Question

what is the area of δvwx? area = square units

Explanation:

Step1: Find the base length

The coordinates of \( X \) are \((-8, 0)\) and \( W \) are \((5, 0)\)? Wait, no, looking at the graph, \( W \) is at \( (5, 0) \)? Wait, no, the grid: \( X \) is at \(-8\) on x-axis, \( W \) is at \( 5 \)? Wait, no, the x-coordinate of \( W \) is 5? Wait, no, the graph shows \( W \) at \( x = 5 \)? Wait, no, the grid lines: from \( X(-8, 0) \) to \( W(5, 0) \)? Wait, no, let's check the x-coordinates. \( X \) is at \( x = -8 \), \( W \) is at \( x = 5 \)? Wait, no, the vertical line from \( V \) to \( W \): \( V \) is at \( (5, 10) \)? Wait, no, the grid: each square is 1 unit. So \( X \) is at \((-8, 0)\), \( W \) is at \((5, 0)\)? Wait, no, the distance between \( X \) and \( W \) on the x-axis: from \( -8 \) to \( 5 \), the length is \( 5 - (-8) = 13 \)? Wait, no, maybe I misread. Wait, \( W \) is at \( x = 5 \)? Wait, the graph: \( W \) is at \( (5, 0) \), \( V \) is at \( (5, 10) \), so \( VW \) is vertical, length 10. \( X \) is at \( (-8, 0) \), so the base \( XW \) is the distance from \( -8 \) to \( 5 \) on x-axis: \( 5 - (-8) = 13 \)? Wait, no, maybe \( W \) is at \( x = 5 \), \( X \) at \( x = -8 \), so the base length \( b = 5 - (-8) = 13 \)? Wait, no, maybe I made a mistake. Wait, the triangle is \( \triangle VWX \), with \( X \) at \((-8, 0)\), \( W \) at \((5, 0)\), and \( V \) at \((5, 10)\). So it's a right triangle with base \( XW \) and height \( VW \).

Wait, \( X \) is at \((-8, 0)\), \( W \) is at \((5, 0)\), so the length of \( XW \) is \( 5 - (-8) = 13 \)? Wait, no, \( 5 - (-8) = 13 \)? Wait, \( -8 \) to \( 0 \) is 8 units, \( 0 \) to \( 5 \) is 5 units, so total 13? Wait, but \( VW \) is from \( (5, 0) \) to \( (5, 10) \), so height is 10. Then area of triangle is \( \frac{1}{2} \times base \times height \).

Wait, maybe I misread the coordinates. Let's check again: \( X \) is at \( (-8, 0) \), \( W \) is at \( (5, 0) \)? Wait, the x-axis: from \( -10 \) to \( 10 \), each grid is 1. So \( X \) is at \( x = -8 \), \( W \) is at \( x = 5 \), so the distance between \( X \) and \( W \) is \( 5 - (-8) = 13 \)? Wait, no, \( 5 - (-8) = 13 \), yes. And \( V \) is at \( (5, 10) \), so \( VW \) is vertical, length 10 (from y=0 to y=10). So the triangle is right-angled at \( W \), so base \( XW = 13 \), height \( VW = 10 \). Then area is \( \frac{1}{2} \times 13 \times 10 = 65 \)? Wait, but maybe I made a mistake in the x-coordinate of \( W \). Wait, the graph shows \( W \) at \( x = 5 \)? Wait, the vertical line from \( V \) to \( W \): \( V \) is at \( (5, 10) \), \( W \) at \( (5, 0) \), so \( x = 5 \). \( X \) is at \( (-8, 0) \), so the base is \( 5 - (-8) = 13 \), height is 10. So area is \( \frac{1}{2} \times 13 \times 10 = 65 \).

Wait, but maybe the x-coordinate of \( W \) is 5? Let me confirm. The grid: each square is 1 unit. So from \( X(-8, 0) \) to \( W(5, 0) \), the horizontal distance is \( 5 - (-8) = 13 \). The vertical distance from \( W \) to \( V \) is \( 10 - 0 = 10 \). So area of triangle is \( \frac{1}{2} \times base \times height = \frac{1}{2} \times 13 \times 10 = 65 \).

Step2: Calculate the area

Using the formula for the area of a triangle, \( A = \frac{1}{2} \times base \times height \). Here, base \( = 13 \) (distance from \( X \) to \( W \)), height \( = 10 \) (distance from \( W \) to \( V \)). So \( A = \frac{1}{2} \times 13 \times 10 = 65 \).

Answer:

65