Question

find the length of the longest altitude of a triangle of whose sides are 35,54 and 61 CM.

Answer 1

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Answer 2

Answer:

53.66562 cm

Step-by-step explanation:

a = 35

b = 54

c = 61

Area of triangle using Heron's formula : $A = \sqrt{s(s-a)(s-b)(s-c)}$

Where $s=\frac{a+b+c}{2}$

Substitute the values

$s=\frac{35+54+61}{2}$

$s=75$

$A = \sqrt{75(75-35)(75-54)(75-61)}$

$A =939.1485$

Shortest length = 35 cm

Area of triangle = $\frac{1}{2} \times base \times height$

$939.1485=\frac{1}{2} \times 35 \times height$

$\frac{939.1485 \times 2}{35}=Height$

$53.66562=Height$

Hence the length of the longest altitude is 53.66562 cm