2 mathematics Q.

1) Let r be the radius of the incircle :Area of △ = Perimeter x The radius of the incircle / 2By Hero's formula :√ [ (9/2) (9/2 - 2) (9/2 - 3) (9/2 - 4) ] = (9/2) r / 2√ [ (9/2) (5/2) (3/2) (1/2) ] = (9/2) r / 2√ (135/16) = 9r / 43√15 / 4 = 9r / 4r = √15 / 3 The radius of the incircle = √15 / 3 cmThe area of the incircle = π (√15 / 3)² = 5π/3 cm²
2)6√3 / (7 + √13)= 6√3(7 - √13) / [(7 + √13) (7 - √13)]= (42√3 - 6√39) / (49 - 13)= (42√3 - 6√39) / 36= (7√3 - √39) / 6

2011-08-13 23:59:45 補充：
1) Corrections :

Let r be the radius of the incircle :

Area of △ = Perimeter x The radius of the incircle / 2

By Heron's formula :

√ [ (9/2) (9/2 - 2) (9/2 - 3) (9/2 - 4) ] = 9 r / 2

√ [ (9/2) (5/2) (3/2) (1/2) ] = 9 r / 2

2011-08-13 23:59:50 補充：
√ (135/16) = 9r / 2

3√15 / 4 = 9r / 2

r = √15 / 6

The radius of the incircle = √15 / 6 cm

The area of the incircle = π (√15 / 6)² = 5π/12 cm²