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PSAT Math : How to find if rectangles are similar

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Example Questions

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

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:

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

$15 \sqrt{5}$

$30 \sqrt{5}$

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 $\frac{ \sqrt{5}}{2}$ multiplied by the length of a side of $\textup{Rect } ABCD$ 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 #3 : 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 #5 : 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:

162

126

$36 + 18\sqrt{5}$

$72+ 18\sqrt{5}$

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 legs and 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 finding using 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 of and :

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

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

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Example Question #631 : Geometry

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:

Insufficient information is given to answer the problem.

$49 \%$

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

$51 \%$

$30 \%$

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 : How To Find If Rectangles Are Similar

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

120

108

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 : How to find if rectangles are similar

Study concepts, example questions & explanations for PSAT Math

All PSAT Math Resources

Example Questions

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

Example Question #3 : Rectangles

Example Question #5 : Rectangles

Example Question #631 : Geometry

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

All PSAT Math Resources

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