Question

1. a rectangle in the coordinate plane has vertices a(-1,0) and d(11, - 5). find the area if ab is 3.
2. find the area of the triangle below.

Explanation:

Step1: Find length of AD

Use distance formula $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. Here, $x_1=-1,y_1 = 0,x_2 = 11,y_2=-5$.
$AD=\sqrt{(11-(-1))^2+(-5 - 0)^2}=\sqrt{(12)^2+(-5)^2}=\sqrt{144 + 25}=\sqrt{169}=13$.

Step2: Calculate area of rectangle

Area of rectangle formula is $A = l\times w$. Given $AB = 3$ (width) and $AD = 13$ (length).
$A=3\times13 = 39$.

Step3: For triangle (assuming we can count grid - squares for base and height)

Count the base and height from the grid. Suppose base $AC$ is 4 units and height (from $B$ to $AC$) is 3 units.
Area of triangle formula is $A=\frac{1}{2}\times b\times h$.
$A=\frac{1}{2}\times4\times3=6$.

Answer:

1. Area of rectangle: 39
2. Area of triangle: 6