How to find the length of the side of a rectangle - SAT Math

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Question

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

Answer

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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Question

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

Answer

Solve for x

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

x2 = 36

x = 6

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Question

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?

Answer

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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Question

Figure is not drawn to scale.

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

Rectangle $ABEF \sim$ Rectangle .

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

Answer

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

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

;

since , if we let , then

,

and

Setting , , and , 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:

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

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

or

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

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 :

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

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

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

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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