### All PSAT Math Resources

## Example Questions

### Example Question #71 : Quadrilaterals

If the area of a rhombus is 24 and one diagonal length is 6, find the perimeter of the rhombus.

**Possible Answers:**

16

8

12

24

20

**Correct answer:**

20

The area of a rhombus is found by

*A* = 1/2(*d*_{1})(*d*_{2})

where *d*_{1} and *d*_{2} are the lengths of the diagonals. Substituting for the given values yields

24 = 1/2(*d*_{1})(6)

24 = 3(*d*_{1})

8 = *d*_{1}

Now, use the facts that diagonals are perpendicular in a rhombus, diagonals bisect each other in a rhombus, and the Pythagorean Theorem to determine that the two diagonals form 4 right triangles with leg lengths of 3 and 4. Since 3^{2} + 4^{2} = 5^{2}, each side length is 5, so the perimeter is 5(4) = 20.

### Example Question #1 : Rhombuses

Note: Figure NOT drawn to scale.

Calculate the perimeter of Quadrilateral in the above diagram if:

**Possible Answers:**

Insufficient information is given to answer the question.

**Correct answer:**

, so Quadrilateral is a rhombus. Its diagonals are therefore perpendicular to each other, and the four triangles they form are right triangles. Therefore, the Pythagorean theorem can be used to determine the common sidelength of Quadrilateral .

We focus on . The diagonals are also each other's bisector, so

By the Pythagorean Theorem,

26 is the common length of the four sides of Quadrilateral , so its perimeter is .

### Example Question #1 : How To Find The Area Of A Rhombus

A rhombus has a side length of 5. Which of the following is NOT a possible value for its area?

**Possible Answers:**

24

15

10

30

25

**Correct answer:**

30

The area of a rhombus will vary as the angles made by its sides change. The "flatter" the rhombus is (with two very small angles and two very large angles, say 2, 178, 2, and 178 degrees), the smaller the area is. There is, of course, a lower bound of zero for the area, but the area can get arbitrarily small. This implies that the correct answer would be the largest choice. In fact, the largest area of a rhombus occurs when all four angles are equal, i.e. when the rhombus is a square. The area of a square of side length 5 is 25, so any value bigger than 25 is impossible to acheive.

### Example Question #1 : How To Find The Length Of The Diagonal Of A Rhombus

In Rhombus , . If is constructed, which of the following is true about ?

**Possible Answers:**

obtuse and scalene

is acute and isosceles, but not equilateral

is acute and equilateral

is obtuse and isosceles, but not equilateral

is acute and scalene

**Correct answer:**

is acute and equilateral

The figure referenced is below.

Consecutive angles of a rhombus are supplementary - as they are with all parallelograms - so

A diagonal of a rhombus bisects its angles, so

A similar argument proves that .

Since all three angles of measure , the triangle is acute. It is also equiangular, and, subsequently, equilateral.

### Example Question #1 : Quadrilaterals

In Rhombus , . If is constructed, which of the following is true about ?

**Possible Answers:**

is acute and scalene

is right and scalene

is acute and isosceles, but not equilateral

is right and isosceles, but not equilateral

is acute and equilateral

**Correct answer:**

is acute and isosceles, but not equilateral

The figure referenced is below.

The sides of a rhombus are congruent by definition, so , making isosceles. It is not equilateral, since , and an equilateral triangle must have three angles.

Also, consecutive angles of a rhombus are supplementary - as they are with all parallelograms - so

A diagonal of a rhombus bisects its angles, so

Similarly,

This makes acute.

The correct response is that is acute and isosceles, but not equilateral.

### Example Question #1 : Quadrilaterals

Quadrilateral *ABCD* contains four ninety-degree angles. Which of the following must be true?

I. Quadrilateral *ABCD* is a rectangle.

II. Quadrilateral *ABCD* is a rhombus.

III. Quadrilateral *ABCD* is a square.

**Possible Answers:**

I only

I and II only

II and III only

II only

I, II, and III

**Correct answer:**

I only

Quadrilateral *ABCD* has four ninety-degree angles, which means that it has four right angles because every right angle measures ninety degrees. If a quadrilateral has four right angles, then it must be a rectangle by the definition of a rectangle. This means statement I is definitely true.

However, just because *ABCD* has four right angles doesn't mean that it is a rhombus. In order for a quadrilateral to be considered a rhombus, it must have four congruent sides. It's possible to have a rectangle whose sides are not all congruent. For example, if a rectangle has a width of 4 meters and a length of 8 meters, then not all of the sides of the rectangle would be congruent. In fact, in a rectangle, only opposite sides need be congruent. This means that *ABCD* is not necessarily a rhombus, and statement II does not have to be true.

A square is defined as a rhombus with four right angles. In a square, all of the sides must be congruent. In other words, a square is both a rectangle and a rhombus. However, we already established that *ABCD* doesn't have to be a rhombus. This means that *ABCD* need not be a square, because, as we said previously, not all of its sides must be congruent. Therefore, statement III isn't necessarily true either.

The only statement that has to be true is statement I.

The answer is I only.

### Example Question #1 : How To Find The Area Of A Trapezoid

A trapezoid has a base of length 4, another base of length *s*, and a height of length *s*. A square has sides of length *s*. What is the value of *s* such that the area of the trapezoid and the area of the square are equal?

**Possible Answers:**

**Correct answer:**

In general, the formula for the area of a trapezoid is (1/2)(*a* + *b*)(*h*), where *a* and *b* are the lengths of the bases, and *h* is the length of the height. Thus, we can write the area for the trapezoid given in the problem as follows:

area of trapezoid = (1/2)(4 + *s*)(*s*)

Similarly, the area of a square with sides of length *a* is given by *a*^{2}. Thus, the area of the square given in the problem is *s*^{2}.

We now can set the area of the trapezoid equal to the area of the square and solve for *s*.

(1/2)(4 + *s*)(*s*) = *s*^{2}

Multiply both sides by 2 to eliminate the 1/2.

(4 + *s*)(*s*) = 2*s*^{2}

Distribute the *s* on the left.

4*s* + *s*^{2} = 2*s*^{2}

Subtract *s*^{2} from both sides.

4*s* = *s*^{2}

Because *s* must be a positive number, we can divide both sides by *s*.

4 = *s*

This means the value of *s* must be 4.

The answer is 4.

### Example Question #1 : How To Find The Area Of A Trapezoid

Note: Figure NOT drawn to scale.

The white region in the above diagram is a trapezoid. What percent of the above rectangle, rounded to the nearest whole percent, is blue?

**Possible Answers:**

**Correct answer:**

The area of the entire rectangle is the product of its length and width, or

.

The area of the white trapezoid is one half the product of its height and the sum of its base lengths, or

Therefore, the blue polygon has area

.

This is

of the rectangle.

Rounded, this is 70%.

### Example Question #1 : Trapezoids

Refer to the above diagram. .

Give the area of Quadrilateral .

**Possible Answers:**

**Correct answer:**

, since both are right; by the Corresponding Angles Theorem, , and Quadrilateral is a trapezoid.

By the Angle-Angle Similarity Postulate, since

and

(by reflexivity),

,

and since corresponding sides of similar triangles are in proportion,

, the larger base of the trapozoid;

The smaller base is .

, the height of the trapezoid.

The area of the trapezoid is

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

A circle with a radius 2 in is inscribed in a square. What is the perimeter of the square?

**Possible Answers:**

24 in

16 in

12 in

28 in

32 in

**Correct answer:**

16 in

To inscribe means to draw inside a figure so as to touch in as many places as possible without overlapping. The circle is inside the square such that the diameter of the circle is the same as the side of the square, so the side is actually 4 in. The perimeter of the square = 4s = 4 * 4 = 16 in.