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 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: $A=\sqrt{94.5(94.5 - 88)(94.5 - 40)(94.5 - 61)}$.
$A=\sqrt{94.5\times6.5\times54.5\times33.5}$.
$A=\sqrt{94.5\times(6.5\times54.5\times33.5)}$.
$A=\sqrt{94.5\times(354.25\times33.5)}$.
$A=\sqrt{94.5\times11867.375}$.
$A=\sqrt{1121467.9375}\approx1059.0$ square inches.

Answer:

$1059.0$ square inches