Question

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.

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:

\(648.7\) \(cm^{2}\)