Question

Determine whether a triangle can be formed with the given side lengths. If so, use Heron's formula to find the area of the triangle. a = 240 b = 121 c = 302

Answer 1

These sides will form a triangle because the longest side (302) is less than the sum of the other 2 sides.  http://www.1728.org/trianinq.htm

Using Heron's formula, we first have to determine the semi-perimeter of the triangle.  (perimeter divided by 2)
(240 + 121 + 302) / 2 = 331.50 = semi-perimeter (s)

Now, to use Heron's formula:
area = square root (s • (s - a) • (s - b) • (s - c))
area = square root (331.5 • (331.5 - a) • (331.5 - b) • (331.5 - c))
area = square root (331.5 • (331.5 - 240) • (331.5 - 121) • (331.5 - 302))
area = square root (331.5 • (331.5 - 240) • (331.5 - 121) • (331.5 - 302))
area = square root (331.5 • (91.50) • (210.5) • (29.5))
area = square root ( 188,355,689.44 )
area = 13,724.27

Source for Heron's Formula:
http://www.1728.org/triang.htm

Answer 2

Answer:

The area of triangle formed by given side lengths is 13724.27.

Step-by-step explanation:

A triangle can be formed If the sum of the other two sides (except largest side) is greater than the largest side that is a+b>c.

Here, given side lengths:

a = 240 

b = 121

c = 302

Here, largest side is c = 302

so taking sum of other two sides,

a+b = 240+121 = 361> 302

Hence, triangle can be formed with a, b and c as given lengths.

Calculating area using Heron's formula,

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

where s is the semi perimeter

Formula to find semi perimeter is

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

So, first we find the semi perimeter by using given sides,

$s=\frac{240+121+302}{2}$

$s=\frac{663}{2}$

$s=331.5$

Now put the value of s,a,b and c in the Heron's formula,

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

$Area=\sqrt{331.5(331.5-240)(331.5-121)(331.5-302)}$

$Area=\sqrt{331.5\times91.5\times210.5\times29.5}$

$Area=\sqrt{188355689.438}$

$Area=13724.2737308$

Therefore, the area of triangle is approx. 13724.27.