Question

a rectangular flag and a triangular flag are being made. the same amount of fabric is used for each flag, so both have the same area.

• the rectangular flag is shown on the grid.
• the height of the triangular flag is shown on the grid

use the given height to create a triangle that has the same area as the rectangle.
note: with the polygon tool selected graph the vertices of the triangle in order ending with the first vertex.

Explanation:

Step1: Find area of rectangle

The rectangle has a height (vertical side) from \( y = 6 \) to \( y = 10 \), so height \( h = 10 - 6 = 4 \). The width (horizontal side) from \( x = 0 \) to \( x = 4 \) (assuming grid units, since it spans 4 columns), so width \( w = 4 \). Area of rectangle \( A_{rect} = w \times h = 4 \times 4 = 16 \)? Wait, no, wait: Wait, looking at the grid, the rectangle is from \( x=0 \) to \( x=4 \) (since it's 4 units wide) and \( y=6 \) to \( y=10 \) (4 units tall). Wait, no, actually, the vertical side: from \( y=6 \) to \( y=10 \) is \( 10 - 6 = 4 \) units. Horizontal side: from \( x=0 \) to \( x=4 \) (since it's 4 grid squares wide). So area \( A = 4 \times 4 = 16 \)? Wait, no, maybe I miscounted. Wait, the rectangle's top is at \( y=10 \), bottom at \( y=6 \), so height is \( 10 - 6 = 4 \). Left at \( x=0 \), right at \( x=4 \) (since it's 4 units wide). So area \( A_{rect} = 4 \times 4 = 16 \)? Wait, no, maybe the width is 4 (from x=0 to x=4) and height is 4 (from y=6 to y=10), so area 16. Now, the triangular flag has the same area. The formula for the area of a triangle is \( A_{tri} = \frac{1}{2} \times base \times height \). We need \( A_{tri} = A_{rect} = 16 \). Wait, but the height of the triangle is given? Wait, the blue vertical segment is from \( y=1 \) to \( y=5 \), so height of triangle \( H = 5 - 1 = 4 \)? Wait, no, the triangle's height is the vertical length. Wait, the problem says "the height of the triangular flag is shown on the grid" – the blue segment from (0,1) to (0,5), so height \( H = 5 - 1 = 4 \). So we need \( \frac{1}{2} \times base \times H = A_{rect} \). So \( \frac{1}{2} \times base \times 4 = 16 \) (wait, no, earlier rectangle area: wait, maybe I messed up the rectangle's dimensions. Wait, let's re-examine the grid. The rectangle is from \( x=0 \) to \( x=4 \) (4 units) and \( y=6 \) to \( y=10 \) (4 units), so area \( 4 \times 4 = 16 \). The triangle's height is 4 (from y=1 to y=5). So \( \frac{1}{2} \times base \times 4 = 16 \) → \( base \times 2 = 16 \) → \( base = 8 \). So the base of the triangle should be 8 units long. So we can place the triangle with base from, say, (0,1) to (8,1) and the third vertex at (0,5) (or (8,5)), but wait, the height is vertical? Wait, no, the height of the triangle is the perpendicular distance from the base to the opposite vertex. If the base is horizontal, then the height is vertical. So if the base is along the x-axis from (0,1) to (8,1), then the height is the vertical distance from (0,5) to the base (y=1), which is \( 5 - 1 = 4 \), which matches. Then area of triangle is \( \frac{1}{2} \times 8 \times 4 = 16 \), same as rectangle. So the vertices of the triangle would be (0,1), (8,1), (0,5), and back to (0,1).

Step2: Plot the triangle vertices

Using the polygon tool, plot the first vertex at (0,1), second at (8,1), third at (0,5), then back to (0,1) to close the polygon.

Answer:

The triangle has vertices at (0,1), (8,1), (0,5) (and back to (0,1)). (Note: The actual plotting would be done on the grid, but the key is determining the base length based on area equality.)