A Very Nice Geometry Problem | 2 Different Methods

[@hongningsuen1348]

Method using tangent property and Pythagoras theorem: 1. Let O be centre of larger circle and T be point of tangency for both circles. As tangent at T is perpendicular to both CT and OT, OCT must be a straight line. 2. Let R be radius of larger circle. In right-angled triangle OAC, OC^2 = AC^2 + OA^2 (Pythagoras theorem) OC = R -3, AC = 3, OA = R - 4 Hence (R - 3)^2 = 3^2 + (R - 4)^2 R = 8 3. Area of shaded region = (1/2)(8^2)pi = (3^2)pi = 23pi

[@jimlocke9320]

In Method #1, at about 4:30, we know that OB = OD = R, where R = radius of semicircle. We can readily determine that OC = R - 3 and OA = R - 4, then apply the Pythagorean theorem to ΔOAC, so OC² = AC² + OA², (R - 3)² = (3)² + (R - 4)², which, doing the algebra, produces the solution R = 8. This method seems to me to be much simpler than the one used in the video.

[@santiagoarosam430]

Si "r" es el radio y "O" el centro del semicírculo → En el triángulo rectángulo CAO: 3²+(r-4)²=(r-3)²→ r=8 → Área sombreada =π[(r²/2)-3²] =23π ud². Gracias y un saludo cordial.

[@toninhorosa4849]

Great solution teacher!!! 👏👏👏👏👏

[@himo3485]

radius of larger semicircle : R (R-4)²+3²=(R-3)² R²-8R+16+9=R²-6R+9 2R=16 R=8 Shaded area = 8*8*π*1/2 - 3*3*π = 23π

[@michaeldoerr5810]

The answer is 23pi. I actually think that BOTH methods can just be streamlined using just the Pythagorean triplet criterion using the 3-4-5 combination. And just using that I mentally guessed the quantities without all of the extra steps. And all of these extra steps are in case you do not have a Pythagorean triplet right???

[@johnbrennan3372]

AO=r-4 so OC=r-3 since OC -AO=1. In the triangle CAO, (r-3)^2= (r-4)^2 + (3)^2 etc.

[@cosmosapien597]

Why you need to specify that OC-AO=1 ?

[@johnbrennan3372]

@@cosmosapien597. We know AO=R-4 but OC=AO+1 therefore OC=R-3.

[@cosmosapien597]

@@johnbrennan3372 we know that OC=r-3 from the diagram itself, as OC=OD-CD, that is, big radius minus small radius. So I thought why on earth are you saying OC-AO=1 as it is not needed. It is not even apparent, you need to derive it first.

[@johnbrennan3372]

@@cosmosapien597 it’s just another way of saying it.

[@RealQinnMalloryu4]

(3)^2=9 (4)^2= 16 {9+16}=25 180°CD/25= 7.5 (CD ➖ 7CD+5) .

[@vcvartak7111]

Second method nicely explored

[@giuseppemalaguti435]

√((R-4)^2+3^2)+3=R...R=8

[@devondevon4366]

64 pi - 9 pi I forget it was a semi-circle, not a full one Draw a line from C to the circle's cente to form a right triangle. and thus the sides 3, r-4, and r-3 r-3 is the hypotenuse Hence, 3^2+ (r-4)^2 = (r-3)^2 9 + r^2 + 16 - 8r = r^2 + 9 - 6r 25 - 8r = 9- 6r (the r^2 cancels each other) 16 = 2r 8 = r the radius of the circle = 8 Since the radius of the small circle is given at 3 Then the shaded region is the difference between the two areas Hence, 8^2 pi - 3^2 pi 64 pi - 9pi Answer The thing it is a semi-circle, not a full circle So the area of the semi-circle is half the area of the circle or 32 pi, not 64 pi So I should have put 32 pi - 9pi instead of 64pi - 9p

[@andryvokubadra2644]

Sebaiknya gambar ½ lingkaran berdimeter 16, lalu lingkaran kecil berdiameter 6 dengan posisi seperti gambar yaitu sisi lingkaran kecil menyinggung titik A dan D. Buktikan bahwa AB = 4, AO = 4 terbukti 🤔🤔🤔 Karena saya meragukan hasil ini. Terlalu banyak asumsi. Memang matematika tidak pernah salah, namun asumsi bisa saja meleset.

[@Irtsak]

If you start in reverse, i.e. in the semicircle with a diameter equal to 16, you make a circle with a diameter equal to 6, which is tangent to the semicircle and to the diameter of the semicircle, then you prove that AB=4. This does not mean that the reverse is also true !!! The reverse would apply, if you started from the conclusion of the exercise and with equivalent actions you ended up in the hypoyhesis.

[@andryvokubadra2644]

​@@Irtsak Sebelum dibuktikan secara eksperimen, bisa saja BO ≠ DO. Saya awalnya berfikir juga bahwa garis lurus imajiner DO = BO, namun saya tidak berani. Takutnya keliru. Harus dibuktikan langsung bukan teoritis. Karena yang kita tahu hanya AC & AB. . . Jika DC = 3 saya paham dan bisa diterima. Juga kita bisa hitung BC. Yg perlu dibuktikan secara eksperimen adalah BC = CO atau BO = DO. Kapan-kapan saya akan bereksperimen dengan data ini.

[@Irtsak]

@@andryvokubadra2644 Unfortunately you write in Indonesian. I am translating into English, but the translation cannot render your words. Αs a mathematician I speak the strict mathematical language, while you speak intuitively. Reminds me a Sofia Copola's movie " lost in translation". Αν γράψω κι εγώ στα Ελληνικά , θα γίνει ο πύργος της Βαβέλ 😊

[@brettgbarnes]

(R - 3)² = (R - 4)² + 3² Duh!