### All SSAT Upper Level Math Resources

## Example Questions

### Example Question #791 : Geometry

The length and width of a rectangle are and . Give its perimeter in terms of .

**Possible Answers:**

**Correct answer:**

A rectangle has perimeter , the length and the width. Substitute and in the perimeter formula, and simplify.

### Example Question #791 : Geometry

A rectangle has length 30 inches and width 25 inches. Which of the following is true about its perimeter?

**Possible Answers:**

Its perimeter is between 4 and 5 feet.

Its perimeter is between 10 and 11 feet.

Its perimeter is between 9 and 10 feet.

Its perimeter is between 5 and 6 feet.

Its perimeter is between 8 and 9 feet.

**Correct answer:**

Its perimeter is between 9 and 10 feet.

In inches, the perimeter of the rectangle can be calculated by substituting in the following formula:

The perimeter is 110 inches.

Now divide by 12 to convert to feet:

This makes the perimeter 9 feet 2 inches, which is between 9 feet and 10 feet.

### Example Question #801 : Geometry

Perimeter of a rectangle is 36 inches. If the width of the rectangle is 3 inches less than its length, give the length and width of the rectangle.

**Possible Answers:**

inches

inches

inches

inches

inches

**Correct answer:**

inches

Let:

The perimeter of a rectangle is , where is the length and is the width of the rectangle. The perimeter is known so we can set up an equation in terms of and solve it:

So we can get:

inches

inches

### Example Question #802 : Geometry

The length and width of a rectangle are and , respectively. Give its perimeter in terms of .

**Possible Answers:**

**Correct answer:**

The perimeter of a rectangle is , where is the length and is the width of the rectangle. In order to find the perimeter we can substitute the and in the perimeter formula:

### Example Question #31 : Perimeter Of Polygons

The length of a rectangle is and the width of this rectangle is meters shorter than its length. Give its perimeter in terms of .

**Possible Answers:**

**Correct answer:**

The length of the rectangle is known, so we can find the width in terms of :

The perimeter of a rectangle is , where is the length and is the width of the rectangle.

In order to find the perimeter we can substitute the and in the perimeter formula:

### Example Question #32 : Perimeter Of Polygons

A rectangle has a length of inches and a width of inches. Which of the following is true about the rectangle perimeter if ?

**Possible Answers:**

Its perimeter is more than 8 feet.

Its perimeter is between 7.2 and 7.4 feet.

Its perimeter is less than 7 feet.

Its perimeter is between 8 and 9 feet.

Its perimeter is between 7 and 8 feet.

**Correct answer:**

Its perimeter is between 7 and 8 feet.

Substitute to get and :

The perimeter of a rectangle is , where is the length and is the width of the rectangle. So we have:

inches

Now we should divide the perimeter by 12 in order to convert to feet:

feet

So the perimeter is 7 feet and 6 inches, which is between 7 and 8 feet.

### Example Question #1 : How To Find The Perimeter Of A Rectangle

Which of these polygons has the same perimeter as a rectangle with length 55 inches and width 15 inches?

**Possible Answers:**

A regular pentagon with sidelength two feet

A regular hexagon with sidelength two feet

The other answer choices are incorrect.

A regular octagon with sidelength two feet

A regular heptagon with sidelength two feet

**Correct answer:**

The other answer choices are incorrect.

The perimeter of a rectangle is twice the sum of its length and its width; a rectangle with dimensions 55 inches and 15 inches has perimeter

inches.

All of the polygons in the choices are regular - that is, all have congruent sides - and all have sidelength two feet, or 24 inches, so we divide 140 by 24 to determine how many sides such a polygon would need to have a perimeter equal to the rectangle. However,

,

so there cannot be a regular polygon with these characteristics. All of the choices fail, so the correct response is that none are correct.

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