.

[@irtnyc]

Is it 10.6 or is it 10.537? And more importantly, why?

[@borysszacki8002]

@irtnyc: are you serious or you ask for many comments?

[@oahuhawaii2141]

Nope. There's no √5 in the solution. The area is π*27/8 m² ≈ 10.602875... m²

[@Grizzly01-vr4pn]

You are writing √5 where you should be writing √3

[@stephenash777]

Well I impressed myself. It's only been 45 years since I did this in high school, but I still remember SOH CAH TOA, and was able to think thru the logic on my own.

[@sethheasley9538]

Man I'd listen to a ton of audiobooks if she was the narrator. These videos are great.

[@sparkyfromel]

Hurrah , nearly got this one !

[@Doc_Fartens]

Gosh, I hadn't thought about the acronym SOHCAHTOA since high school 15-20 years ago. This brings back memories!

[@DavidAnderson-m5c]

Sufferin' SOHCAHTOA!

[@situational.analysis]

KRAKATOA!

[@richard17lucas]

We were taught (many many years ago) Can All Housewives (CAH) Shop On Horseback (SOH) To Oblige Albatrosses (TOA) And it has stuck with me ever since

[@quercusquercus532]

Haha. I guess nobody needs Trig quite like a Mariner, eh? I couldn't help myself. Sun Over Head! We seek the Global Opposite, With Pirate Metal's din! We'll put the "sex" in "sextant", lads! We'll sail and then we'll Sin! Crew Are Hungry? A sound is heard: Adjacent bird, They hatch a fowl plot! A sudden jolt, an angled bolt, An albatross is shot. Tie Of Albatross... "They groaned, they stirred, they all uprose!" His life becomes his brig. He shoulda picked a soldier's life, And never studied Trig! Eternal pleas, from bended knees, To die, as any man, By KrakaTOA? Kraken claw? In Hell, he'd get a Tan! (While TAN indeed is infinite, Unending like the sky, To dodge this beast, complete your feast, And don't eat half a Pi!)

[@GerardHammond]

Fantastic. I almost had this one!

[@simomsymos4111]

great presentation!

[@frederickpell5254]

I also used the properties of a 60, 30, 90 triangle (1, 2, root 3) to get the radius.

[@wolf1066]

Awesome, that's pretty much the logic I would've used to work it out - seems I haven't forgotten as much geometry as I had thought - knew that the point where the arc met the triangle was a tangent and therefore perpendicular to the radius line at that point, giving a right angle triangle, remembered the rules about all interior angles adding up to 180 degrees and that an equilateral triangle has 60-degree angles - so determined the triangle with a hypotenuse of 3 had angles of 30, 60 and 90.

[@barneyDcaller]

My approach is that 3 = 2x or x = 1.5 Since its a 30-60-90 right triangle with the hypotenuse of 3, therefore the shorter leg is 1.5, then the other leg is 1.5 × √3 which is the radius of the circle

[@miroremenar1704]

Thank you for the problem. Altough I have quite mixed feelings - solved it without trigonometry, using only the slope of the line (side of the triangle) and finding the distance of point(x=0, y=0) from that line (the radius). Downside of that proces - it took me ages and I made about 200 mistakes while it was correct🤣. Definitely time to regain some forgotten skills!

[@schitanstefan2696]

Thanks for the video! Very well explained! By the way, could you please let me know what app you are using?

[@MrOj53]

One subject in school interests me a lot, it was mathematics. I have followed a couple of your mathematical problems, they remind me of my studies of logic. What I experience from your calculations is that counting has a first use-by date. Praktis, Praktis, Praktis And praktisk

[@HolySoliDeoGloria]

Any student at this level should know (or be able to derive within seconds) the ratios of the various sides of a 30-60-90 triangle (2:1:√3), thereby either avoiding the use of sine or easily calculating the value of the sine of 60 degrees. Great problem and video!

[@alistairbaird3711]

Takes me back to O-level maths. Happy days

[@arehomann-danielsen3137]

Hi Susanne and thank you for a great channel. I was wondering which program you are using to present your math-problems.

[@ПетрЛонзингер]

Good puzzle, have to think a bit to solve it.

[@jangroterlinden569]

solved it with Pythagoras. First got the hight of the triangle, from the upper horizontal to the lower corner. Than r and and x (which is the length between the upper left corner to the tangetial point of the circel) also with Pythagoras.

[@budgey0065]

Wow, you make me want to go back to school to learn math! 😂

[@mariehart4294]

Nice video!

[@timdavies710]

thanks again.

[@Jptoutant]

another interesting one!

[@williammann6198]

You are an awesome woman. God Bless!

[@taijiquanzhe]

As others have done, I used pythagoas (and memory too) to tell what the radius value would be. Since the sides of the 30/60 right triangle are 1,2,√3. That lead me to the value of 27π/8 m² To break that down into smaller easier to follow steps, I would calculate the height of the triangle first. That would be 6^2-3^2 = 36-9 =√27. (I could simplify that, but I am going to use that term again to get the radius. Since we know that the adjacent side is half the hypotenuse for a 60degree right triangle, the radius must be (√27)/2. We need to square that for the radius term in the area equation. That gives 27/4. So putting that into the area of a semicircle, we would have π x (27/4)/2 = 27π/8. That also gives 10.6. This also avoids the trig function, which could put some people off I guess. Excellent video as always. I enjoy the presentation style and it always comes across incredibly well.

[@Tamessah_560]

I've solved it with the area of half the equilateral triangle : A = 1/2BH A=(1/2) * 3 * 3(sqrt(3)) A = (9*(sqrt(3)))/2 After that I changed the base value to 6 so that the height (which is perpendicular to the base) is the same as the radius of the circle , therefore : A = (9*(sqrt(3))) / 2 = (6H)/2 = 3H 9*(sqrt(3))/2 = 3H H = (9*(sqrt(3))/2 )/3 = 3*(sqrt(3))/2 H approximately equals to 2.6 which is the same as the radius , then substitute in the half circle area formula : (r^2* pi )/2 ((2.6)^2 * pi)/2 = 10.6 Anyone who has another solution please share it with me!

[@commonmancrypto1648]

=(PI() * (SQRT(6^2 - 3^2) / 2)^2) / 2 = 10.6028 Area of the semicircle

[@ronvonbargen8411]

3 x cos 30°=2.598 2.598²=6.75 x pi =21.205/2 =10.602

[@quigonkenny]

Draw a line from the center of the semicircle to the right side point of tangency. This creates a 30-60-90 triangle with the right corner. The sides of a 30-60-90 triangle are of the ratio 2:√3:1 going from hypotenuse to smallest leg, and as we can see that the hypotenuse corresponds to half the length (3m) of the top side of the equilateral triangle, we can use the ratio with the longer leg of the 30-60-90 triangle to find the radius. 3/2 = r/√3 3√3 = 2r r = 3√3/2 A = πr²/2 = π(3√3/2)²/2 A = (27π/4)/2 [ A = 27π/8 ≈ 10.60m² ]

[@ericwickeywoodworkersurfbo6135]

Trigonometry is the most fun ever.

[@willemkelles7942]

Solution without calculator: As SIN (30”) = sqrt (3) / 2, R^2 becomes 27/4 and A becomes pi*27/8.

[@surrealistidealist]

How do you figure out 27/4 without a calculator?

[@WastrelWay]

@@surrealistidealist Long division 🙂

[@difi6581]

Sin(30°)=0,5 sin(60°)=sqrt(3)/2

[@willemkelles7942]

Hi, in the small triangle she draws, the small angle is 30 degrees, the longest side is 3m and the second longest side is R. So we have cos 30’ =sqrt(2)/3=R/3. Solve this to R gives R=3*sqrt(3)/2. So R^2=9*3/2=27/2.

[@willemkelles7942]

see below.

[@IsabinMarius]

If you know your standard side relations of a 30° right triangle then this is easily calculable in your head without resorting to sin values. The long side (the radius you're looking for) is: r=(√3)/2 or about 0.866 multiplied by the hypotenuse, in this case 3. The rest really is just plugging it into the area formula.

[@don911donny9]

Nice little challenge, I just had to remember to set my calculator to Deg, not Radians, and out popped the correct answer 😀

[@Recipe-Review]

Just spent the last 15min wondering why I didn't get 10.6 on my calculator. Then I read a comment and set mine to degrees 👍 it's been a lot of years since I ventured into these sorts of math problems. But well worth the brain shake. Thanks.

[@mechaileh1]

The most beautiful math teacher in the entire universe. Just that.

[@mikeschinkel]

Could you do a video to elaborate on SOH, CAH, TOA? I have an engineering degree yet don’t remember ever covering that in school.

[@razsego]

Draw a vertical line that divides the triangle into two equal right triangles. Pythagoras theorem gives us the length of the line: sqrt(6^2-3^2)=sqrt(27) This triangle is congruent with the one drawn in the video, which means that the ratio of the sides match: 6/3 = sqrt(27)/r = 2, ---> r = sqrt(27)/2 The area of the circle is pi*r^2 /2 = pi*(sqrt(27)/2)^2 /2 = pi*(27/4) /2 = pi*27/8, or approximately 10,6m^2

[@neilhabermehl6187]

I think an exact solution using 1, 2, sqrt(3) would be more elegant as opposed to a numerical approximation from the calculator.

[@bryanalexander1839]

One solution: Construct the tangent line as in the video and an altitude from the center of the semi-circle diameter to the non-collinear vertex of the triangle. The resulting triangle with vertex, diameter center, and tangent point is similar to the triangle that is one half of the equilateral triangle (as split by the altitude) and similar to the one in the video (*), since they have the same measure of angles, namely, 30°, 60°, and 90°. Using the properties of 30-60-90 triangles, the triangle where the altitude is the hypotenuse has leg r opposite to 30°, so the altitude is 2r and 2r=6*√(3)/2 using those same properties on a triangle with hypotenuse 6 and leg 2r opposite 60°. Thus the radius is r=3*√(3)/2 and area of the semi-circle is A=(1/2)*π*[3*√(3)/2]^2=27π/8. (*) I did not use the triangle in the video, but, given that it has r opposite 60° with hypotenuse 3, it could be used as another way to demonstrate via proportionality of similar triangles that the altitude is 2r when that leg is opposite 60° and the hypotenuse is 6. Final note: sin(0°)=cos(90°)=√(0/4)=0 sin(30°)=cos(60°)=√(1/4)=1/2 sin(45°)=cos(45°)=√(2/4)=√(2)/2 sin(60°)=cos(30°)=√(3/4)=√(3)/2 sin(90°)=cos(0°)=√(4/4)=1

[@dv2915]

You have found an approximate solution. In this kind of problems you are supposed to give an exact solution, which in this case is 27π/8.

[@toby9999]

I'm pretty sure she knows that

[@richardashby6981]

Use the Pythagorean theorem to get the length of the imaginary line that splits the triangle down the middle (~5.2), then use the theorem of similar right triangles to get the radius of the semicircle: R / 3 = 5.2 / 6 R = (3 × 5.2) ÷ 6 R = 2.6 Area = πR² ÷ 2 = 10.6 Voila! No trig needed.

[@rolandmichels1]

{*} I did it another way, without using any sine, cosine or tangent formula. Only Pythagoras' theorem. First of all I made a circular pattern of the triangle + semi circle around the bottom point of the triangle. Number is 6. Then I completed the semi circles into full circles (not really necessary, but easier to understand). So I have 6 circles, all touching the 2 neighbous. In the centre I drew another circle of the same diameter, touching all of the other 6 circles. (It is common knowledge, that if you have 7 equal coins, for instance, you can put 1 in the centre and the other six around that, and all will touch their neighbours.) Now let D be the diameter of the circles. Then I divided one of the triangles (6 - 6 - 6) in two equal halves. Looking at one half, one side = 6, one side = 6/2 = 3 and the ohter side we call 'a'. Now 'a' is one of the rectangular sides, whereas the side of 3 is the other one. Pythagoras: a^2 + 3^2 = 6^2. Or a^2 = 6^2 - 3^2 = 36 - 9 = 27. So 'a' = SQRT (27) = SQRT (9) * SQRT (3) = 3 * SQRT (3). Now, if we look at our drawing of 7 circles and 6 triangles, consider the distance between the centre circle and 1 of the others. It is clear that this distance is the value of 'a', being 1/2*D + 1/2*D = D. So 'a' = D, so D also equals 3 * SQRT (3). Any circle with diameter D has an area of pi / 4 * D^2 (**}. So the area of the full circle equals pi / 4 * (3 * SQRT (3))^2 = pi /4 * 9 * 3 = 27 * pi /4. So the area of the semi circle equals half of that: A = 27 * pi / 4 /2 =27 * pi /8. (about 10,6). {*} Note that I don't mention 'm', the length of the side of the triangle, because it is irrelevant. Unless you have to plough the area. {**} As a mechanical engineer, I mostly use A = pi /4 * D^2 instead of the, more mathematically used, A = pi * r^2, because its quite hard to measure the radius of a round piece of stock, whereas it is very easy to measure it's diameter. Grüße aus Venray, NL.

[@AlanMcYou]

Completing the circle with a matching triangle gives one parallelogram with the circle touching both parallel sides. Now one only needs to find the separation of the parallelogram sides. Pythagoras gives us a diameter of SQRT(27). The rest is easy.

[@chadwiza1538]

What device/program do you use for writing like that? iPad and OneNote?

[@Водитель_КрАЗа]

В прямоугольном треугольнике напротив угла 30° лежит катет, равный половине гипотенузы. По теореме Пифагора r² = 3² - 1,5² = 6,75 A = 3,375π

[@ianmoses3140]

What software are you using?

[@kubagornowicz]

I used two triangles made from spliting big triangle in half - than one of them is one you used, and both have R as one side. Than I calculated side which is the height of big triangle. Than other two sides left where x and z (where z=6-x and x=6-z). Than I had two pythagorean theorems, in one I substituted z for 6-x and got quadratic equation which gave me 6,75 π / 2 = 10,6... Wow I'm so proud of my self! :)

[@benv6875]

That's basically what I did; forgot all my trig anyway. See my coment above. Split the triangle in half, then use pythagoras on 3 triangles, with radius forming 2 triangles, you will have 2 variables and two equations but it resolves to the correct answer.

[@bontrom8]

More fun.. i enjoyed soh cah toa and finding more angles inside the larger space like 120° between tangents

[@daniellerosalie2155]

Area of half circle: Pir^2/2 30-60-90 triangle 180/3=60 degrees We take sin(60)x Sin(60)=r/3 3sun(60)=r A=((pi3sin60)^2)/2 =10.6 m^2

[@Jbig1430]

Once you established it was a special right triangle 30-60-90 why would you do the trig.

[@gelbkehlchen]

Solution: This is an equilateral triangle with angles in the corners of 60°. From the center M of the semicircle to the right point of contact B is the radius r of the semicircle. C = right corner point of the triangle. MBC is the famous right-angled triangle with angles 30° - 90° - 60°. MC = 3 = hypotenuse, BC = short leg = 3/2 = 1.5. Pythagoras: MB = long leg = r = √(3²-1.5²) = √6.75 Red area = red semicircle = π*6.75/2 ≈ 10.6029

[@cantablocal6768]

I just multiplied 3 times the cos of 30 degrees = 2.598 radius to get 10.6028m2 area of the half circle while you were still explaining how to get the area of a circle.

[@psmitty3715]

Susanne, your Awesome.

[@spdas5942]

2*1/2*6r=sqrt3/4*6^2=> r=9*sqrt3/6=3*sqrt3/2. So1/2pi.r^2=1/2*(3*sqrt3/2)^2=27*pi/8. ❤ it.

[@aswinkumar4051]

Can you please something from statistics and probability?

[@Robert...Schrey]

chatgpt recommends using point-to line distance formula.

[@babotond]

you can use similar triangles to calculate the radius, dont need sin(). A = 27/8 * π

[@elsenored562]

1 : 2 : sqrt(3) proportions, because it's a right triangle with a 60° angle.

[@montynorth3009]

Just another method.😊 Height of triangle = sq.rt. 27. Similar triangles. r / 3 = sq.rt. 27 / 6. r = sq.rt. 27 / 2. Red area = 1 / 2 x Pi x 27 / 4. 27 Pi / 8. 10.603.

[@surrealistidealist]

How did you figure out that the height of the triangle is the square root of 27?

[@montynorth3009]

@@surrealistidealist Pythagoras' Theorem after joining the bottom apex point up to the middle point of the 6m line. Triangle sides 3, 6, & height applies.

[@VishnuVichu-dx7rp]

Thankyou

[@alexyo5402]

That triangle is 30 60 90, so the small one is half of the hypotenuse, 1.5 Apply Pythagoras 9=2.25+squared R Square R=6.75 Half the area is pi*6.75/2=10.6028

[@Водитель_КрАЗа]

Square R=6.75

[@roland3et]

Here's another solution using the good old 'Pythagoras': short side q of the little right-angle triangle is q = 3sin(30°) = 3×0.5 = 1.5 r² = 3²-1.5² = 9-2.25 = 6.75 Area As of semi-circle: As = ½πr² = ½π × 6.75 = 3.375π As ≈ 10.6 [m²] 🙂👻 P. S. Don't use the equal sign at the very end, Susanne - the calculator almost always delivers just a rough estimate, "Pi mal Daumen" so to say 😉...

[@andyonions7864]

THis is the way I did it. Sin(30) = 0.5 should be well known to most students of trigonometry. I like the way that the square route isn't needed too.

[@scottleish]

No need to even hit the sin button to know it’s 1.5 on the short side. It’s just 4 equilateral triangles inside a bigger one. Can be deduced visually.

[@andydavis6054]

Wow 😮

[@m.h.6470]

Solution: Since it is an equilateral triangle, we can calculate its area with A = √3/4 * a² = √3/4 * 6² = 9√3. If we divide the triangle in the middle, we get two right angle triangle, with the radius of the semicircle as the height of those triangle. This is true, because the hypotenuse of the triangles is tangential to the semicircle and as such it is perpendicular to its radius AND because halving the big triangle means that the right angle is exactly at the center of the semicircle. Now, since we have the area of the big triangle and the small triangle are exactly half of it, we can use 6 as the base, and the radius as its height to calculate the radius: 9√3/2 = 1/2 * 6 * r |/3 r = 3√3/2 As such, the semicircle has an area of: A = πr²/2 A = π(3√3/2)²/2 A = 27π/8 A ≅ 10.603 [m²]

[@m.h.6470]

After the video: Oh boy... WAY to complicated in the video. Trig is absolutely not necessary here!

[@cherairiali-zf1ps]

happy new year.

[@kobi2187]

cool. this should be great for my kids. i should either translate or help them learn English fast, since so much knowledge is locked up without this language.

[@cyruschang1904]

Triangle area = (6(3√3)/2) m^2 = ((6r)/2 + (6r)/2) m^2 (9√3) m = 6r r = ((3√3)/2)m Semi circle area = π(r^2)/2 = π(27/8) m^2

[@gregnixon1296]

Heron’s formula * (2/3) = Area of semi-circle.

[@robertroemer4233]

Is 6m the diameter of the semicircle or the length of one side of the triangle?

[@mikechappell4156]

The length of the side of the triangle.

[@Grizzly01-vr4pn]

If it were the diameter of the semicircle then the video would present a pretty trivial question.

[@daveincognito]

I ended my math journey with trigonometry. I didn't get something about it, but I'm not sure what it was I didn't get.

[@benv6875]

You can do it without trig using Pythagoras and two variables.

[@toby9999]

This is a really tough one. I was also trying to find the "touching point", but couldn't, and I couldn't follow your reasoning either.

[@thecarman3693]

3:49 Knowing that the cos(60) is 0.5 I would have used that to get the adjacent side as 1.5m, then solved for R. R = SQRT(9 - 2.25) = 2.6 (approx.) Pythagoras to the rescue!

[@mohssenkh6422]

Or calculate h which is 3 sq root 3 × cos 60 and you have r.

[@chandrashekharchavan8156]

By using Geometry only we could have also done like this. 1. This triangle is equilateral triangle therefore ar=√3×(6)×6)÷4=9√3. 2. Construct line from point opp. to semi circle to centre of semi circle so that we can get ar of two triangles 3. Ar= (0.5×6×r)+(0.5×6×r)=9√3 18√3=6r+6r=12r r=18√3÷12=3√3÷2 4. Ar of semi circle 0.5πr×r =0.5π×(3√3÷2)×(3√3÷2) =(27÷4)π =6.75π meters. sq.

[@nomadkeller8612]

To find the leg of the 30 60 90 triangle why not use the special triangle relationship without SOH CAH TOA which is simply 3radical3 all over 2 ???