find the area of a triangle
sides- 9,40,41
Answers
AB = 9 (c)
BC = 40 (a)
AC = 41 (a)
s = (a + b + c) / 2
= (9 + 40 + 41) / 2
= 90 / 2
= 45
By Heron's formula
Area of ∆ ABC = √ s (s-a) (s-b) (s-c)
= √ 45 (45 - 9) (45 - 40) (45 - 41)
= √ 45 × 36 × 5 × 4
= √ 3 × 3 × 5 × 2 × 2 × 3 × 3 × 5 × 2 × 2
= 3 × 3 × 2 × 2 × 5
= 9 × 20
= 180 unit²
Area of ∆ABC = 180 unit²
hope it's helpful to uhh.....!!!!
Step-by-step explanation:
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Triangle
Solve for
AB = 9 (c)
BC = 40 (a)
AC = 41 (a)
s = (a + b + c) / 2
= (9 + 40 + 41) / 2
= 90 / 2
= 45
By Heron's formula
Area of ∆ ABC = √ s (s-a) (s-b) (s-c)
= √ 45 (45 - 9) (45 - 40) (45 - 41)
= √ 45 × 36 × 5 × 4
= √ 3 × 3 × 5 × 2 × 2 × 3 × 3 × 5 × 2 × 2
= 3 × 3 × 2 × 2 × 5
= 9 × 20
= 180 unit²
Area of ∆ABC = 180 unit²
hope it's helpful to uhh.....!!!!
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