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PSAT Math : Rectangles

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

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

Possible Answers:

$5\ feet$

$4\ feet$

$6\ feet$

$7\ feet$

$8\ feet$

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.

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

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

24 centimeters

15 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 3xand 4x.

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

(3x)(4x) = 12x²= 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+宽度²= c²

9²+ 12²=c²

81 + 144 = c²

225 = c²

c= 15 centimeters

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

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

Sat_mah_166_02

Possible Answers:

Correct answer:

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.

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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:

Correct answer:

Explanation:

Solve for x

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

x²= 36

x = 6

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Example Question #201 :Plane Geometry

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:

12 meters

14 meters

15 meters

16 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

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Example Question #1 :Rectangles

Rectangles

Note: Figure NOT drawn to scale.

In the above figure,

$\textup{Rect } ABCD \sim \textup{Rect } DBEF$ .

AB = 10, AD = 5 .

Give the perimeter of $\textup{Rect } DBEF$ .

Possible Answers:

$15 \sqrt{5}$

$30 \sqrt{5}$

$\frac{15}{2} \sqrt{5}$

Correct answer:

$15 \sqrt{5}$

Explanation:

We can use the Pythagorean Theorem to find:

$DB = \sqrt{(AD)^{2}+(AB)^{2}}$

$DB = \sqrt{5^{2}+10^{2}} = \sqrt{25+100} = \sqrt{125} = \sqrt{25} \cdot \sqrt{5} =5 \sqrt{5}$

The similarity ratio of $\textup{Rect } DBEF$ to $\textup{Rect } ABCD$ is

$\frac{DB}{AB} = \frac{5\sqrt{5}}{10} = \frac{ \sqrt{5}}{2}$

so $\frac{ \sqrt{5}}{2}$ 乘的边的长度 $\textup{Rect } ABCD$ is the length of the corresponding side of $\textup{Rect } DBEF$ . We can subsequently multiply the perimeter of the former by $\frac{ \sqrt{5}}{2}$ to get that of the latter:

$\frac{ \sqrt{5}}{2} (10 + 5 + 10 + 5) = \frac{ \sqrt{5}}{2} (30) = 15 \sqrt{5}$

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Example Question #2 :Rectangles

Rectangles

Note: Figure NOT drawn to scale.

In the above figure,

$\textup{Rect } ABCD \sim \textup{Rect } DBEF$ .

AB = 8, AD = 4 .

Give the area of $\textup{Rect } DBEF$ .

Possible Answers:

Insufficient information is given to determine the area.

Correct answer:

Explanation:

Corresponding sidelengths of similar polygons are in proportion, so

$\frac{DF}{AD}= \frac{DB}{AB}$

AB = 8, AD = 4 , so

$\frac{DF}{4}= \frac{DB}{8}$

$DF= \frac{DB}{8} \cdot 4$

$DF= \frac{DB}{2}$

We can use the Pythagorean Theorem to find:

$DB = \sqrt{(AD)^{2}+(AB)^{2}}$

$DB = \sqrt{4^{2}+8^{2}} = \sqrt{16+64 } = \sqrt{80} = \sqrt{16} \cdot \sqrt{5} = 4 \sqrt{5}$

$DF= \frac{DB}{2}= \frac{ 4 \sqrt{5}}{2} = 2\sqrt{5}$

The area of $\textup{Rect } DBEF$ is

$DB \cdot DF = 4 \sqrt{5} \cdot 2\sqrt{5} = 8 \cdot 5 = 40$

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Example Question #3 :Rectangles

Rectangles

Note: Figure NOT drawn to scale.

In the above figure,

$\textup{Rect } ABCD \sim \textup{Rect } DBEF$ .

AB = 12, AD = 6 .

Give the area of Polygon ABEFD .

Possible Answers:

$36 + 18\sqrt{5}$

162

$72+ 18\sqrt{5}$

126

Correct answer:

126

Explanation:

Polygon ABEFD can be seen as a composite of right $\Delta ABD$ and $\textup{Rect } DBEF$ , so we calculate the individual areas and add them.

The area of $\Delta ABD$ is half the product of legsand A D :

$\frac{1}{2} \cdot AB \cdot AD = \frac{1}{2} \cdot 12 \cdot 6 = 36$

Now we find the area of $\textup{Rect } DBEF$ . We can do this by first findingusing the Pythagorean Theorem:

$DB = \sqrt{(AD)^{2}+(AB)^{2}}$

$DB = \sqrt{6^{2}+12^{2}} = \sqrt{36+144} = \sqrt{180} = \sqrt{36} \cdot \sqrt{5} = 6 \sqrt{5}$

The similarity of $\textup{Rect } DBEF$ to $\textup{Rect } ABCD$ implies

$\frac{DF}{AD}= \frac{DB}{AB}$

$\frac{DF}{6}= \frac{6\sqrt{5}}{12}$

$DF= \frac{6 \cdot 6\sqrt{5}}{12} = \frac{3 6\sqrt{5}}{12} = 3\sqrt{5}$

The area of $\textup{Rect } DBEF$ is the product ofand:

$DB \cdot DF = 6 \sqrt{5} \cdot 3 \sqrt{5}= 6\cdot 3 \cdot \sqrt{5} \cdot \sqrt{5}= 18 \cdot 5 = 90$

Now add: 90 + 36 = 126 , the correct response.

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Example Question #1 :Rectangles

Rectangles

Note: Figure NOT drawn to scale.

Refer to the above figure.

$\textup{Rect }ABGF \sim \textup{Rect }ACDE$

AB = 7 and BC = 3 .

What percent of $\textup{Rect }ACDE$ has been shaded brown ?

Possible Answers:

$30 \%$

$49 \%$

Insufficient information is given to answer the problem.

$57 \frac{1}{7} \%$

$51 \%$

Correct answer:

$51 \%$

Explanation:

AB = 7 and AC = AB + BC = 7 + 3 = 10 , so the similarity ratio of $\textup{Rect }ACDE$ to $\textup{Rect }ABGF$ is 10 to 7. The ratio of the areas is the square of this, or

$10^{2}:7^{2}$

100:49

Therefore, $\textup{Rect }ABGF$ comprises $\frac{49}{100} = 49 \%$ of $\textup{Rect }ABGF$ , and the remainder of the rectangle - the brown region - is 51% of $\textup{Rect }ABGF$ .

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Example Question #1 :Rectangles

Rectangles

Note: figure NOT drawn to scale.

Refer to the above figure.

$\textup{Rect }ABGF \sim \textup{Rect }ACDE$

AB = 12 , BC = 3 , AE = 10 .

Give the area of $\textup{Rect }ABGF$ .

Possible Answers:

216

108

120

Correct answer:

Explanation:

$\textup{Rect }ABGF \sim \textup{Rect }ACDE$ , so the sides are in proportion - that is,

$\frac{AF}{AB}= \frac{AE}{AC}$

AC = AB + BC = 12 + 3 = 15

Set

AB = 12 , AC = 15 , AE = 10 and solve for:

$\frac{AF}{12}= \frac{10}{15}$

$AF = \frac{10}{15} \cdot 12 = 8$

$\textup{Rect }ABGF$ has area

$AB \cdot AF = 12 \cdot 8 = 96$

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PSAT Math : Rectangles

Study concepts, example questions & explanations for PSAT Math

All PSAT Math Resources

Example Questions

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

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

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

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

Example Question #201 :Plane Geometry

Example Question #1 :Rectangles

Example Question #2 :Rectangles

Example Question #3 :Rectangles

Example Question #1 :Rectangles

Example Question #1 :Rectangles

All PSAT Math Resources

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