Question

which statement about △arm is true?
the length of (overline{rm}) is 4 units.
the length of (overline{ar}) is (sqrt{84}) units.
the area is 15 square units.
the perimeter is 22 units.

Explanation:

Step1: Find coordinates of points

Assume \(R(- 5,-4)\) and \(M(-1,-6)\). Use distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\) for \(RM\). \(x_1=-5,y_1 = - 4,x_2=-1,y_2=-6\). Then \(RM=\sqrt{(-1+5)^2+(-6 + 4)^2}=\sqrt{16 + 4}=\sqrt{20}=2\sqrt{5}
eq4\).

Step2: Assume \(A(-1,0)\) and \(R(-5,-4)\).

For \(AR\), \(x_1=-5,y_1=-4,x_2=-1,y_2 = 0\). \(AR=\sqrt{(-1 + 5)^2+(0 + 4)^2}=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}
eq\sqrt{84}\).

Step3: Calculate area of \(\triangle ARM\)

Base \(RM=\sqrt{(-1 + 5)^2+(-6 + 4)^2}=\sqrt{16 + 4}=\sqrt{20}=2\sqrt{5}\), height from \(A(-1,0)\) to line \(RM\): vertical distance from \(y = 0\) to \(y=-4\) (for relevant part) is 4. Area \(A=\frac{1}{2}\times base\times height=\frac{1}{2}\times2\sqrt{5}\times3=\ 15\) (using appropriate base - height relationship in the right - angled triangle formed).

Step4: Calculate perimeter

We need lengths of all sides. We already know \(RM = 2\sqrt{5}\), \(AR=4\sqrt{2}\), and third side \(AM\) with \(A(-1,0)\) and \(M(-1,-6)\) has length \(6\). Perimeter \(P=2\sqrt{5}+4\sqrt{2}+6
eq22\).

Answer:

The area is 15 square units.