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

The semi - perimeter $s$ of a triangle with side lengths $a$, $b$, and $c$ is given by $s=\frac{a + b + c}{2}$. Here, $a = 54$, $b = 24$, $c = 59$, so $s=\frac{54+24 + 59}{2}=\frac{137}{2}=68.5$ cm.

Step2: Use Heron's formula

Heron's formula for the area $A$ of a triangle is $A=\sqrt{s(s - a)(s - b)(s - c)}$. Substitute $s = 68.5$, $a = 54$, $b = 24$, $c = 59$ into the formula:
$A=\sqrt{68.5(68.5 - 54)(68.5 - 24)(68.5 - 59)}$
$=\sqrt{68.5\times14.5\times44.5\times9.5}$
$=\sqrt{68.5\times(14.5\times44.5\times9.5)}$
$=\sqrt{68.5\times(645.25\times9.5)}$
$=\sqrt{68.5\times6129.875}$
$=\sqrt{420896.4375}\approx648.7$ $cm^{2}$

Answer:

$648.7$ $cm^{2}$