Perimeter of triangle: P = a + b + c P = 13+14+15 = 42 cm Area of triangle: A² = P/2 . (P/2-a) . (P/2-b). (P/2-c) A = 84 cm² A = Σ (½.b.h) A = ½R (a + b + c) A = ½R (13+14+15) A = 21.R R = A/21 = 84/21 R = 4 cm Area of circle: A = π R² A = 16π cm² ( Solved √ )
Hi i think there could be more ways to solve, i know one. By knowing that a + b =13, b + c = 14, and a+c = 15, solving the equations you get a= 7 (bottom leftmost segment) Next by cosine rule you can obtain the angle that contains bottom line and left line. So, radius of the circle becomes a tmes sine of that angle, divided by thye cosine Thanks
in radius be r then s = (a + b + c)/2 r * s = √ ( s ( s - a)( s - b)( s - c)) r = = √ ( ( s - a ) ( s - b)( s - c)/s) Desired area π r ^2 = π ( s - a )( s - b)( s - c)/s
Perimeter of triangle:
P = a + b + c
P = 13+14+15 = 42 cm
Area of triangle:
A² = P/2 . (P/2-a) . (P/2-b). (P/2-c)
A = 84 cm²
A = Σ (½.b.h)
A = ½R (a + b + c)
A = ½R (13+14+15)
A = 21.R
R = A/21 = 84/21
R = 4 cm
Area of circle:
A = π R²
A = 16π cm² ( Solved √ )
Hi i think there could be more ways to solve, i know one. By knowing that a + b =13, b + c = 14, and a+c = 15, solving the equations you get a= 7 (bottom leftmost segment) Next by cosine rule you can obtain the angle that contains bottom line and left line. So, radius of the circle becomes a tmes sine of that angle, divided by thye cosine Thanks
Very nice. Method 2 is clever. Who would think that the given triangle comprises two pythagorean triplets?
Please, solve only ntse questions.! 🙏
radius=4 unit. Area of green circle=16.pi sq unit.when pi=22/7
in radius be r then s = (a + b + c)/2
r * s = √ ( s ( s - a)( s - b)( s - c))
r = = √ ( ( s - a ) ( s - b)( s - c)/s)
Desired area
π r ^2
= π ( s - a )( s - b)( s - c)/s
method 2, BD vertical to AC is not proved strictly.
Combine the 2 right angled triangles. You will get triangle ABC.
Area of the green circle is 16pi ❤🔥❤🔥❤🔥
This question is in class 10 ncert ch - circle...🧐
16π