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Answer :
We are given the vertices of triangle [tex]\( \triangle KLM \)[/tex] as
[tex]\[
K(-5,\, 18), \quad L(10,\, -2), \quad \text{and} \quad M(-5,\, -10).
\][/tex]
Below is a step-by-step solution for parts (a), (b), and (c):
──────────────────────────────
Step 1. Find the Length of Each Side
We use the distance formula between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex]:
[tex]\[
\text{distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.
\][/tex]
(a) Find the lengths of sides [tex]\(KL\)[/tex], [tex]\(LM\)[/tex], and [tex]\(KM\)[/tex].
1. Length of side [tex]\(KL\)[/tex] (between [tex]\(K(-5,18)\)[/tex] and [tex]\(L(10,-2)\)[/tex]):
[tex]\[
\begin{aligned}
KL &= \sqrt{(10 - (-5))^2 + (-2 - 18)^2} \\
&= \sqrt{(15)^2 + (-20)^2} \\
&= \sqrt{225 + 400} \\
&= \sqrt{625} \\
&= 25.
\end{aligned}
\][/tex]
2. Length of side [tex]\(LM\)[/tex] (between [tex]\(L(10,-2)\)[/tex] and [tex]\(M(-5,-10)\)[/tex]):
[tex]\[
\begin{aligned}
LM &= \sqrt{(-5 - 10)^2 + (-10 - (-2))^2} \\
&= \sqrt{(-15)^2 + (-8)^2} \\
&= \sqrt{225 + 64} \\
&= \sqrt{289} \\
&= 17.
\end{aligned}
\][/tex]
3. Length of side [tex]\(KM\)[/tex] (between [tex]\(K(-5,18)\)[/tex] and [tex]\(M(-5,-10)\)[/tex]):
[tex]\[
\begin{aligned}
KM &= \sqrt{(-5 - (-5))^2 + (-10 - 18)^2} \\
&= \sqrt{(0)^2 + (-28)^2} \\
&= \sqrt{0 + 784} \\
&= 28.
\end{aligned}
\][/tex]
──────────────────────────────
Step 2. Find the Perimeter of the Triangle
The perimeter [tex]\(P\)[/tex] is the sum of the side lengths:
[tex]\[
\begin{aligned}
P &= KL + LM + KM \\
&= 25 + 17 + 28 \\
&= 70.
\end{aligned}
\][/tex]
──────────────────────────────
Step 3. Find the Area of the Triangle
We use the formula for the area of a triangle given its vertices:
[tex]\[
\text{Area} = \frac{1}{2} \left| x_1\bigl(y_2-y_3\bigr) + x_2\bigl(y_3-y_1\bigr) + x_3\bigl(y_1-y_2\bigr) \right|.
\][/tex]
Labeling the vertices as:
[tex]\[
(x_1, y_1) = (-5,\, 18), \quad (x_2, y_2) = (10,\, -2), \quad (x_3, y_3) = (-5,\, -10),
\][/tex]
we substitute into the area formula:
[tex]\[
\begin{aligned}
\text{Area} &= \frac{1}{2} \Big| (-5)((-2)-(-10)) + 10\bigl((-10)-18\bigr) + (-5)\bigl(18-(-2)\bigr) \Big| \\
&= \frac{1}{2} \Big| (-5)(8) + 10(-28) + (-5)(20) \Big| \\
&= \frac{1}{2} \Big| -40 - 280 - 100 \Big| \\
&= \frac{1}{2} (420) \\
&= 210.
\end{aligned}
\][/tex]
──────────────────────────────
Final Answers
(a) The side lengths are:
[tex]\[
KL = 25, \quad LM = 17, \quad KM = 28.
\][/tex]
(b) The perimeter of [tex]\( \triangle KLM \)[/tex] is:
[tex]\[
70.
\][/tex]
(c) The area of [tex]\( \triangle KLM \)[/tex] is:
[tex]\[
210.
\][/tex]
[tex]\[
K(-5,\, 18), \quad L(10,\, -2), \quad \text{and} \quad M(-5,\, -10).
\][/tex]
Below is a step-by-step solution for parts (a), (b), and (c):
──────────────────────────────
Step 1. Find the Length of Each Side
We use the distance formula between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex]:
[tex]\[
\text{distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.
\][/tex]
(a) Find the lengths of sides [tex]\(KL\)[/tex], [tex]\(LM\)[/tex], and [tex]\(KM\)[/tex].
1. Length of side [tex]\(KL\)[/tex] (between [tex]\(K(-5,18)\)[/tex] and [tex]\(L(10,-2)\)[/tex]):
[tex]\[
\begin{aligned}
KL &= \sqrt{(10 - (-5))^2 + (-2 - 18)^2} \\
&= \sqrt{(15)^2 + (-20)^2} \\
&= \sqrt{225 + 400} \\
&= \sqrt{625} \\
&= 25.
\end{aligned}
\][/tex]
2. Length of side [tex]\(LM\)[/tex] (between [tex]\(L(10,-2)\)[/tex] and [tex]\(M(-5,-10)\)[/tex]):
[tex]\[
\begin{aligned}
LM &= \sqrt{(-5 - 10)^2 + (-10 - (-2))^2} \\
&= \sqrt{(-15)^2 + (-8)^2} \\
&= \sqrt{225 + 64} \\
&= \sqrt{289} \\
&= 17.
\end{aligned}
\][/tex]
3. Length of side [tex]\(KM\)[/tex] (between [tex]\(K(-5,18)\)[/tex] and [tex]\(M(-5,-10)\)[/tex]):
[tex]\[
\begin{aligned}
KM &= \sqrt{(-5 - (-5))^2 + (-10 - 18)^2} \\
&= \sqrt{(0)^2 + (-28)^2} \\
&= \sqrt{0 + 784} \\
&= 28.
\end{aligned}
\][/tex]
──────────────────────────────
Step 2. Find the Perimeter of the Triangle
The perimeter [tex]\(P\)[/tex] is the sum of the side lengths:
[tex]\[
\begin{aligned}
P &= KL + LM + KM \\
&= 25 + 17 + 28 \\
&= 70.
\end{aligned}
\][/tex]
──────────────────────────────
Step 3. Find the Area of the Triangle
We use the formula for the area of a triangle given its vertices:
[tex]\[
\text{Area} = \frac{1}{2} \left| x_1\bigl(y_2-y_3\bigr) + x_2\bigl(y_3-y_1\bigr) + x_3\bigl(y_1-y_2\bigr) \right|.
\][/tex]
Labeling the vertices as:
[tex]\[
(x_1, y_1) = (-5,\, 18), \quad (x_2, y_2) = (10,\, -2), \quad (x_3, y_3) = (-5,\, -10),
\][/tex]
we substitute into the area formula:
[tex]\[
\begin{aligned}
\text{Area} &= \frac{1}{2} \Big| (-5)((-2)-(-10)) + 10\bigl((-10)-18\bigr) + (-5)\bigl(18-(-2)\bigr) \Big| \\
&= \frac{1}{2} \Big| (-5)(8) + 10(-28) + (-5)(20) \Big| \\
&= \frac{1}{2} \Big| -40 - 280 - 100 \Big| \\
&= \frac{1}{2} (420) \\
&= 210.
\end{aligned}
\][/tex]
──────────────────────────────
Final Answers
(a) The side lengths are:
[tex]\[
KL = 25, \quad LM = 17, \quad KM = 28.
\][/tex]
(b) The perimeter of [tex]\( \triangle KLM \)[/tex] is:
[tex]\[
70.
\][/tex]
(c) The area of [tex]\( \triangle KLM \)[/tex] is:
[tex]\[
210.
\][/tex]
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