Quadrilateral ABCD has diagonals that are congruent and perpendicular to each other. What classification can be given to ABCD? (
1 answer:
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Answer:
it depends
Step-by-step explanation:
If 17 is the hypotenuse
if you are looking for the hypotenuse
Because it is extremely hard to find the area of this figure all together, it would be in our best interest to split this figure up into three different pieces: the two horizontal rectangles, and the verticle rectangle. We can find the area of all three and add them up. Be aware that there are two different ways that you can break this figure up, As shown in the attachments. I will be using the first image (the one with the tall horizontal rectangles, NOT the almost-squares).
So, we see that we have enough information to solve for the area of the left-most rectangle. Area = lw. 10 x 4 = 40, so the area is 40. Next, we have to notice, that the horizontal rectangles are also the same, so both of the areas of the two horizontal rectangles are 40.
Now, we can find the middle rectangle. We know that the length of the entire thing is 18, but it is taken up by 8 (4+4) of the horizontal triangles, so 18-8=10, so the length Is 10. We also know that the height of the horizontal rectangles is 10, so 10-3=7. Our dimensions for the rectangle are 10x7 or 70 square units. If we add them all together, 40+40+70=150.
The area is 150 square units
