Question

Find the area of a triangle whose two sides are 12 inches and 14 inches long, and has a perimeter of 34 inches.

Answer 1

Answer:

$\huge\boxed{A=3\sqrt{255}\ in^2\approx47.91\ in^2}$

Step-by-step explanation:

We have two sides

$a=12in;\ b=14in$

and the preimeter

$P=34in$

We can calculate the length of the third side:

$c=P-a-b$

substitute

$c=34-12-14=8\ (in)$

Use the Heron's formula:

$A=\sqrt{p(p-a)(p-b)(p-c)$

where

$p=\dfrac{P}{2}$

substitute:

$p=\dfrac{34}{2}=17\ (in)\\\\A=\sqrt{17(17-12)(17-14)(17-8)}=\sqrt{(17)(5)(3)(9)}\\\\=\sqrt{9}\cdot\sqrt{(17)(5)(3)}=3\sqrt{255}\ (in^2)\approx47.91\ (in^2)$