in $\triangle def$, $d = 54$ cm, $e = 24$ cm and $f = 59$ cm. find the area of $\triangle def$ to the…

in $\triangle def$, $d = 54$ cm, $e = 24$ cm and $f = 59$ cm. find the area of $\triangle def$ to the nearest 10th of a square centimeter.

in $\triangle def$, $d = 54$ cm, $e = 24$ cm and $f = 59$ cm. find the area of $\triangle def$ to the nearest 10th of a square centimeter.

Answer

Explanation:

Step1: Calculate the semi - perimeter

Let (s=\frac{d + e + f}{2}), where (d = 54), (e=24), and (f = 59). Then (s=\frac{54+24 + 59}{2}=\frac{137}{2}=68.5) cm.

Step2: Use Heron's formula

The area (A=\sqrt{s(s - d)(s - e)(s - f)}). Substitute the values: (A=\sqrt{68.5(68.5 - 54)(68.5 - 24)(68.5 - 59)}). First, calculate the differences: (68.5-54 = 14.5), (68.5 - 24=44.5), (68.5 - 59 = 9.5). Then (A=\sqrt{68.5\times14.5\times44.5\times9.5}). (68.5\times14.5 = 993.25), (993.25\times44.5=44299.625), (44299.625\times9.5 = 420846.4375). So (A=\sqrt{420846.4375}\approx648.7) (cm^{2}).

Answer:

(648.7) (cm^{2})