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SAT Math : Rectangles

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Example Questions

Example Question #201 : Sat Mathematics

Find the length of the diagonal of a rectangle whose sides are 8 and 15.

Possible Answers:

$\sqrt{17}$

$\sqrt{23}$

Correct answer:

Explanation:

To solve. simply use the Pythagorean Theorem where a=8 and b=15 .

Thus,

$c=\sqrt{a^2+b^2}=\sqrt{8^2+15^2}=\sqrt{289}=17$

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Example Question #2 : How To Find The Length Of The Diagonal Of A Rectangle

Prism

The above figure depicts a cube, each edge of which has length 18. Give the length of the shortest path from Point A to Point B that lies completely along the surface of the cube.

Possible Answers:

$18 \sqrt{2}$

$18\sqrt{5}$

$18\sqrt{6}$

$18\sqrt{3}$

Correct answer:

$18\sqrt{5}$

Explanation:

The shortest path is along two of the surfaces of the prism. There are three possible choices - top and front, right and front, and rear and bottom - but as it turns out, since all faces are (congruent) squares, all three paths have the same length. One such path is shown below, with the relevant faces folded out:

Prism 2

The length of the path can be seen to be equal to that of the diagonal of a rectangle with length and width 18 and 36, so its length can be found by applying the Pythagorean Theorem. Substituting 18 and 36 for and :

$c = \sqrt{a^{2}+b^{2}}$

$= \sqrt{18^{2}+36^{2}}$

$= \sqrt{324+1,296}$

$= \sqrt{1,620 }$

Applying the Product of Radicals Rule:

$c = \sqrt{1,620 } = \sqrt{324} \cdot \sqrt{5 } = 18 \sqrt{5 }$ .

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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 #2 : 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 #4 : Rectangles

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:

16 meters

14 meters

13 meters

15 meters

12 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 #33 : Quadrilaterals

Rectangles a

Figure is not drawn to scale.

The provided figure is a rectangle divided into two smaller rectangles, with

Rectangle $ABEF \sim$ Rectangle CDEB .

Which expression is equal to the length of $\overline{FE}$ ?

Possible Answers:

$12+ 2 \sqrt{5}$

$12 + 4\sqrt{5}$

$12+\sqrt{5}$

Correct answer:

$12 + 4\sqrt{5}$

Explanation:

Since Rectangle ABEF is similar to Rectangle CDEB , it follows that corresponding sides are in proportion. Specifically,

$\frac{FE}{BE} = \frac{BE}{DE}$

FE + DE = FD ;

since FD = 24 , if we let t = FE , then

t+ DE = 24 ,

and

DE= 24 - t

Setting BE = 8 , FE = t , and DE= 24 - t , the proportion statement becomes

$\frac{t}{8} = \frac{8}{24-t }$

Cross-multiplying, we get

$t(24-t) = 8 \cdot 8$

Simplifying, we get

$24t-t ^{2}= 64$

Since this is quadratic, all terms must be moved to one side:

$24t-t ^{2} - 24t + t ^{2} = 64 - 24t + t ^{2}$

$0= 64 - 24t + t ^{2}$

$t ^{2} - 24t + 64 = 0$

The solutions to quadratic equation

$ax^{2} + bx + c = 0$

can be found by way of the quadratic formula

$x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

Set a = 1, b= -24, c= 64 :

$x = \frac{-(-24) \pm \sqrt{(-24)^{2}-4(1)(64)}}{2(1)}$

$= \frac{ 24 \pm \sqrt{576-256}}{2}$

$= \frac{ 24 \pm \sqrt{320}}{2}$

Simplifying the radical using the Product of Radicals Property, we get

$\frac{ 24 \pm \sqrt{64}\cdot \sqrt{5}}{2}$

$=\frac{ 24 \pm8\sqrt{5}}{2}$

Splitting the fraction and reducing:

$\frac{ 24 }{2} \pm \frac{ 8\sqrt{5}}{2}$

$=12 \pm 4\sqrt{5}$

This actually tells us that the lengths of the two segments $\overline{FE}$ and $\overline{ED}$ have the lengths $12 + 4\sqrt{5}$ and $12 -4\sqrt{5}$ . $\overline{FE}$ is seen in the diagram to be the longer, so we choose the greater value, $12 + 4\sqrt{5}$ .

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Example Question #1 : Rectangles

A rectangle has a width of 2x. If the length is five more than 150% of the width, what is the perimeter of the rectangle?

Possible Answers:

10(x + 1)

6x² + 5

5x + 10

6x² + 10x

5x + 5

Correct answer:

10(x + 1)

Explanation:

Given that w = 2x and l = 1.5w + 5, a substitution will show that l = 1.5(2x) + 5 = 3x + 5.

P = 2w + 2l = 2(2x) + 2(3x + 5) = 4x + 6x + 10 = 10x + 10 = 10(x + 1)

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

Find the perimeter of a rectangle with width 7 and length 9.

Possible Answers:

$\frac{63}{2}$

Correct answer:

Explanation:

To solve, simply use the formula for the perimeter of a rectangle.

Substitute in the width of seven and the length of nine.

Thus,

P=2(w+l)=2*(7+9)=2*16=32

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Example Question #2 : How To Find The Perimeter Of A Rectangle

Find the perimeter of a rectangle whose side lengths are 1 and 2.

Possible Answers:

Correct answer:

Explanation:

To solve, simply use the formula for the perimeter of a rectangle. Thus,

P=2(w+l)=2*(1+2)=2*3=6

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Example Question #2 : How To Find The Perimeter Of A Rectangle

Find the perimeter of a rectangle with width 6 and length 9.

Possible Answers:

Correct answer:

Explanation:

To solve, simply use the formula for the perimeter.

P=2(w+l)

P=2*(6+9)=2*15=30

Another way to solve this problem is to add up all of the sides. Remember that even though only two values are given, a rectangle has 4 sides. Thus,

P=6+6+9+9=30

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SAT Math : Rectangles

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #201 : Sat Mathematics

Example Question #2 : 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 #2 : How To Find The Length Of The Side Of A Rectangle

Example Question #4 : Rectangles

Example Question #33 : Quadrilaterals

Example Question #1 : Rectangles

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

Example Question #2 : How To Find The Perimeter Of A Rectangle

Example Question #2 : How To Find The Perimeter Of A Rectangle

All SAT Math Resources

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