PSAT Math - How to find if rectangles are similar

Rectangles

Note: Figure NOT drawn to scale.

In the above figure,

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

Give the area of Polygon .

$36 + 18\sqrt{5}$

$72+ 18\sqrt{5}$

Explanation

Polygon 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 :

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

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: , the correct response.

Rectangles

Note: Figure NOT drawn to scale.

In the above figure,

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

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

Insufficient information is given to determine the area.

Explanation

Corresponding sidelengths of similar polygons are in proportion, so

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

, 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

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

Rectangles

Note: Figure NOT drawn to scale.

In the above figure,

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

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

$15 \sqrt{5}$

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

$30 \sqrt{5}$

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:

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

Rectangles

Note: Figure NOT drawn to scale.

Refer to the above figure.

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

and .

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

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

Insufficient information is given to answer the problem.

Explanation

and , 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

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

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 $\textup{Rect }ABGF$ .