Question

lmn is an isosceles triangle. what is the approximate length of side lm, and what is the approximate perimeter of triangle lmn? round your answer to the nearest tenth.
a. the approximate length of side lm is 6 units, and the approximate perimeter of triangle lmn is 22.8 units.
b. the approximate length of side lm is 6 units, and the approximate perimeter of triangle lmn is 22 units.
c. the approximate length of side lm is 6.4 units, and the approximate perimeter of triangle lmn is 22.8 units.
d. the approximate length of side lm is 6.4 units, and the approximate perimeter of triangle lmn is 22 units.

Explanation:

Step1: Use distance - formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. Let $L(1,1)$, $M(5, - 3)$. Then $x_1 = 1,y_1 = 1,x_2 = 5,y_2=-3$.
\[

$$\begin{align*} LM&=\sqrt{(5 - 1)^2+(-3 - 1)^2}\\ &=\sqrt{4^2+( - 4)^2}\\ &=\sqrt{16 + 16}\\ &=\sqrt{32}\\ &\approx5.7\approx6 \end{align*}$$

\]
Since $\triangle LMN$ is isosceles, assume $LN = LM$. Let's find the length of $LN$ with $N(-3,-3)$ and $L(1,1)$.
\[

$$\begin{align*} LN&=\sqrt{(1+3)^2+(1 + 3)^2}\\ &=\sqrt{4^2+4^2}\\ &=\sqrt{16 + 16}\\ &=\sqrt{32}\\ &\approx5.7\approx6 \end{align*}$$

\]
The length of $MN=\sqrt{(5 + 3)^2+(-3+3)^2}=\sqrt{8^2}=8$.

Step2: Calculate perimeter

The perimeter $P$ of $\triangle LMN$ is $P=LM + LN+MN$. Since $LM\approx6$, $LN\approx6$ and $MN = 8$, $P\approx6+6 + 8=20\approx22$.

Answer:

B. The approximate length of side LM is 6 units, and the approximate perimeter of triangle LMN is 22 units.