A quadrilateral has vertices at (0,1), (3,4), (4,3) and (3,0). Its perimeter can be expressed in the form a\sqrt2+b\sqrt{10} with A and B integers. What is the sum of A and B?

Respuesta :

Answer:

The sum is 6

Step-by-step explanation:

Given

[tex]Vertices: (0,1), (3,4), (4,3), (3,0)[/tex]

[tex]Perimeter: a\sqrt{2} + b\sqrt{10}[/tex]

Required

[tex]a + b[/tex]

The first step is to name each points, as follows

[tex]A: (0,1)\\B: (3,4)\\C: (4,3)\\D: (3,0)[/tex]

Next is to calculate the distance between each consecutive point

We'll calculate the distance AB, BC, CD and DA

Distance between points is calculated as thus;

[tex]d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}[/tex]

Calculating distance AB

[tex]A: (0,1)\\B: (3,4)[/tex]

Here,

[tex]x_1 = 0; x_2 = 3[/tex]

[tex]y_1 = 1; y_2 = 4[/tex]

[tex]AB = \sqrt{(0 - 3)^2 + (1 - 4)^2}[/tex]

[tex]AB = \sqrt{(- 3)^2 + (-3)^2}[/tex]

[tex]AB = \sqrt{9+9}[/tex]

[tex]AB = \sqrt{18}[/tex]

Expand 18 as 9 * 2

[tex]AB = \sqrt{9 * 2}[/tex]

Split surds

[tex]AB = \sqrt{9} * \sqrt{2}[/tex]

Take square root of 9

[tex]AB = 3 * \sqrt{2}[/tex]

[tex]AB = 3 \sqrt{2}[/tex]

Calculating distance BC

[tex]B: (3,4)\\C: (4,3)[/tex]

Here,

[tex]x_1 = 3; x_2 = 4[/tex]

[tex]y_1 = 4; y_2 = 3[/tex]

[tex]BC = \sqrt{(3 - 4)^2 + (4 - 3)^2}[/tex]

[tex]BC = \sqrt{(-1)^2 + (1)^2}[/tex]

[tex]BC = \sqrt{1 + 1}[/tex]

[tex]BC = \sqrt{2}[/tex]

Calculating distance CD

[tex]C: (4,3)\\D: (3,0)[/tex]

Here,

[tex]x_1 = 4; x_2 = 3[/tex]

[tex]y_1 = 3; y_2 = 0[/tex]

[tex]CD = \sqrt{(4 - 3)^2 + (3 - 0)^2}[/tex]

[tex]CD = \sqrt{(1)^2 + (3 )^2}[/tex]

[tex]CD = \sqrt{1 + 9}[/tex]

[tex]CD = \sqrt{10}[/tex]

Calculating distance CD

[tex]D: (3,0)\\A: (0,1)[/tex]

Here,

[tex]x_1 = 3; x_2 = 0[/tex]

[tex]y_1 = 0; y_2 = 1[/tex]

[tex]DA = \sqrt{(3 - 0)^2 + (0 - 1)^2}[/tex]

[tex]DA = \sqrt{(3)^2 + (- 1)^2}[/tex]

[tex]DA = \sqrt{9 + 1}[/tex]

[tex]DA = \sqrt{10}[/tex]

At this point, the perimeter can then be calculated

[tex]Perimeter = AB + BC + CD + DA[/tex]

[tex]Perimeter = 3 \sqrt{2} + \sqrt{2}\ + \sqrt{10} + \sqrt{10}[/tex]

[tex]Perimeter = 4 \sqrt{2} + 2\sqrt{10}[/tex]

From the given parameters;

[tex]Perimeter: a\sqrt{2} + b\sqrt{10}[/tex]

This implies that;

[tex]a\sqrt{2} + b\sqrt{10} = 4 \sqrt{2} + 2\sqrt{10}[/tex]

By comparison;

a = 4 and b = 2

Hence;

[tex]a + b = 4 + 2[/tex]

[tex]a +b = 6[/tex]