Question

1. which expression is equivalent to the perimeter of δghj?

a. ( 5 + sqrt{65} + sqrt{58} )
b. ( sqrt{56} + sqrt{65} + sqrt{32} )
c. ( 5 + sqrt{58} + sqrt{32} )
d. ( sqrt{65} + sqrt{24} + sqrt{58} )
(accompanied by a coordinate grid image showing points g, h, j)

Explanation:

First, we need to find the coordinates of points \( G \), \( H \), and \( J \) (wait, actually \( \triangle GHJ \)? Wait, the problem says \( \triangle GHJ \)? Wait, the graph has \( G \), \( H \), \( J \). Let's get their coordinates. Let's assume each grid is 1 unit.

Looking at the graph:

• \( G \): Let's see, x-coordinate: -3, y-coordinate: 5 (so \( G(-3, 5) \))
• \( J \): x-coordinate: 0, y-coordinate: -2 (so \( J(0, -2) \))
• \( H \): x-coordinate: 4, y-coordinate: 1 (so \( H(4, 1) \))

Now, we need to find the lengths of \( GH \), \( HJ \), and \( GJ \) (wait, the triangle is \( \triangle GHJ \)? Wait, the problem says \( \triangle GHJ \)? Wait, the options have terms like \( \sqrt{65} \), \( \sqrt{58} \), etc. Let's calculate each side using the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).

Step 1: Length of \( GJ \)

Points \( G(-3, 5) \) and \( J(0, -2) \)
\( \Delta x = 0 - (-3) = 3 \)
\( \Delta y = -2 - 5 = -7 \)
Length \( GJ = \sqrt{(3)^2 + (-7)^2} = \sqrt{9 + 49} = \sqrt{58} \) Wait, no, wait: Wait, maybe I misread the points. Wait, maybe the triangle is \( \triangle GHJ \), but let's check again. Wait, maybe \( J \) is a typo, but the options have 5, so maybe one side is horizontal or vertical. Wait, maybe \( G \) is at (-3,5), \( J \) is at (0,-2)? No, maybe \( H \) is at (4,1), \( J \) is at (0,-2), and \( G \) is at (-3,5). Wait, maybe another side: Let's check \( HJ \): points \( H(4,1) \) and \( J(0,-2) \)
\( \Delta x = 0 - 4 = -4 \)
\( \Delta y = -2 - 1 = -3 \)
Length \( HJ = \sqrt{(-4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \). Ah! So \( HJ = 5 \).

Then \( GH \): points \( G(-3,5) \) and \( H(4,1) \)
\( \Delta x = 4 - (-3) = 7 \)
\( \Delta y = 1 - 5 = -4 \)
Length \( GH = \sqrt{(7)^2 + (-4)^2} = \sqrt{49 + 16} = \sqrt{65} \)

Then \( GJ \): points \( G(-3,5) \) and \( J(0,-2) \)
\( \Delta x = 0 - (-3) = 3 \)
\( \Delta y = -2 - 5 = -7 \)
Length \( GJ = \sqrt{(3)^2 + (-7)^2} = \sqrt{9 + 49} = \sqrt{58} \)

Wait, but the triangle is \( \triangle GHJ \), so perimeter is \( GH + HJ + GJ = \sqrt{65} + 5 + \sqrt{58} \), which is option A: \( 5 + \sqrt{65} + \sqrt{58} \).

Wait, let's confirm again:

• \( HJ \): from \( H(4,1) \) to \( J(0,-2) \): horizontal distance 4 (from x=4 to x=0: 4 units left), vertical distance 3 (from y=1 to y=-2: 3 units down). So by Pythagoras, \( 3^2 + 4^2 = 9 + 16 = 25 \), so \( \sqrt{25} = 5 \). Correct, so \( HJ = 5 \).

• \( GH \): from \( G(-3,5) \) to \( H(4,1) \): horizontal distance 7 (from x=-3 to x=4: 7 units right), vertical distance 4 (from y=5 to y=1: 4 units down). So \( 7^2 + 4^2 = 49 + 16 = 65 \), so \( \sqrt{65} \). Correct.

• \( GJ \): from \( G(-3,5) \) to \( J(0,-2) \): horizontal distance 3 (from x=-3 to x=0: 3 units right), vertical distance 7 (from y=5 to y=-2: 7 units down). So \( 3^2 + 7^2 = 9 + 49 = 58 \), so \( \sqrt{58} \). Correct.

Thus, perimeter is \( 5 + \sqrt{65} + \sqrt{58} \), which is option A.

Answer:

A. \( 5 + \sqrt{65} + \sqrt{58} \)