### All High School Math Resources

## Example Questions

### Example Question #173 : Plane Geometry

Two similiar triangles have a ratio of perimeters of .

If the smaller triangle has sides of 3, 7, and 5, what is the perimeter of the larger triangle.

**Possible Answers:**

**Correct answer:**

Adding the sides gives a perimeter of 15 for the smaller triangle. Multipying by the given ratio of , yields 52.5.

### Example Question #1 : Triangles

Two sides of an isosceles triangle are 20 and 30. What is the difference of the largest and the smallest possible perimeters?

**Possible Answers:**

0

10

30

The answer cannot be determined

15

**Correct answer:**

10

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.

### Example Question #174 : Plane Geometry

Two similiar triangles exist where the ratio of perimeters is 4:5 for the smaller to the larger triangle. If the larger triangle has sides of 6, 7, and 12 inches, what is the perimeter, in inches, of the smaller triangle?

**Possible Answers:**

18

25

23

20

**Correct answer:**

20

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

### Example Question #131 : Geometry

If a = 7 and b = 4, which of the following could be the perimeter of the triangle?

I. 11

II. 15

III. 25

**Possible Answers:**

I and II Only

II and III Only

I, II and III

I Only

II Only

**Correct answer:**

II Only

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.