Answer:
8. 46 units
9. a, b, c, f
Explanation:
8. The distance formula tells you the distance between two points (x1, y1) and (x2, y2) is ...
d = √((x2-x1)² +(y2-y1)²)
Then the length of RS is ...
d = √((2 -(-4))² + (1 -7)²) = √(36+64) = 10
and the length of ST is ...
d = √((14-2)² +(6-1)²) = √(144+25) = 13
These two adjacent sides of a parallelogram make up half the perimeter, so the perimeter is ...
P = 2(10 +13) = 2·23 = 46 . . . units
Please note that your familiarity with the Pythagorean triples (3, 4, 5) and (5, 12, 13) could make this a problem you can do mentally. The vector RS has components (6, 8), so is double the legs of the (3, 4, 5) right triangle. Hence its length is 2·5 = 10, as we computed.
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9. Here, it is helpful to know that a right triangle with legs of length 1 will have a hypotenuse of √(1²+1²) = √2, and one with legs 1 and 2 will have a hypotenuse of √(1²+2²) = √5. These fun facts can save some effort computing lengths in the given figure.
Let's call the choices by letter names a through f. Then we can consider them one at a time:
a. The midpoint of AB is (A+B)/2 = ((-3+1)/2, (1+4)/2) = (-1, 5/2) (yes)
b. ED is 2 units horizontally and 4 units vertically, so is twice the hypotenuse of the 1×2 triangle above. Its length is 2√5. (yes)
c. AE is a vertical segment from y=-5 to y=1, a length of 6. (yes)
d. AB is 2 units horizontally and 2 units vertically, so is twice the hypotenuse of the 1×1 triangle above. Its length is 2√2 ≈ 2.83. DC extends 4 units horizontally and 6 units vertically. Both of these lengths are greater than 2.8, so DC is longer than AB. (no)
e. BD has a vertical extent of 5, and CD has a vertical extent of 6. The total length of these two segments must be more than 11, so the perimeter of BCD must be greater than 9.3. (no)
f. AD we know is 2√2 ≈ 2.83; DE we know is 2√5 ≈ 4.47; AE = 6, so the perimeter of ADE is 2.83 + 4.47 + 6 = 13.3 (yes)
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When considering part "e", we took advantage of the fact that the hypotenuse of a right triangle is longer than the longest side. (It will be shorter than the sum of the side lengths.) This is another fun fact that can simplify some problems, such as this one.