Determine the area of the parallelogram with vertices (4, 6, 2), (4, 7, 2), (9, 7, 2), and (9, 8, 2). Use the square root symbol
'√' where needed to give an exact value for your answer.
1 answer:
Answer:
5 square units
Step-by-step explanation:
The parallelogram lies entirely in the plane z=2, so we can treat this as a 2-dimensional problem.
In the order given, the vertices do not define a parallelogram. Two of the sides are parallel, but the other two sides cross each other, for a net area of zero.
If we swap the order of the last two vertices, we get a parallelogram that has a base of 1 unit and a height of 5 units. Its area is ...
A = bh = 1·5 = 5 . . . square units
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Answer:
Step-by-step explanation:
Note that the entire segment RT is the combined lengths of RS and ST. So:
We know that RT is 108, RS is (9y+5), and ST is (3y+7). Substitute:
On the right, combine like terms:
Add:
Subtract 12 from both sides. The right cancels:
Divide both sides by 12:
So, the value of y is 8.
To find RS and ST, substitute 8 for y. So:
Substitute 8 for y:
Multiply:
Add:
Do the same for ST:
Substitute 8 for y:
Multiply:
Add: