Question

what is the area of δstu?
coordinate grid graph with points u, s, t
area = square units

Explanation:

Step1: Identify coordinates of points

From the graph, we can find the coordinates of the vertices of the triangle:
- Point \( U \): Let's assume the grid has each square as 1 unit. Looking at the graph, \( U \) is at \( (-9, 6) \) (wait, actually, let's check the x - axis. Wait, the distance between \( U \) and \( S \): \( S \) is at \( (9,6) \)? Wait, no, let's count the horizontal distance between \( U \) and \( S \). Wait, \( U \) is at \( x=-9 \)? Wait, no, the x - coordinate of \( U \): from the origin (0,0), moving left 9 units? Wait, no, let's look at the vertical and horizontal lengths. Wait, actually, the base of the triangle (the horizontal side \( US \)): let's find the length of \( US \). The y - coordinate of \( U \) and \( S \) is 6. The x - coordinate of \( U \) is - 9? Wait, no, the distance between \( U \) and \( S \): from \( x=-9 \) to \( x = 9 \)? Wait, no, let's count the number of grid squares. Wait, actually, the horizontal distance between \( U \) and \( S \): if \( U \) is at \( (-9,6) \) and \( S \) is at \( (9,6) \), then the length of \( US \) is \( 9 - (-9)=18 \)? No, that can't be. Wait, maybe I made a mistake. Wait, looking at the graph, the vertical side \( UT \): \( U \) is at \( (-9,6) \) and \( T \) is at \( (-9,-7) \)? Wait, no, let's check the y - coordinates. Wait, \( U \) is at \( y = 6 \), \( T \) is at \( y=-7 \)? No, the distance between \( U \) and \( T \): from \( y = 6 \) to \( y=-7 \), the length is \( 6-(-7)=13 \)? No, that's not right. Wait, maybe the coordinates are: \( U(-9,6) \), \( S(9,6) \), \( T(-9,-6) \)? Wait, no, let's look at the graph again. Wait, the triangle is a right - triangle? Because \( US \) is horizontal and \( UT \) is vertical. So, the length of \( US \): the x - coordinate of \( U \) is - 9, x - coordinate of \( S \) is 9? Wait, no, the distance between \( U \) and \( S \) is the difference in x - coordinates (since y - coordinates are the same). Wait, if \( U \) is at \( (-9,6) \) and \( S \) is at \( (9,6) \), then \( US=9 - (-9)=18 \)? No, that's too long. Wait, maybe the x - coordinate of \( U \) is - 9, and \( S \) is at \( 9 \)? Wait, no, the grid lines: from - 10 to 10 on x - axis. Wait, \( U \) is at \( x=-9 \), \( y = 6 \); \( S \) is at \( x = 9 \), \( y = 6 \); \( T \) is at \( x=-9 \), \( y=-6 \). Then the length of \( US \) (the base) is \( 9-(-9)=18 \)? No, that can't be. Wait, maybe I misread the coordinates. Wait, let's count the number of units between \( U \) and \( S \) horizontally. From \( x=-9 \) to \( x = 9 \), that's 18 units? And the vertical side \( UT \): from \( y = 6 \) to \( y=-6 \), that's 12 units? Wait, no, if \( T \) is at \( (-9,-6) \), then the length of \( UT \) is \( 6-(-6)=12 \). Then the area of a right - triangle is \( \frac{1}{2}\times base\times height \). Wait, but maybe the coordinates are \( U(-9,6) \), \( S(9,6) \), \( T(-9,-6) \). Then base \( US = 9 - (-9)=18 \), height \( UT=6 - (-6)=12 \). Then area \(=\frac{1}{2}\times18\times12 = 108 \)? No, that seems too big. Wait, maybe I made a mistake in the coordinates. Wait, let's look at the graph again. Wait, the y - coordinate of \( U \) is 6, \( T \) is at \( y=-6 \), so the vertical distance between \( U \) and \( T \) is \( 6 - (-6)=12 \) units. The horizontal distance between \( U \) and \( S \): \( U \) is at \( x=-9 \), \( S \) is at \( x = 9 \), so horizontal distance is \( 9-(-9)=18 \) units. But that would make the area \( \frac{1}{2}\times18\times12 = 108 \). But maybe the coordinates are different. Wait, maybe \( U \) is at \( (-9,6) \), \( S \) is at \( (9,6) \), and \( T \) is at \( (-9,-6) \). Alternatively, maybe the x - coordinate of \( U \) is - 9, \( S \) is at 9, so the length of \( US \) is 18, and the height (vertical side) is 12. Then area is \( \frac{1}{2}\times18\times12 = 108 \). Wait, but maybe I made a mistake. Wait, let's check the grid. Each square is 1 unit. So from \( U(-9,6) \) to \( S(9,6) \): the number of units in x - direction is \( 9 - (-9)=18 \). From \( U(-9,6) \) to \( T(-9,-6) \): the number of units in y - direction is \( 6-(-6)=12 \). So the area of the right - triangle is \( \frac{1}{2}\times base\times height=\frac{1}{2}\times18\times12 = 108 \). Wait, but maybe the coordinates are \( U(-9,6) \), \( S(9,6) \), \( T(-9,-6) \). So that's a right - triangle with legs 18 and 12. Then area is \( \frac{1}{2}\times18\times12 = 108 \).

Wait, maybe I misread the coordinates. Let's re - examine: The x - axis: from - 10 to 10. The y - axis: from - 10 to 10. Point \( U \): x=-9, y = 6. Point \( S \): x = 9, y = 6. Point \( T \): x=-9, y=-6. So the horizontal side \( US \): length is \( 9-(-9)=18 \). The vertical side \( UT \): length is \( 6-(-6)=12 \). Then area of triangle \( \triangle STU \) (which is a right - triangle) is \( \frac{1}{2}\times base\times height=\frac{1}{2}\times18\times12 = 108 \).

Wait, but maybe the coordinates are different. Let's check again. Wait, the line from \( T \) to \( S \): it passes through the origin. So the slope of \( TS \) is \( \frac{6 - (-6)}{9 - (-9)}=\frac{12}{18}=\frac{2}{3} \). And the line from \( T(-9,-6) \) to \( S(9,6) \) has the equation \( y=\frac{2}{3}x \), which passes through (0,0), so that's correct. So the triangle is a right - triangle with legs along the vertical (from \( U \) to \( T \)) and horizontal (from \( U \) to \( S \)) lines. So the length of \( US \) is the horizontal distance between \( U \) and \( S \), which is \( 9-(-9)=18 \), and the length of \( UT \) is the vertical distance between \( U \) and \( T \), which is \( 6-(-6)=12 \). Then the area is \( \frac{1}{2}\times18\times12 = 108 \).

Step2: Calculate the area

The formula for the area of a right - triangle is \( A=\frac{1}{2}\times base\times height \). Here, the base (length of \( US \)) is \( 18 \) units and the height (length of \( UT \)) is \( 12 \) units.
\[ A=\frac{1}{2}\times18\times12 \]
First, calculate \( 18\times12 = 216 \). Then, \( \frac{1}{2}\times216=108 \).

Answer:

108