Question

the area of triangle efg is 34.8 square units. what is the perimeter of triangle efg to the nearest unit?
29 units
30 units
41 units
60 units
trigonometric area formula. area = \frac{1}{2}ab\sin(c)

Explanation:

Step1: Use the area formula to check given sides and angle

Given area formula $A=\frac{1}{2}ab\sin(C)$. Let $a = 10$, $b = 12$ and $C = 56^{\circ}$. Then $A=\frac{1}{2}\times10\times12\times\sin(56^{\circ})=60\times\sin(56^{\circ})\approx60\times0.829=49.74$ which is inconsistent with the given area of 34.8 square - units. Let's assume the sides are $a$ and $b$ and the included - angle $C$. We know $A = 34.8=\frac{1}{2}ab\sin(C)$.

Step2: Use the Law of Cosines to find the third side

Let the sides of the triangle be $a = 10$, $b = 12$. By the Law of Cosines, $c^{2}=a^{2}+b^{2}-2ab\cos(C)$. First, from the area formula $34.8=\frac{1}{2}\times10\times12\times\sin(C)$, we can solve for $\sin(C)$:
\[

$$\begin{align*} 34.8&=60\sin(C)\\ \sin(C)&=\frac{34.8}{60}=0.58 \end{align*}$$

\]
Using $\sin^{2}C+\cos^{2}C = 1$, we get $\cos(C)=\pm\sqrt{1 - 0.58^{2}}=\pm\sqrt{1 - 0.3364}=\pm\sqrt{0.6636}\approx\pm0.8146$. Since the angle in a triangle is between $0^{\circ}$ and $180^{\circ}$, and from the area formula we know the angle is non - degenerate, and considering the context of the triangle, we use $\cos(C)\approx0.8146$.
Then, by the Law of Cosines $c^{2}=10^{2}+12^{2}-2\times10\times12\times0.8146=100 + 144-195.504 = 48.496$, so $c=\sqrt{48.496}\approx7$.

Step3: Calculate the perimeter

The perimeter $P=a + b + c$. Substituting $a = 10$, $b = 12$ and $c\approx7$, we get $P=10 + 12+7=29$.

Answer:

29 units