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:

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Answer:

$648.8$ square centimeters