What is the area of a parallelogram whose vertices are a(−1, 12) , b(13, 12) , c(2, −5) , and d(−12, −5) ?

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What is the area of a parallelogram whose vertices are a(−1, 12) , b(13, 12) , c(2, −5) , and d(−12, −5) ?

Check the picture below.

you can pretty much count the units for the base and height off the grid.

[tex]What is the area of a parallelogram whose vertices are a(−1, 12) , b(13, 12) , c(2, −5) , and d(−12,[/tex]

A = B x H

B = 12 + 2 = 14 or 1 + 13 = 14

H = 12 + 5 = 17

14 x 17 = 238 units^2

AB = 14, that is our base length, then the height is the 12 to -5, or 17 in length, then multiply those 2 numbers 17 x 14, to get 238

It should be 238 for the area

The answer is actually 238 units squared.

Step-by-step explanation:

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This is Pancho... 1 thanks equals one friend for him

The area of the parallelogram with the given vertices=238 square units.

Step-by-step explanation:

Let CD be the base of the parallelogram ABCD is given by distance formula

CD=[tex]\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} =\sqrt{(-12-2)^2+(-5-(-5))^2} =\sqrt{196} =14[/tex]units[or you can count distance by using graph as ordinates of C and D are same to we can count the distance of x from C to D]

Now consider P a point on CD such that AP become the height of the parallelogram which is given by (from the graph attached)=17 [as abscissa of P and A are same to we can count the distance of y from A to P]

Area of parallelogram = base × height=14×17=238 square units.

[tex]What is the area of a parallelogram whose vertices are a(−1, 12) , b(13, 12) , c(2, −5) , and d(−12,[/tex]

Looking at the first two vertices, A and B, we see that the x values have a difference of (-1 - 13) units, or 14 units. This gives us the base of the parallelogram. Next, we look at the difference between the y values between B and C to find the height. In this case, they are (12 - [-5] ) units, or 17 units. The area of a parallelogram is simply base multiplied by height, which is (14 * 17), or 238 square units.

1. 238 2. 50 ftÂ˛ The area of a parallelogram is bh where b is the base, and h is the height. Since line segments AB and CD are conveniently horizontal, I'll use the length of line segment CD as the base (which is 14), and the distance between line segments AB and CD as the height (which is 17). So 14 * 17 = 238 So the area of the parallelogram is 238. As for estimating the area of the polygon in the drawing, first, look at the overall length and width. The figure covers an area 8 units wide and 8 units tall except for the corners. So the upper limit on it's size is 64. Now look at the upper left hand corner. A bit over 3 square units isn't covered. So the new upper limit to your estimate is 64 - 3 = 61 units. Look at the upper right corner. Looks like about 3.5 units aren't covered there. So the new estimate becomes 61-3.5 = 57.5. Looking at the lower left corner let's us subtract another 4 units giving 57.5 - 4 = 53.5. Lower right corner shows another 4 units or so uncovered, so 53.5 - 4 = 49.5. Now look at the available choices of 25, 35, 50, and 65 to see what's closest. And that's obviously 50. So the answer is 50 ftÂ˛.

238

Step-by-step explanation:

1- formula; 2 - to find the length of the base CD; 3 - to find the length of the height AH; 4 - for calculate the area according to the formula, defined below.

[tex]What is the area of a parallelogram whose vertices are a(−1, 12) , b(13, 12) , c(2, −5) , and d(−12,[/tex]

238 square units. The area of a parallelogram is the base multiplied by the height. You can use any of its four sides as the base, so pick the one that is easiest to deal with. Examining the parallelogram, you'll notice that line segments AB and CD are both parallel to the x axis which makes it extremely easy to calculate the height which is 12 - (-5) = 17. The length of AB is 13 - (-1) = 14. So the area of the parallelogram is 14 * 17 = 238