It could also be solved this way: Radius Of in-circle for a Right angle ∆ is R= a(Cos§+Sin§-1)/2; where a is hypotenuse and § is any one of acute angle. Here R is given if we will find § then we could calculate a from above equation. From figure mentioned in video CE:CF=4:1;CE=4k,CF=k;EF=3k=√(R+r)^2+(R-r)^2=4 that means k=4/3 (Here r is radius of small circle). let §=angle ACB so tan §/2=AF/CF=3/4 So tan§=2tan§/2/(1-tan^2(§/2))=24/7; So Sin§=24/25,Cos§=7/25 So in radius R=a(24/25+7/25-1)/2=3a/25 But R=4 so a=100/3 Area of ∆ABC =1/2*AB*BC=1/2*100/3 Sin§*100/3 Cos§=1/2*100/3*100/3*24/25*7/25=448/3

@mibsaamahmed

19 күн бұрын

Which formula did you use in the first step

@RahulKumar-id5cq

19 күн бұрын

@@mibsaamahmed Suppose perpendicular legs of Right angle ∆(Right angle at B) are AB and BC. let point of tangency of in-Circle be E and F respectively for AB and BC. Drop perpendicular to AB and BC from Centre(let O) of in-Circle . From Figure it could be observed that: AB-R+BC-R=AC; So R=(AB+BC-AC)/2; let Angle ACB=§ then AB=AC Sin§ ;BC=AC Cos§; So R=AC(Sin§+Cos§-1)/2 .

I got the same answer, 448/3. I did the same as you through determining X = 4/3, but then went trigonometric after that. Since 3/4 = 1/(4/3) = tan(C/2), then tan(C) = 2*(3/4)/[1-(3/4)^2] = 24/7. Now, AB = 4 + 4*cot (A/2). Since tan(C) = 24/7 and B = 90, tan(A) = 1/tan(C) = 7/24. Tan(A/2) = u is determined by 7/24 = 2*u / [1- u^2] --> 7 - 7u^2 = 48u --> 7u^2 + 48u - 7 = 0 --> u = [-48 + sqrt(48^2 + 4*7^2)] / (2*7) = [-48 + 50]/14 = 1/7 --> cot (A/2) = 7 --> AD = 4*7 = 28 --> AB = 4+28 = 32 --> [ABC] = 32*(4+4+4/3)/2 = 32*(28/3)/2 = 896/6 = 448/3. Another way to determine AB is by BC*tan(C) = (28/3) * (24/7) = 4*8 = 32. I feel safer with my first method.

7:30 Your deduction that point Q (based on angle bisector) is aligned with points PC is too slight and incomplete.

@Grizzly01-vr4pn

12 күн бұрын

No, I don't agree. The justification begins at 5:59 It's all there, as required.

At 9:45, Math Booster has determined that PC is a straight line and bisects

Well presented. This was an interesting question and at the same time required detailed explanation. You have explained it so well. It needs a lot patience and dedication.You have done a great job 💯

This is shockingly similar to one of the problems that you showed in your channel. The only difference is that you had to apple the circle thrm/angle bisector thrm twice and the Pythagorean Thrm. I also liked the way you showed a calculator free method of calculating y. And I think that if you use the angle bisector thrm, you know that you used it correctly if you established similarity of sides due to a right angle, right? Please clarify if you feel that I have missed something.

Points assigned: Large circle center: O Small circle center: P Point of tangency between circles: T Point of tangency on AB: M Points of tangency on BC: N (circle O), J (circle P) Points of tangency on CA: K (circle P), L (circle O) Draw OM and ON. As radii of circle O, OM = ON = 4. As AB and BC are tangent to circle O at M and N respectively, ∠OMB = ∠BMO = 90°. As ∠MBN = 90° as well, then ∠NOM must equal 90°, so OMBN is a square with side length 4. Draw OP. As the point of tangency between two circles is collinear with the two center points, T is on OP. Therefore OP = OT+PT = 4+1 = 5. Draw PS, where S is the point on ON where PS is perpendicular to ON and parallel to BC. BC is tangent to circles O and P at N and J respectively, so ON and PJ are both perpendicular to BC. Therefore, PSNJ is a rectangle and SN = PJ = 1. As ON = 4, OS = 4-1 = 3. Triangle ∆OSP: OS² + PS² = OP² 3² + PS² = 5² PS² = 25 - 9 = 16 PS = √16 = 4 As NJ = PS, NJ = 4. Draw OL and PV, where V is the point on OL where PV is parallel to CA and perpendicular to OL. Using the same steps as above with PS, PV and KL are also equal to 4. JC and CK are tangents to circle P that intersect at C, so JC = CK. Draw PC. As PJ = PK = 1, JC = CK, and PC is common, ∆PJC and ∆CKP are congruent triangles. NC and CL are tangents to circle O that intersect at C, so NC = CL. As ON = OL = 4, NC = CL, and OC is common, ∆ONC and ∆CLO are congruent triangles. Additionally, as NJ = KL, PC is collinear with OC, so ∆ONC, ∆CLO, ∆PJC, and ∆CKP are all similar. Let JC = CK = x. ON/PJ = NC/JC 4/1 = (x+4)/x x + 4 = 4x 3x = 4 x = 4/3 BC = BN + NJ + JC BC = 4 + 4 + 4/3 = 28/3 LA and AM are tangents to circle O that intersect at A, so LA = AM. Let LA = AM = y. Triangle ∆ABC: BC² + AB² = CA² (28/3)² + (y+4)² = (y+4+4/3)² = (y+16/3)² 784/9 + y² + 8y + 16 = y² + 32y/3 + 256/9 32y/3 - 8y + 256/9 - 928/9 = 0 32y - 24y - 672/3 = 0 8y - 224 = 0 y = 224/8 = 28 AB = 28 + 4 = 32 [ABC] = bh/2 = (28/3)(32)/2 [ABC] = 14(32)/3 = 448/3 ≈ 149.333 sq units

@RahulKumar-id5cq

19 күн бұрын

I would like to suggest u if u will do nomenclature of points as done by math booster(as much possible )it could be very easy for others to understand what u have written.

@ShreyasKote

16 күн бұрын

Hatts of to you