Question

The sides of triangular field are 65 m,156 m and 169 m . find the area of the field and the length of the shortest altitude

Answer 1

Answer:

let the sides be a,b&c

semiperimeter (s) = (65+156+169)/2

= 390/2

= 195

area = √s(s-a)(s-b)(s-c)

= √195(195-65)(195-156)(195-169)

= √ 195(130)(39)(26)

= √ 25704900

= 5070m²

area of ∆ = 1/2 × base × height

Or,5070 = 1/2 × 169(largest base)× altitude

smallest altitude = 5070÷169/2

= 60m

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Answer 2

Step-by-step explanation:

60 meter²

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