In a triangle, the lengths of the sides are x, y, and z, where x, y, and z are integers su… — ACT

In a triangle, the lengths of the sides are $x$ , $y$ , and $z$ , where $x$ , $y$ , and $z$ are integers such that $x < y < z$ . If $x = 7$ and the perimeter of the triangle is $32$ , what is the number of possible integer values for $z$ ?

Cevap: 3

Cevap

There are exactly 3 possible integer values for the side length z.

The correct answer is 3. By expressing the second side as

y = 25 - z

and applying the ordering constraint

7 < 25 - z < z

, we determine that

12.5 < z < 18

. Applying the Triangle Inequality Theorem (

7 + y > z

) yields the restriction

z < 16

. Combining these conditions restricts the integer values of

z

\{13, 14, 15\}

, which counts to exactly 3 possible values.

Adım Adım Çözüm

Express the side length

y

in terms of

z

y = 25 - z

The perimeter of the triangle is the sum of the side lengths:

x + y + z = 32

. Substituting

x = 7

gives

7 + y + z = 32

, which simplifies to

y = 25 - z

Apply the given inequality constraint

x < y < z

to find initial bounds for

z

12.5 < z < 18

Substituting

x = 7

and

y = 25 - z

into

x < y < z

yields

7 < 25 - z < z

. The left inequality

7 < 25 - z

simplifies to

z < 18

. The right inequality

25 - z < z

simplifies to

25 < 2z

, or

z > 12.5

Apply the Triangle Inequality Theorem to establish the final constraint on

z

z < 16

Since

z

is the longest side, the sum of the two shorter sides must be strictly greater than

z

x + y > z

. Substituting

x = 7

and

y = 25 - z

gives

7 + 25 - z > z

, which simplifies to

32 > 2z

, or

z < 16

Combine all constraints and count the valid integer values for

z

3 possible values (

13, 14, 15

)

Combining the bounds from the steps gives

12.5 < z < 16

. The integers satisfying this inequality are

13

14

, and

15

, which gives a total of 3 possible integer values.

Anahtar Kavram

Triangle Inequality Theorem and algebraic constraints on side lengths

Soru

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