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You can calculate semi-perimeter of the triangle = (9 + 3√3)/2 & know area of triangle = 9(/2)√3. Radius is area/semi-perimeter. This eliminates all the constructions.

wow you’re a great maths tutor keep it up!!

actually really informative !

Καλησπέρα σας από Patras(Greece). Mία άλλη λύση: αφού βρούμε τις πλευρές του ορθογωνίου τριγώνου, βρίσκουμε το εμβαδόν το ως το μισό γινόμενο των κάθετων πλευρών: Ε=(9*sqrt3)/2. Mετά, από τον τύπο Ε=s*r, όπου r η ακτίνα του κύκλου και s=(a+b+c)/2 (a,b,c οι πλευρές του τριγώνου) βρίσκουμε την ακτίνα του κύκλου r=[sqrt3*(3-sqrt3)]/2. Tέλος, από τον τύπο Ε_1=π*r^2 βρίσκουμε το εμβαδόν του κύκλου όσο ακριβώς και ο Μagazith. 🤦‍♂

^=read as to the power *=read as square root As per question h=6 Sin 30=p/h So, p/h=(1/2) h=2p So, 2p=6 P=6/2=3......eqn1 b=*(h^2-p^2) =*{(6^2)-(3^2)} =*(36-9)=*27=3.*3 Now, h=6, p=3, b=3.*3 Semiperimeter S=(h+p+b)/2 So, S=(6+3+3.*3)/2=(9+3.*3)/2 =3(3+*3)/2 Area of triangle =(1/2)×p×b =(1/2)×3×3.*3 =(9.*3)/2 Radios r=area triangle /S So, r={(9.*3)/2}÷{3(3+*3)/2} =(9.*3)/{3(3+*3)} 3.*3/(3+*3) ={3.*3(3-*3)}/{(3+*3)(3-*3)} ={3.*3(3-*3)}/6 ={*3(3-*3)}/2 =(3.*3 -3)./2............May be

How to derive formula for area of triangle with given side lengths and radius of inscribed circle Connect vertices with incenter You will have three smaller triangles and notice that radius is height of each triangle an the base is side of original triangle If you sum areas of these three triangles you will get formula

cos(30)=sqrt(3)/2 cos(30)=x/6 sqrt(3)/2 = x/6 x=3sqrt(3) sin(30)=1/2 1/2 = y/6 y = 3 1/2*3*3sqrt(3)=1/2(3+6+3sqrt(3))r 3sqrt(3) = (3+sqrt(3))r 3sqrt(3)(3-sqrt(3)) = 6r sqrt(3)/2(3-sqrt(3)) = r 3/2(sqrt(3) - 1) = r Area = 9/2(2-sqrt(3))pi

A similar question was asked in grade 10 cbse board

Solve it right

Low noice

شكرا

6*1/2=3 3*√3=3√3 3-r+3√3-r=6 2r=3√3-3 r=(3√3-3)/2 Area of blue circle = (3√3-3)/2*(3√3-3)/2*π = (36-18√3)π/4 = (18-9√3)π/2

This was pretty fun to solve glad I got it right :)