Exercise 144 Page 336

a If we scale down the triangle as far as we can and get two sides that match two sides in either a 3-4-5 triangle or a 5-12-13 triangle, we know the triangle is a Pythagorean Triple.

In the given triangle the hypotenuse is 15 units and one leg is 12 units. These lengths are multiples of 3, which means we can scale down the triangle.

Now we know that the given triangle is a dilated version of a 3-4-5 triangle, where the unknown side in the given triangle corresponds to the leg in a 3-4-5 triangle that has a length of 3. Now we can write and solve an equation for x. x/3= 3 ⇔ x=9 units

b Again, like in Part A we will scale down the triangle as much as possible to investigate which type of Pythagorean Triple it might be. Both legs are multiples of 2, which means we can divide both sides by this factor.

As we can see, the given triangle is a dilated version of a 5-12-13 triangle, where the unknown side in the given triangle corresponds to the hypotenuse in a 5-12-13 triangle that has a length of 13. We can write and solve an equation for x. x/2= 13 ⇔ x=26 units

c By the Triangle Inequality Theorem we know that the sum of two sides of a triangle has to be less than the third side. Here we have two scenarios to consider.

The side that is 17 units is the longest.
The side that is y units is the longest.

In the first scenario by the Triangle Inequality Theorem we know that the sum of y and the side that is 10 units must be greater than the side that is 17 units. y+10>17 ⇔ y>7 units In the second scenario, the sum of the sides that is 10 and 17 units must be greater than y. 10+17>y ⇔ y< 27 units Therefore, the third side has to be within the following interval. 7 units< y < 27 units

d Like in Part C, by the Triangle Inequality Theorem we know that the sum of two sides of a triangle has to be less than the third side. We have two scenarios to consider.

The side that is 17 units is the longest.
The side that is y units is the longest.

In the first scenario we know by the Triangle Inequality Theorem that the sum of y and the side that is 15 units must be greater than the side that is 17 units. y+15>17 ⇔ y>2 units In the second scenario, the sum of the sides that is 17 and 15 units must be greater than y. 15+17>y ⇔ y< 32 units Therefore, the third side has to be within the following interval. 2 units< y < 32 units

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Exercise 144 Page 336

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