Question

in $delta wxy$, $w = 4.7$ cm, $x = 9.2$ cm and $y = 8.8$ cm. find the area of $delta wxy$ to the nearest 10th of a square centimeter.

Explanation:

Step1: Calculate semi - perimeter

Let \(s=\frac{w + x + y}{2}\), where \(w = 4.7\), \(x = 9.2\), \(y = 8.8\). Then \(s=\frac{4.7+9.2 + 8.8}{2}=\frac{22.7}{2}=11.35\) cm.

Step2: Use Heron's formula

The area \(A=\sqrt{s(s - w)(s - x)(s - y)}\). Substitute \(s = 11.35\), \(w = 4.7\), \(x = 9.2\), \(y = 8.8\) into the formula:
\(A=\sqrt{11.35(11.35 - 4.7)(11.35 - 9.2)(11.35 - 8.8)}\)
\(=\sqrt{11.35\times6.65\times2.15\times2.55}\)
\(=\sqrt{11.35\times(6.65\times2.15\times2.55)}\)
\(=\sqrt{11.35\times(14.3975\times2.55)}\)
\(=\sqrt{11.35\times36.713625}\)
\(=\sqrt{417.69964375}\approx20.4\) \(cm^{2}\)

Answer:

\(20.4\) \(cm^{2}\)