Question

area of a triangle is 60 square units. if 2 of its sides are 8 units and 17 units find the 3 side​

Answer 1

Let the third side be x units.

Then 's' = (8+17+x)/2 = (25=x)/2

Therefore, by Heron's formula,

$\sqrt{s(s-a)(s-b)(s-c)}$

= $\sqrt{(25+x)(25+x-8)(25+x-17)(25+x-x)}$

= $\sqrt{(25)(17+x)(8+x)(25)}$

$25\sqrt{(17+x)(8+x)}$

Now,

$25\sqrt{(17+x)(8+x)} = 60\\\\\sqrt{(17+x)(8+x)}=\frac{12}{5} \\\\\\(17+x)(8+x)=\frac{144}{12}$

now, either x = $\frac{144}{12} -17 or, x = \frac{144}{12} -8$

⇒ x = -8 or, x = 4

Since a side can never be negative, the side will be 4 units. (ANS)

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Answer 2

Step-by-step explanation:

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