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

The semi - perimeter $s=\frac{w + x + y}{2}=\frac{4.7+9.2 + 8.8}{2}=\frac{22.7}{2}=11.35$ cm.

Step2: Use Heron's formula

The area $A=\sqrt{s(s - w)(s - x)(s - y)}$.
Substitute the values: $A=\sqrt{11.35(11.35 - 4.7)(11.35 - 9.2)(11.35 - 8.8)}$.
First, calculate the differences: $11.35−4.7 = 6.65$, $11.35−9.2 = 2.15$, $11.35−8.8 = 2.55$.
Then, $A=\sqrt{11.35\times6.65\times2.15\times2.55}$.
$11.35\times6.65 = 75.4775$, $75.4775\times2.15 = 162.276625$, $162.276625\times2.55 = 413.80539375$.
$A=\sqrt{413.80539375}\approx20.3$ $cm^{2}$.

Answer:

$20.3$ $cm^{2}$