Which of these sets of side lengths are Pythagorean triples? Check all that apply.

2 answers:

7 0

Some primitive triples are ...
(3, 4, 5)
(5, 12, 13)
(7, 24, 25)
(9, 40, 41)
One interesting characteristic of these is that the sum of the last two numbers is the square of the first number.

Any multiple of these will be a Pythagorean triple.

Now consider your list.
a) (10, 24, 26) = 2×(5, 12, 13) . . . IS a Pythagorean Triple
b) 2×(7, 24, 25) = (14, 48, 50), so (14, 48, 49) is NOT a Pythagorean Triple
c) 3×(3, 4, 5) = (9, 12, 15), so (9, 12, 16) is NOT a Pythagorean Triple
d) (9, 40, 41) . . . IS a Pythagorean Triple
e) 5×(3, 4, 5) = (15, 20, 25) . . . IS a Pythagorean Triple

The sets of side lengths that are Pythagorean Triples are ...
(10, 24, 26)
(9, 40, 41)
(15, 20, 25)

miss Akunina [59]3 years ago

7 0

Answer:

10, 24, 26

9, 40, 41

15, 20, 25

Step-by-step explanation:

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9514 1404 393

Answer:

7 square units

Step-by-step explanation:

There are several ways the area of triangle EBD can be found.

find the lengths EB, BD, DE and use Heron's formula (messy due to roots of roots being involved).
define point G at the lower left corner and subtract the areas of ∆DEG and BCD from trapezoid BCGE.
figure the area from the coordinates of the vertices.
use Pick's theorem and count the dots.

We choose the latter.

Pick's theorem says the area of a polygon can be found as ...

A = i + b/2 -1

where i is the number of grid intersection points interior to the polygon, b is the number of grid points intersected by the border.

The attached figure shows the lines EB, BD, and DE intersect one point in addition to the vertices. So, b=4. A count of the red dots reveals 6 interior points (i=6). So, the area is ...

A = 6 + (4/2) -1 = 7

The area of ∆EBD is 7 square units.