Question

compute the perimeter of the rectangle using the distance formula. (round to the nearest integer)
a 33
b 34
c 52
d 69
e 70

Explanation:

Step1: Find the coordinates of the rectangle's vertices

Assume the vertices of the rectangle are \( (x_1,y_1)\),\( (x_2,y_2)\),\( (x_3,y_3)\),\( (x_4,y_4)\). From the graph (assuming standard grid - based coordinates), let's say two adjacent vertices \(A(x_1,y_1)=(6, - 10)\) and \(B(x_2,y_2)=(12,-16)\).

Step2: Use the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)

For the length \(l\) (distance between two adjacent vertices):

$$
LATEXBLOCK0
$$

For the width \(w\) (distance between another pair of adjacent vertices, assuming perpendicular sides). Let's say \(A(6,-10)\) and \(C(x_3,y_3)=(18,-10)\)

$$
LATEXBLOCK1
$$

Step3: Use the perimeter formula \(P=2(l + w)\)

$$
LATEXBLOCK2
$$

(Note: If we assume more accurate coordinate - picking. Let's assume four vertices of the rectangle: \(A(6,-10)\), \(B(12,-16)\), \(C(18,-10)\), \(D(12,-4)\)
For \(AB\): \(d_{AB}=\sqrt{(12 - 6)^2+(-16 + 10)^2}=\sqrt{36+36}=\sqrt{72}\approx8.5\)
For \(BC\): \(d_{BC}=\sqrt{(18 - 12)^2+(-10 + 16)^2}=\sqrt{36 + 36}=\sqrt{72}\approx8.5\)
For \(CD\): \(d_{CD}=\sqrt{(12 - 18)^2+(-4+10)^2}=\sqrt{36 + 36}=\sqrt{72}\approx8.5\)
For \(DA\): \(d_{DA}=\sqrt{(6 - 12)^2+(-10 + 4)^2}=\sqrt{36+36}=\sqrt{72}\approx8.5\). \(P = 4\times\sqrt{72}\approx4\times8.5 = 34\))

Answer:

B. 34