Calculate the Radius of the semicircle | Two Methods | (Math skills explained) |

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Learn how to find the Radius of the semicircle inscribed in a right triangle. Important Geometry skills are also explained: circle theorem; Pythagorean Theorem; similar triangles; Two-tangents theorem. Step-by-step tutorial by PreMath.com
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14 окт 2024

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Комментарии : 66

@ClubCrafteria Год назад

Nice share

@PreMath Год назад

Thanks for liking ❤️ You are awesome. Keep it up 👍

@wackojacko3962 Год назад

First method is more fun! Reinforcing the concept of ratios of similar triangles. So cool! 🙂

@PreMath Год назад

Excellent! You are awesome. Keep it up 👍

@nassernasser879 11 месяцев назад

Awesome!

@ybodoN Год назад

Mirror the diagram and apply the general formula for the radius (r) of a circle inscribed in any triangle: _r = √((s − a) (s − b) (s − c) / s)_ where _a, b_ and _c_ are the sides of the triangle and _s_ is the semiperimeter. In our case, a = 30, b = c = 39 so s = 54. Then r = √((54 - 30)(54 - 39)(54 - 39) / 54) = √(5400 / 54) = 10.

@zbanovic Год назад

Exactly my idea also... Short and fast...

@manuelgonzales2570 Год назад

Excellent! Thank you!

@quigonkenny 7 месяцев назад

By Two Tangent Theorem, DC = CB. This means AC = 24+15 = 39. Triangle ∆CBA: 15² + BA² = 39² BA² = 1521 - 225 = 1296 BA = √1296 = 360 Draw the radius OD. As OA, AD, and ∠OAD are shared between them, and ∠ADO is a right angle, ∆ADO and ∆CBA are similar. Triangle ∆ADO: DO/AD = CB/BA r/24 = 15/36 = 5/12 r = 24(5/12) = 10

@D.P9 Год назад

4th method -- by using and drawing symmetrical fig Then we get an isosceles triangle such that their their sides are 39, 39 and v30 cm Now radius of semicircle = area/semi perimeter = 1/2 × 30×36/(39+39+30) = 10 cm. Answer

@PreMath Год назад

Bravo! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

@marioalb9726 Год назад

AB = h Pytagorean theorem: h² = (24+15)² - 15² h = 36 cm Intersecting chords theorem: 24² = h..(h -2r) 24² = h² - 2hr 2.h.r = h² - 24² r = ½h - 24²/2h r = 36/2 - 24²/2.36 r = 10 cm ( Solved √ )

@MarieAnne. Год назад

A third method is to calculate area of △ABC in 2 ways. Once we've found side AB = 36 as shown at 3:14, then we get: Area(△ABC) = 1/2 (AB)(BC) = 1/2 (36)(15) = 270 Connect O to D and connect O to C Since OD is radius, it is perpendicular to tangent line AC. So we have 3 right triangles: △AOD, △BOC, △COD that have bases 24, 15 and 15 respectively, and have height = r. Area(△ABC) = Area(△AOD) + Area(△BOC) + Area(△COD) 270 = 24r/2 + 15r/2 + 15r/2 = 12r + 15r = 27r r = 10

@MrPaulc222 Месяц назад

I would do this by calculating triangle area then having three triangles. Two-tangents means hypotenuse is 39. 39^2 (1521) - 15^2 (225) = 1296. sqrt(1296) = 36 There triangles: AOD, area 12r, ODC, area 7.5r, and OBC 7.5r. Total of 27r = (15*36)/2 27r = 15*18 27r = 270. r = 10

@harikatragadda Год назад

Reflect the triangle to complete the circle. Calculating the area of the triangle ½R(30+39+39) = ½*30*36 R= 10

@PreMath Год назад

Thanks for sharing! Cheers! You are awesome. Keep it up 👍

@unknownidentity2846 Год назад

Knowing the relation between the area of a triangle, its perimeter and the radius of its inscribed circle can make things a lot easier.🙂 Best regards from Germany

@jimlocke9320 Год назад

Even simpler, once we have determined the lengths AC = 39 and AB = 36, construct OC. The area of ΔABC = (1/2)(15)(36), area ΔAOC = (1/2)(39)R and area ΔBOC = (1/2)(15)R. However the sum of areas of ΔAOC and ΔBOC equals the area of ΔABC. So (1/2)(39)R + (1/2)(15)R = (1/2)(15)(36) (39/2)R + (15/2)R = 270 (54/2)R = 270 (27)R = 270 R = 10

@vrocher93 Год назад

DC=BC OD=AD*TAN(SIN-1(BC/AC)) => 24*TAN(SIN-1(15/39))=10

@raya.pawley3563 Год назад

Thank you

@D.P9 Год назад

5th method Join C to O Now area of ∆AOC + area of ∆ COB = area of ∆ABC 1/2 × 39 × r + 1/2 × 15×r = 1/2 × 36 × 15 r = 10 answer

@PreMath Год назад

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

@D.P9 Год назад

@@PreMath thank you sir I am just trying to learn from you Please keep sharing such a golden concept and question too

@Ahmed-mn1ri 8 месяцев назад

حل سهل ورائع🤩

@santiagoarosam430 Год назад

CB=15=CD》AB^2=(24+15)^2 - 15^2》AB=36》(36-r)^2=24^2+r^2》r=10 Gracias y un saludo cordial.

@tombufford136 Год назад

At a quick glance, DC=BC, AB^2=AC^2-BC^2. AB =sqrt(39*39-225)=sqrt(13*3*13*3-5*3*5*3)=sqrt(9*(144))=36. AB=36. AO= 36-R.Then AD^2+R^2=AO^2. Then 24^2+R^2=(36-R)^2. Expand this, 576+R^2=1296-72R+R^2. 72R=720 and R=10. Radius = 10.

@davidstecchi9501 Год назад

Once you established the triangles were similar, since the first triangle was a Pythagorean triple, it would follow that the second s also, thereby eliminating half of the work needed to find the radius.

@devondevon4366 Год назад

10 Answer First Method DC = 15 (tangent-circle theorem) Hence AC = 39 (24 + 15) This is a 5, 12, 13 triangle scaled up by 3; hence, AB =36 Draw a right triangle from D to O (tangent-circle theorem). Hence A) = 36 -R Using Pythagorean 24^2 + r^2 = (36-r)^2 576 + r^2= 1276 -72r + r^2 72 r = 1276 - 576 72 r = 720 r=10 Second method Since AB= 36 and ADO is similar to ABC then r /24 = 15/36 36 r = 15 * 24 3r= 15*2 r = 5*2 r =10

@williamwingo4740 Год назад

Here's a more algebraic solution. The big triangle is integer Pythagorean, 15-36-39; so we have all the sides immediately. Next, consider right triangle AOD (as shown at 3:19) with legs r and 24, and hypotenuse 36 -- r. Invoking Pythagoras:: (36 -- r)^2 -- r^2 = 24^2. Multiplying the leftmost term out: 36^2 -- 72 r + r^2 -- r^2 = 24^2; the r^2 terms cancel each other out--as seems to happen pretty often in this type of problem--and we have 36^2 -- 72 r = 24^2; multiplying everything out, 1296 -- 72 r = 576; collecting terms and rearranging: 72 r = 1296 -- 576 = 720; dividing both sides by 72 and [drum roll......] r = 10. voila! And FWIW, triangle AOD is also integer Pythagorean, 10-24-26. Carpe Diem! 🤠

@murdock5537 Год назад

Nice, many thanks, Sir! tan⁡(φ) = 15/36 = 5/12 = r/24 → r = 10

@soli9mana-soli4953 Год назад

Your first method is the best one! Once known DC and AB in some way, an alternative method to find radius could be with tangent secant theorem, tool that I like so much AB : AD = AD : AC 36 : 24 = 24 : (36 - 2r)

@D.P9 Год назад

Sir finally I solved it by most easy approach in which no pen and paper are required. Most easy method - (BY INTERNAL ANGLE BISECTOR CONCEPT) O is incentre Hence BO/OA = BC/AC BO/OA = 15/39 = 5/13 AND BA = 36 unit BO = r = 5/18 × 36 = 10 unit answer ✓✓

@sergiofernandez116 Год назад

Angle BAC = Angle OAD. If angle BAC is P, then SinP =15/39. Also, SinP=r/(36-r) therefore 15/39=r/(36-r). Solve for r.

@unknownidentity2846 Год назад

Let's find out the radius: . .. ... .... ..... According to the two tangent theorem we have: CD = CB = 15 ⇒ AC = AD + CD = 24 + 15 = 39 Since ABC is a right triangle, we can apply the Pythagorean theorem: AB² + BC² = AC² AB² + 15² = 39² (AB/3)² + 5² = 13² (AB/3)² + 25 = 169 (AB/3)² = 144 AB/3 = 12 ⇒ AB = 36 Now we have a closer look at the interior angles of the triangles ABC and ADO: ABC: α(A), 90°(B), 90°−α(C) ADO: α(A), 90°(D), 90°−α(O) Therefore the two triangles are similar and we can conclude: AC : BC : AB = AO : DO : AD ⇒ AC / BC = AO / DO AC / BC = AO / DO 39 / 15 = (AB − R) / R 13 / 5 = (36 − R) / R (13/5)R = 36 − R (13/5)R + R = 36 (18/5)R = 36 ⇒ R = 10 As an alternative, we can apply the Pythagorean theorem again: AD² + DO² = AO² 24² + R² = (36 − R)² 576 + R² = 1296 − 72R + R² 72R = 720 ⇒ R = 10 Best regards from Germany

@PreMath Год назад

Super ❤️ Thanks for sharing! Cheers! You are awesome. Keep it up 👍

@abstragic4216 Год назад

DC=15, so AC =39, so AB=36 (5-12-13 triangle). OD/AD = BC/AB (similar triangles). OD/24 = 15/36 so OD = 10.

@misterenter-iz7rz Год назад

Although CD is not given, it must be equal to CB for equal tangents, then AB is sqrt(39^2-15^2)=36, now let r be the radius, so r^2+24^2=(36-r)^2, thus 72r=36^2-24^2=720, and then r=10.😊

@PreMath Год назад

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

@D.P9 Год назад

Sir it can be solved in one another method 3rd method Use 5, 12, 13 triplets Sides are 15, 36 and 39 Now inradius = p b /(P + h ) = 15 ×36/(15+39) = 540/54 = 10

@PreMath Год назад

Thanks for sharing! Cheers! You are awesome. Keep it up 👍

@ИванПоташов-о8ю Год назад

Area ABC=Area ABO+Area BCO=0,5*39r+0,5*15r=27r, On the other hand, Area ABC=0,5*15*36=270. So, 27r=270 and r=10

@PreMath Год назад

Thanks for sharing! Cheers! You are awesome. Keep it up 👍

@AmirgabYT2185 7 месяцев назад

r=15√39/13≈7,21

@alster724 Год назад

Method 2 is easier. I used that technique since I find the Pythagorean Theorem more convenient.

@angeluomo Год назад

I used the second method.

@leewilliam3417 Год назад

Mmmmm😊

@devondevon4366 Год назад

@자유천지 2 месяца назад

24*15/36=10

@arnavkange1487 Год назад

why u not calculate DC length ?

@ybodoN Год назад

No need to calculate DC since BC is given and, by the two tangent theorem, DC = BC 😉

@温油-b8x Год назад

用一个公切线，再用个相似成比例，不就求出来了吗？

@PreMath Год назад

Thanks

@misterenter-iz7rz Год назад

DC is unknown?

@rajendrab.shetty321 Год назад

Two tangents drawn to the circle from the same point are equal.

@ashishmchauhan1 Год назад

Radius 10

@mariorossi5857 Год назад

Pythagorean theorem: (x+2r)² = (24+15)² - 15² x+2r = 36 cm (Segment AB) Tangent secant theorem : (x + 2r).x = 24² 36 . x = 576 x = 16 cm r = 10 cm ( Solved ✓ )