An isosceles triangle has a base 9.6 units long. If the congruent side lengths have measures to the first decimal place, what is the shortest possible length of the sides?
If they were laid out flat, they'd each be 4.8 units long. So, they must be more than that, and since 4.8 is not an acceptable answer, 4.9 is what we have to go with.
Well, it's clear that the side lengths have to be longer than 9.6 units. Otherwise, we'd have a degenerate triangle. But how long can they be?
Let's take a trip down imagination lane. Imagine the side lengths were magically equal to 9.6 units. Now, imagine you're trying to balance the triangle on its base.
What happens? Well, the triangle would flop over like a pancake! 🥞 So, it's safe to say that the side lengths have to be longer than 9.6 units.
Now, how short can they be? Well, they can't be zero, because then we wouldn't have a triangle at all! So, the shortest possible length of the sides is greater than zero and longer than 9.6 units.
I hope that clears things up! 🤡
If an isosceles triangle has a base of 9.6 units long, then the two congruent side lengths must be greater than 9.6 units in order to form a triangle.
Since the side lengths must be to the first decimal place, the shortest possible length would be 9.7 units.
To find the shortest possible length of the sides of an isosceles triangle, we need to consider the triangle inequality theorem. According to the theorem, the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.
In an isosceles triangle, the base is one of the congruent sides. Let's denote the length of each congruent side as "x." According to the problem, the base has a length of 9.6 units.
Applying the triangle inequality theorem, we have the following equation:
x + x > 9.6
This simplifies to:
2x > 9.6
Dividing both sides of the equation by 2, we get:
x > 4.8
Therefore, the length of each congruent side, to the first decimal place, must be greater than 4.8 units.