Question

5 find the exact perimeter and area of the figure. * (25 points) image of a coordinate grid with a quadrilateral then the multiple - choice options: $p = 2sqrt{10}$ units; $a = 4sqrt{10}$ units², $p = 10$ units; $a = 40$ units², $p = 10$ units; $a = 4sqrt{10}$ units², $p = 4sqrt{10}$ units; $a = 10$ units²

Explanation:

Step1: Identify the figure (a parallelogram)

The figure is a parallelogram. For a parallelogram, perimeter is \(2\times(\text{side}_1 + \text{side}_2)\), area is \(\text{base}\times\text{height}\).

Step2: Find side lengths using distance formula

Let's assume the coordinates (from grid). Let's take points: Let's say the vertices are \((1,2)\), \((4,3)\), \((5,0)\), \((2,-1)\) (approx from grid). Using distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).

For one side: between \((1,2)\) and \((4,3)\): \(d=\sqrt{(4 - 1)^2+(3 - 2)^2}=\sqrt{9 + 1}=\sqrt{10}\).

Another side: between \((4,3)\) and \((5,0)\): \(d=\sqrt{(5 - 4)^2+(0 - 3)^2}=\sqrt{1 + 9}=\sqrt{10}\)? Wait, no, maybe better to see it's a parallelogram with two pairs of equal sides. Wait, actually, let's check the base and height.

Wait, maybe the figure is a parallelogram with base and height. Alternatively, use the shoelace formula for area.

But let's check the options. The perimeter options: \(4\sqrt{10}\) would be \(2\times(2\sqrt{10})\) no, wait, if each of the four sides? Wait, no, parallelogram has two pairs. Wait, maybe the figure has four sides, each with length \(\sqrt{10}\)? No, wait, let's re - evaluate.

Wait, let's take the coordinates properly. Let's assume the grid has each square of side 1. Let's list the coordinates of the four vertices (from the graph):

Let’s say the four points are: \(A(1,2)\), \(B(4,3)\), \(C(5,0)\), \(D(2,-1)\)

Distance between \(A\) and \(B\): \(\sqrt{(4 - 1)^2+(3 - 2)^2}=\sqrt{9 + 1}=\sqrt{10}\)

Distance between \(B\) and \(C\): \(\sqrt{(5 - 4)^2+(0 - 3)^2}=\sqrt{1+9}=\sqrt{10}\)? No, that can't be. Wait, maybe I got the points wrong.

Wait, another approach: The figure is a parallelogram. Let's find the length of the sides. Let's consider the horizontal and vertical changes.

Suppose one side has a horizontal change of 3 and vertical change of 1? No, wait, the distance formula: for a side with horizontal difference \(a\) and vertical difference \(b\), length is \(\sqrt{a^{2}+b^{2}}\).

Looking at the options, the perimeter with \(4\sqrt{10}\) and area 10. Let's check area. If we use the shoelace formula. Let's assume the coordinates are \((1,2)\), \((4,3)\), \((5,0)\), \((2,-1)\)

Shoelace formula:

List the coordinates in order: \((1,2)\), \((4,3)\), \((5,0)\), \((2,-1)\), \((1,2)\)

Sum of \(x_i y_{i + 1}\): \(1\times3+4\times0 + 5\times(-1)+2\times2=3 + 0-5 + 4 = 2\)

Sum of \(y_i x_{i + 1}\): \(2\times4+3\times5+0\times2+(-1)\times1 = 8 + 15+0 - 1=22\)

Area=\(\frac{1}{2}\times|2 - 22|=\frac{1}{2}\times20 = 10\)

Now perimeter: distance between \((1,2)\) and \((4,3)\): \(\sqrt{(4 - 1)^2+(3 - 2)^2}=\sqrt{9 + 1}=\sqrt{10}\)

Distance between \((4,3)\) and \((5,0)\): \(\sqrt{(5 - 4)^2+(0 - 3)^2}=\sqrt{1 + 9}=\sqrt{10}\)

Wait, no, that's two sides, but a parallelogram has two pairs. Wait, maybe the other two sides are also \(\sqrt{10}\)? No, that would make a rhombus. Wait, no, maybe I made a mistake in coordinates.

Wait, let's take another set of coordinates. Let's say the four points are \((1,2)\), \((3,3)\), \((5,1)\), \((3,0)\)

Distance between \((1,2)\) and \((3,3)\): \(\sqrt{(3 - 1)^2+(3 - 2)^2}=\sqrt{4 + 1}=\sqrt{5}\) No, not matching.

Wait, the correct option is \(P = 4\sqrt{10}\) units and \(A = 10\) units². Let's verify:

If each of the four sides has length \(\sqrt{10}\), perimeter is \(4\sqrt{10}\). For area, if we consider the base and height. Suppose the base is \(\sqrt{10}\) and height is \(\sqrt{10}\)? No, area is 10. Alternatively, if the figure is a parallelogram with base 5 and height 2? No, 5×2 = 10. Wait, maybe the figure can be divided into triangles or rectangles.

Alternatively, let's use the fact that in the grid, the area of a parallelogram can be calculated as the magnitude of the cross - product of two adjacent sides. If two adjacent sides have vectors \(\vec{u}=(3,1)\) and \(\vec{v}=(1, - 3)\), then the area is \(|3\times(-3)-1\times1|=|-9 - 1| = 10\). The length of each side: \(|\vec{u}|=\sqrt{3^{2}+1^{2}}=\sqrt{10}\), \(|\vec{v}|=\sqrt{1^{2}+(-3)^{2}}=\sqrt{10}\). So perimeter is \(2\times(|\vec{u}|+|\vec{v}|)=2\times(2\sqrt{10}) = 4\sqrt{10}\), and area is 10.

Answer:

\(P = 4\sqrt{10}\text{ units};A = 10\text{ units}^2\) (the last option)