Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary.
-$$\begin{array} { l }
\mathrm { a } = 14.5 \mathrm {~cm} \
\mathrm {~b} = 14.3 \mathrm {~cm} \
\mathrm { c } = 13.1 \mathrm {~cm}
\end{array}$$
A) $87 \mathrm {~cm} ^ { 2 }$
B) $84 \mathrm {~cm} ^ { 2 }$
C) $93 \mathrm {~cm} ^ { 2 }$
D) $90 \mathrm {~cm} ^ { 2 }$

Question

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary.
-$$\begin{array} { l } 
\mathrm { a } = 14.5 \mathrm {~cm} \
\mathrm {~b} = 14.3 \mathrm {~cm} \
\mathrm { c } = 13.1 \mathrm {~cm}
\end{array}$$
A) $87 \mathrm {~cm} ^ { 2 }$
B) $84 \mathrm {~cm} ^ { 2 }$
C) $93 \mathrm {~cm} ^ { 2 }$
D) $90 \mathrm {~cm} ^ { 2 }$

Quizplus · Accepted Answer

To find the area of a triangle with sides of lengths $a$, $b$, and $c$, you can use Heron's formula. First, calculate the semi-perimeter $s = \frac{a + b + c}{2}$. For this triangle, $s = \frac{14.5 + 14.3 + 13.1}{2} = 20.95$. Then, the area $A$ is given by $\sqrt{s(s - a)(s - b)(s - c)}$. Plugging in the values, $A = \sqrt{20.95(20.95 - 14.5)(20.95 - 14.3)(20.95 - 13.1)}$, which approximately equals $84 \mathrm{cm}^2$ when rounded to the nearest tenth.

Find the Area of Triangle ABC with the Given Parts a=14.5 cm b=14.3 cmc=13.1 cm\begin{array} { l } \mathrm { a } = 14.5 \mathrm {~cm} \\\mathrm {~b} = 14.3 \mathrm {~cm} \\\mathrm { c } = 13.1 \mathrm {~cm}\end{array}a=14.5 cm b=14.3 cmc=13.1 cm​

Find the Area of Triangle ABC with the Given Parts $\begin{array} { l } \mathrm { a } = 14.5 \mathrm {~cm} \\\mathrm {~b} = 14.3 \mathrm {~cm} \\\mathrm { c } = 13.1 \mathrm {~cm}\end{array}$