Question

in $delta klm$, $k = 50$ cm, $l = 78$ cm and $m = 52$ cm. find the area of $delta klm$ to the nearest 10th of a square centimeter.

Explanation:

Step1: Calculate the semi - perimeter

The semi - perimeter $s=\frac{k + l+m}{2}=\frac{50 + 78+52}{2}=\frac{180}{2}=90$ cm.

Step2: Apply Heron's formula

The area $A=\sqrt{s(s - k)(s - l)(s - m)}$. Substitute $s = 90$, $k = 50$, $l = 78$, and $m = 52$ into the formula: $A=\sqrt{90(90 - 50)(90 - 78)(90 - 52)}=\sqrt{90\times40\times12\times38}=\sqrt{40\times12\times90\times38}=\sqrt{480\times90\times38}=\sqrt{43200\times38}=\sqrt{1641600}=1281.3$ $cm^{2}$.

Answer:

$1281.3$ $cm^{2}$