Can You Solve this MIT University Entrance Exam?

[@brain_station_videos]

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[@ناصريناصر-س4ب]

There are two similar triangles, one of which is drawn in the semicircle and the other has one side that is the side of the square. From this we can deduce the similarity ratios: 3√2/x = 3√3/6√2, and from this x = 4√3.

[@danielsemendinger4540]

Angles Inscribed in a Semicircle Theorem

[@MrPaulc222]

I paused the video to try this one. It looks like a similar-triangles puzzle. Square's sides are 3. Note that r is the square's diagonal, so 3*sqrt(2). To the right of the square there is a triangle whose sides, in ascending order, are 3, 3*sqrt(2), and x-a. x-a can be calculated as 3^2 + (3*sqrt(2))^2 then taking the square root. 9 + 18 = 27, so x-a = 3*sqrt(3) There is a larger similar triangle produced by Thales Theorem with ascending sides: In ascending order the triangles' sides are now 3, 3*sqrt(2), and 3*sqrt(3) and b, x, 6*sqrt(2) (3*sqrt(3))/(3*sqrt(2)) = (6*sqrt(2))/x Cross multiply: (3*sqrt(2))*(6*sqrt(2)) = (3*sqrt(3))*x 36 = 3*sqrt(3))*x 36/(3*sqrt(3) = x Rationalise: 108*sqrt(3)/27 = x x = 4*sqrt(3). I see you went for intersecting chords. Yes, there are several ways. .

[@paparmar]

Going through the steps, keeping the value of s open, we arrive at x = 4/3 * s * SQRT(3). Check: if s = 3, then x = 4/3 * 3 * SQRT(3) = 4 * SQRT(3). In terms of r: x = 2/3 * r * SQRT(6).

[@JamesMcCullough-lu9gf]

thank you for teaching me a new theorem

[@marioalb9726]

s² = 9 cm² -> s = 3 cm ; R =3√2 cm tanα = s/s√2 = 1/√2 --> cosα= √2/√3 x= 2R cosα= 6√2.√2/√3= 12/√3= 4s/√3 x = 4√3 cm ( Solved √ ) x = f (s) = 4/√3. s

[@tradefortutara9608]

How exciting😊.

[@joso5554]

I absolutely never heard of this chord-chord power theorem.

[@joso5554]

An alternate method uses instead the cosine of the angle between the x line and the horizontal diameter. And the fact that the end of the x line joins the left end of the horizontal diameter with a line that is perpendicular to the x line. This method skips the use of the Chord chord theorem.

[@heenakhandelwal8608]

3:25 x = s√3 + s/√3

[@Fujui]

Keep it up!!

[@arushi1729]

f(s)=4s/sqrt(3)

[@PrithwirajSen-nj6qq]

We may draw another identical square adjacent to the given square Two square will make rectangle. The length of the rectangle is 3*2=6 It is a chord and another chord (x) is given. Now according to intersecting chord theorem ,h *(x -h) =3*3 =9 > 3√3(x -3√3)= 9 > x - 3√3 =9/3√3=√3 > x =4√3

[@zawatsky]

А чего тут думать? Радиус равен диагонали, это корень из двойной площади, т. е. r=3√2. Малый треугольник выходит с катетами 3 и 3√2, его гипотенуза 3√3. Добиваем секущей слева до подобного треугольника (угол опирается на диаметр, значит прямой, а острый угол у них общий). Его гипотенуза - это диаметр, равный двум радиусам. Коэффициент подобия 2r/3√3 - пока оставим так. И на него же надо снова помножить один радиус, т. к. это ещё и больший катет. Т. е. в итоге выйдет 2r²/3√3, т. е. 36/3√3=12/√3=4*3/√3=4√3. Это в России ещё считается «внутри коробки», просто чтобы вы знали.🤭

[@romanm.4763]

To solve the problem I used two similar right triangles and a proportion that ties corresponding edges. One triangle is shown in the video and using by the author. Second one is built on the X and the diameter.

[@NHWY25]

I have a different approach, sir, but sir, i think it is wrong What about Angle inside semicircle property It says,' If an angle is inscribed in a semi-circle, that angle measures 90 degrees' (it was part of my higher school ) Because the chord. Property you use was new for me

[@danielsemendinger4540]

Yes it works. The similar triangle comments below uses the Angles Inscribed in a Semicircle Theorem.

[@ArnavvvvV-b1z]

We can solve this by knowledge of circles !, like write its general equaton x^2+y^2=r^2, here as circle passes through (-9,9) (if centre is origin), r = 9root2, hence we can let that line be y=mx+9, and solve easily, i am preparing for jee, had it been a question in this exam, i am sure more than 20% people would be able to answer it!!, (India) I am sorry the radius is 18...i wrote something else i wanted to write else but idea is same...and ya i have got answer by this method

[@danielsemendinger4540]

If you use the origin as the center of the circle, the equation is x^2 + y^2 = 18; Please check your math, as the radius is 4.24264, (-9.9) is not on the circle.

[@ArnavvvvV-b1z]

@@danielsemendinger4540 i have already solved this with this method and got the answer and only after that i am writing

[@ArnavvvvV-b1z]

@@danielsemendinger4540 I am sorry the radius is 18...i wrote something else i wanted to write else but idea is same...and ya i have got answer by this method Thankyou

[@georgstudnicka9969]

I thougt I know all basic geometry stuff, but I never heard about that chord chord thing! 😮

[@MOAI-yx3og]

This isn’t basic geometry man

[@AKA-f7p]

Similarly will work if you construct a line

[@maxvangulik1988]

r=sqrt(2)s x1=sqrt(3)s (2s^2-9)/sqrt(3)s=x2 x=5s/sqrt(3)-3sqrt(3)/s

[@uwelinzbauer3973]

x = f(s) = s * 4/sqrt(3)

[@lanoviadearana]

the best canal about maths

[@Salsabilla-sv-7]

Y dode

[@JUGNUMEHROTRANEETASPIRANT]

We , at 3:26 , answer is 4s/√(3) = x 😁

[@JUGNUMEHROTRANEETASPIRANT]

Thanks a lot Brain station for Liking (i.e. , appreciating my Answer) 😁😅

[@m.d.abdul.rahman]

एक कमरे में 5 बत्तियां जल रही हैं। अगर तुम 2 बत्तियां बुझा दो, तो कमरे में कितनी बत्तियां बची रहेंगी? विकल्प: (A) 3 (B) 5 (C) 2 (D) 0 Which option is correct comment your answer

[@JigneshGirase-v4k]

5

[@AE-DI]

3 The 2 two lights burnt off

[@wasimahmad-t6c]

9×4=36squroth =6

[@LuisVazquez-ir5lx]

X=(4sqrt(3)/3)*S

[@משהנאון-י6מ]

היכן רשום בשרטוט התחלתי מרובע משיק בנקודה מרכז המעגל כל זה לא שווה כלום

[@heckno3188]

How can u just asume that its the middle of the circle ?

[@toddkes5890]

I think it was stated to be the center of the circle as part of the problem being set up.

[@samdaman2510]

It is part of the problem

[@Omargamingandtech]

It is given

[@Beta_Brain]

(4√3•s)/3

[@SumitVerma-lg3qh]

Easy peasy

[@youssefelyousfi4929]

x=√48 u

[@MartinSun-v2q]

it's like chinese junior high school stuff😂 but we do it like this: draw a line from center that perpendicular to x we can find the radius of the semicircle is 3*SQRT(2), so 0.5x:3*SQRT(2)=3*SQRT(2):SQRT(3^2+(3*SQRT(2)^2) so x=4*SQRT(3) you can calculate like this in your head even without writing anything

[@EVILGAMERYT.]

In India avg 12 yr can solve these stuff too 😂