Notes: `a = 75.4, b = 52, c = 52` Use Heron's Area Formula to find the area of the triangle.

Saturday, 8 August 2015

Given: $\displaystyle{a}={75.4},{b}={52},{b}={52}$

Heron's area formula $\displaystyle{A}=\sqrt{{{s}{\left({s}-{a}\right)}{\left({s}-{b}\right)}{\left({s}-{c}\right)}}}$

where $\displaystyle{s}=\frac{{{a}+{b}+{c}}}{{2}}$

$\displaystyle{s}=\frac{{{75.4}+{52}+{52}}}{{2}}=\frac{{179.4}}{{2}}={89.7}$

$\displaystyle{A}=\sqrt{{{\left({89.7}\right)}{\left({89.7}-{75.4}\right)}{\left({89.7}-{52}\right)}{\left({89.7}-{2}\right)}}}$

$\displaystyle{A}=\sqrt{{{\left({89.7}\right)}{\left({14.3}\right)}{\left({37.7}\right)}{\left({37.7}\right)}}}$

$\displaystyle{A}={1350.22}$

The area is 1350.22 squared units.

Given: $\displaystyle{a}={75.4},{b}={52},{b}={52}$

Heron's area formula $\displaystyle{A}=\sqrt{{{s}{\left({s}-{a}\right)}{\left({s}-{b}\right)}{\left({s}-{c}\right)}}}$

where $\displaystyle{s}=\frac{{{a}+{b}+{c}}}{{2}}$

$\displaystyle{s}=\frac{{{75.4}+{52}+{52}}}{{2}}=\frac{{179.4}}{{2}}={89.7}$

$\displaystyle{A}=\sqrt{{{\left({89.7}\right)}{\left({89.7}-{75.4}\right)}{\left({89.7}-{52}\right)}{\left({89.7}-{2}\right)}}}$

$\displaystyle{A}=\sqrt{{{\left({89.7}\right)}{\left({14.3}\right)}{\left({37.7}\right)}{\left({37.7}\right)}}}$

$\displaystyle{A}={1350.22}$

The area is 1350.22 squared units.

Notes