PSAT Math : Quadrilaterals

Study concepts, example questions & explanations for PSAT Math

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Example Questions

Example Question #21 : Quadrilaterals

A half circle has an area of . What is the area of a square with sides that measure the same length as the diameter of the half circle?

Possible Answers:

144

36

108

81

72

Correct answer:

144

Explanation:

If the area of the half circle is , then the area of a full circle is twice that, or .

Use the formula for the area of a circle to solve for the radius:

36π = πr2

r = 6

If the radius is 6, then the diameter is 12. We know that the sides of the square are the same length as the diameter, so each side has length 12.

Therefore the area of the square is 12 x 12 = 144.

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

The area of square R is 12 times the area of square T. If the area of square R is 48, what is the length of one side of square T?

 

Possible Answers:

4

2

1

16

Correct answer:

2

Explanation:

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.

 

 

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

When the side of a certain square is increased by 2 inches, the area of the resulting square is 64 sq. inches greater than the original square.  What is the length of the side of the original square, in inches?

Possible Answers:

17

15

18

16

14

Correct answer:

15

Explanation:

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 xfrom 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.

 

 

 

Example Question #622 : Geometry

If the area of a square is 25 inches squared, what is the perimeter?

Possible Answers:

10

Not enough information

20

15

25

Correct answer:

20

Explanation:

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 l = \sqrt{25in^{2}} or l=5 in. The perimeter of a square is the sum of the length of all 4 sides or 4 \times 5 in. =20 in.

Example Question #201 : Plane Geometry

What is the length of the diagonal of a rectangle that is 3 feet long and 4 feet wide?

Possible Answers:

5\ feet

4\ feet

7\ feet

8\ feet

6\ feet

Correct answer:

5\ feet

Explanation:

The diagonal of the rectangle is equivalent to finding the length of the hypotenuse of a right triangle with sides 3 and 4. Using the Pythagorean Theorem:

3^{2}+4^{2} = hypotenuse^{2}

25 = hypotenuse^{2}

hypotenuse = 5

Therefore the diagonal of the rectangle is 5 feet.

Example Question #202 : Plane Geometry

The length and width of a rectangle are in the ratio of 3:4. If the rectangle has an area of 108 square centimeters, what is the length of the diagonal?

Possible Answers:

12 centimeters

18 centimeters

9 centimeters

15 centimeters

24 centimeters

Correct answer:

15 centimeters

Explanation:

The length and width of the rectangle are in a ratio of 3:4, so the sides can be written as 3x and 4x.

We also know the area, so we write an equation and solve for x:

(3x)(4x) = 12x= 108.

x2 = 9

x = 3

Now we can recalculate the length and the width:

length = 3x = 3(3) = 9 centimeters

width = 4x = 4(3) = 12 centimeters

Using the Pythagorean Theorem we can find the diagonal, c:

length2 + width2 = c2

92 + 12= c2 

81 + 144 = c2

225 = c2

= 15 centimeters

Example Question #1 : Rectangles

 

 

 

The two rectangles shown below are similar. What is the length of EF?

 Sat_mah_166_02

Possible Answers:

8

6

5

10

Correct answer:

10

Explanation:

When two polygons are similar, the lengths of their corresponding sides are proportional to each other.  In this diagram, AC and EG are corresponding sides and AB and EF are corresponding sides. 

To solve this question, you can therefore write a proportion:

AC/EG = AB/EF ≥ 3/6 = 5/EF

From this proportion, we know that side EF is equal to 10.

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

A rectangle is x inches long and 3x inches wide.  If the area of the rectangle is 108, what is the value of x?

Possible Answers:

3

4

6

12

8

Correct answer:

6

Explanation:

Solve for x

Area of a rectangle A = lw = x(3x) = 3x2 = 108

x2 = 36

x = 6

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

If the area of rectangle is 52 meters squared and the perimeter of the same rectangle is 34 meters. What is the length of the larger side of the rectangle if the sides are integers?

Possible Answers:

14 meters

15 meters

16 meters

12 meters

13 meters

Correct answer:

13 meters

Explanation:

Area of a rectangle is = lw

Perimeter = 2(l+w)

We are given 34 = 2(l+w) or 17 = (l+w)

possible combinations of l + w

are 1+16, 2+15, 3+14, 4+13... ect

We are also given the area of the rectangle is 52 meters squared.

Do any of the above combinations when multiplied together= 52 meters squared? yes 4x13 = 52

Therefore the longest side of the rectangle is 13 meters

 

Example Question #1 : How To Find If Rectangles Are Similar

Rectangles

Note: Figure NOT drawn to scale.

In the above figure,

.

.

Give the perimeter of .

Possible Answers:

Correct answer:

Explanation:

We can use the Pythagorean Theorem to find :

The similarity ratio of  to  is 

so  multiplied by the length of a side of  is the length of the corresponding side of . We can subsequently multiply the perimeter of the former by  to get that of the latter:

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