Question

in $delta qrs$, $q = 88$ inches, $r = 40$ inches and $s = 61$ inches. find the area of $delta qrs$ to the nearest 10th of an square inch.

Explanation:

Step1: Calculate the semi - perimeter

The semi - perimeter $s_{0}=\frac{q + r+ s}{2}=\frac{88 + 40+61}{2}=\frac{189}{2}=94.5$ inches.

Step2: Use Heron's formula

The area $A=\sqrt{s_{0}(s_{0}-q)(s_{0}-r)(s_{0}-s)}$.
Substitute the values: $s_{0}=94.5$, $q = 88$, $r = 40$, $s = 61$.
$A=\sqrt{94.5\times(94.5 - 88)\times(94.5 - 40)\times(94.5 - 61)}$
$=\sqrt{94.5\times6.5\times54.5\times33.5}$
$=\sqrt{94.5\times6.5\times1825.75}$
$=\sqrt{94.5\times11867.375}$
$=\sqrt{1121467.9375}\approx1059.0$ square inches.

Answer:

$1059.0$ square inches