Acute / Obtuse Triangles - Math

The sum of the lengths of two sides of a triangle cannot be less than the length of the third side. 8 + 4 = 12, which is less than 13.

Please remember the Triangle Inequality Theorem, which states that the sum of any two sides of a triangle must be greater than the third side. Therefore, the correct answer is 10 because the sum of 10 and 20 would not be greater than the third side 32.

We know that the perimeter is 36 inches, and one side is 12. This means, the sum of the lengths of the other two sides are 24. The ratio between the two sides is 3:5, giving a total of 8 parts. We divide the remaining length, 24 inches, by 8 giving us 3. This means each part is 3. We multiply this by the ratio and get 9:15, meaning the longest side is 15 inches.

The sum of the lengths of any two sides of a triangle must exceed the length of the third side; therefore, 5+7 > x, which cannot happen if x = 13.

The set of possible lengths is: 7-4 < x < 7+4, or 3 < X < 11.

According to the Triangle Inequality Theorem, the sums of the lengths of any two sides of a triangle must be greater than the length of the third side. Since 10 + 8 is 18, the only length out of the answer choices that is not possible is 19.

Let us call the third side x. According to the Triangle Inequality Theorem, the sum of any two sides of a triangle must be larger than the other two sides. Thus, all of the following must be true:

x + 7 > 9

x + 9 > 7

7 + 9 > x

We can solve these three inequalities to determine the possible values of x.

x + 7 > 9

Subtract 7 from both sides.

x > 2

Now, we can look at x + 9 > 7. Subtracting 9 from both sides, we obtain

x > –2

Finally, 7 + 9 > x, which means that 16 > x.

Therefore, x must be greater than 2, greater than –2, but also less than 16. The only number that satisfies all of these requirements is 12.

The answer is 12.

According to the Triangle Inequality Theorem, the sum of any two sides of a triangle must be greater (not greater than or equal) than the remaining side. Thus, the following inequalities must all be true:

x + 8 > 12

x + 12 > 8

8 + 12 > x

Let's solve each inequality.

x + 8 > 12

Subtract 8 from both sides.

x > 4

Next, let's look at the inequality x + 12 > 8

x + 12 > 8

Subtract 12 from both sides.

x > –4

Lastly, 8 + 12 > x, which means that x < 20.

This means that x must be less than twenty, but greater than 4 and greater than –4. Since any number greater than 4 is also greater than –4, we can exclude the inequality x > –4.

To summarize, x must be greater than 4 and less than 20. We can write this as 4 < x < 20.

The answer is 4 < x < 20.

The sum of the lengths of 2 sides of a triangle must be greater than—but not equal to—the length of the third side. Further, the third side must be longer than the difference between the greater and the lesser of the other two sides; therefore, 20 is the only possible answer.

4+8< 14

6+8= 14

8+14> 20

8+14=22

Rule - the length of one side of a triangle must be greater than the differnce and less than the sum of the lengths of the other two sides.

Given lengths of two of the sides of the Delta ABC are 15 and 5. The length of the third side must be greater than 15-5 or 10 and less than 15+5 or 20.

The question asks what is the least possible integer length of BC, which would be 11

The larger triangle has a perimeter of 25 inches. Therefore, using a 4:5 ratio, the smaller triangle's perimeter will be 20 inches.

Consider the perimeter of a triangle:

P = a + b + c

Since we know a and b, we can find c.

In I:

11 = 7 + 4 + c

11 = 11 + c

c = 0

Note that if c = 0, the shape is no longer a trial. Thus, we can eliminate I.

In II:

15 = 7 + 4 + c

15 = 11 + c

c = 4.

This is plausible given that the other sides are 7 and 4.

In III:

25 = 7 + 4 + c

25 = 11 + c

c = 14.

It is not possible for one side of a triangle to be greater than the sum of both of the other sides, so eliminate III.

Thus we are left with only II.

The trick here is that we don't know which is the repeated side. Our possible triangles are therefore 20 + 20 + 30 = 70 or 30 + 30 + 20 = 80. The difference is therefore 80 – 70 or 10.

Adding the sides gives a perimeter of 15 for the smaller triangle. Multipying by the given ratio of $\frac{7}{2}$, yields 52.5.

The area of a triangle = (1/2)bh where b is base and h is height. 18 = (1/2)4h which gives us 36 = 4h so h =9.

The measure of angle ADB is $30^{circ}$. Since A, B, and C are collinear, and the measure of angle CBD is $60^{circ}$, we know that the measure of angle ABD is $120^{circ}$.

Because the measures of the three angles in a triangle must add up to $180^{circ}$, and two of the angles in triangle ABD are $30^{circ}$ and $120^{circ}$, the third angle, angle ADB, is $30^{circ}$.

This problem requires us to use either the Law of Sines or the Law of Cosines. To figure out which one we should use, let's write down all the information we have in this format:

A = 75° a = 13

B = ? b = 6

C = ? c = ?

Now we can easily see that we have a complete pair, A and a. This tells us that we can use the Law of Sines. (We use the Law of Cosines when we do not have a complete pair).

Law of Sines:

To solve for b, we can use the first two terms which gives us:

Since this is a scalene triangle, all of the interior angles will have different measures. However, it's fundemental to note that in any triangle the sum of the measurements of the three interior angles must equal degrees.

The largest angle is equal to degrees and second interior angle must equal:

Therefore, the final angle must equal:

An obtuse isosceles triangle has one obtuse interior angle and two equivalent acute interior angles. Since the sum total of the interior angles of every triangle must equal degrees, the solution is:

Acute scalene triangles must have three different acute interior angles--which always have a sum of degrees.

Thus, the solution is: