Plane Geometry - 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.

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 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

When two chords of a circle intersect, the measure of the angle they form is half the sum of the measures of the arcs they intercept. Therefore,

Since and form a linear pair, , and .

Substitute and into the first equation:

Proporations can be used to solve for the central angle. Let equal the angle of the sector.

Cross mulitply:

Solve for :

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 , yields 52.5.

We start by dividing the area of square R (48) by 12, to come up with the area of square T, 4. Then take the square root of the area to get the length of one side, giving us 2.

Let x represent the length of the original square in inches. Thus the area of the original square is x2. Two inches are added to x, which is represented by x+2. The area of the resulting square is (x+2)2. We are given that the new square is 64 sq. inches greater than the original. Therefore we can write the algebraic expression:

x2 + 64 = (x+2)2

FOIL the right side of the equation.

x2 + 64 = x2 + 4x + 4

Subtract x2 from both sides and then continue with the alegbra.

64 = 4x + 4

64 = 4(x + 1)

16 = x + 1

15 = x

Therefore, the length of the original square is 15 inches.

If you plug in the answer choices, you would need to add 2 inches to the value of the answer choice and then take the difference of two squares. The choice with 15 would be correct because 172 -152 = 64.

The area of a square is equal to length times width or length squared (since length and width are equal on a square). Therefore, the length of one side is or The perimeter of a square is the sum of the length of all 4 sides or