Exercises on proportional segments

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When the ratio of two line segments is equal to the ratio of two other segments, they are called proportional segments.

A reason between two segments is obtained by dividing the length of one by the other.

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Thus, given four proportional line segments with lengths The, B, w It is d, in that order, we have a proportion:

$\dpi{120} \mathbf{\frac{a}{b} \frac{c}{d}}$

And, by the fundamental property of proportions, we have $\dpi{120} \mathbf{ ad cb}$ .

To learn more, check out a list of exercises on proportional segments, with all questions resolved!

Exercises on proportional segments

Question 1. The segments $\dpi{120} \overline{AB}, \overline{CD}, \overline{EF}\, \mathrm{e}\, \overline{GH}$ are, in that order, proportional segments. Determine the measure of $\dpi{120} \overline{CD}$ knowing that $\dpi{120} \overline{AB} 5$ , $\dpi{120} \overline{EF} 7.5$ It is $\dpi{120} \overline{GH} 13.8$ .

Question 2. Determine $\dpi{120} \overline{BC}$ knowing that $\dpi{120} \frac{\overline{AB}}{7} \frac{\overline{BC}}{4}$ is that:

Question 3. Determine $\dpi{120} \overline{AB}$ knowing that $\dpi{120} \frac{\overline{AB}}{2} \frac{\overline{BC}}{5}$ is that:

Question 4. Determine the lengths of the sides of a triangle that has a perimeter of 52 units and whose sides are proportional to the sides of another triangle with lengths 2, 6, and 5.

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Resolution of question 1

If the segments $\dpi{120} \overline{AB}, \overline{CD}, \overline{EF}\, \mathrm{e}\, \overline{GH}$ are, in that order, proportional segments, then:

$\dpi{120} \frac{\overline{AB}}{\overline{CD}} \frac{\overline{EF}}{\overline{GH}}$

replacing $\dpi{120} \overline{AB} 5$ , $\dpi{120} \overline{EF} 7.5$ It is $\dpi{120} \overline{GH} 13.8$ , We have to:

$\dpi{120} \frac{5}{\overline{CD}} \frac{7,5}{13,8}$

Applying the fundamental property of proportions:

$\dpi{120} \Rightarrow 7.5 \cdot \overline{CD} 69$

$\dpi{120} \Rightarrow \overline{CD} \frac{69}{7.5}$

$\dpi{120} \Rightarrow \overline{CD} 9.2$

Resolution of question 2

We have:

$\dpi{120} \frac{\overline{AB}}{7} \frac{\overline{BC}}{4}$

replacing $\dpi{120} \overline{AB} 11$ , We have to:

$\dpi{120} \frac{11}{7} \frac{\overline{BC}}{4}$

Applying the fundamental property of proportions:

$\dpi{120} \Rightarrow 7\overline{BC} 44$

$\dpi{120} \Rightarrow \overline{BC} \frac{44}{7}$

$\dpi{120} \Rightarrow \overline{BC} \approx 6.28$

Resolution of question 3

We have:

$\dpi{120} \frac{\overline{AB}}{2} \frac{\overline{BC}}{5}$

As $\dpi{120} \overline{AB} + \overline{BC} 21$ , then, $\dpi{120} \overline{AB} 21 - \overline{BC}$ . Substituting in the above expression, we have:

$\dpi{120} \frac{21-\overline{BC}}{2} \frac{\overline{BC}}{5}$

Applying the fundamental property of proportions:

$\dpi{120} \Rightarrow 2\overline{BC} 5(21- \overline{BC})$

$\dpi{120} \Rightarrow 2\overline{BC} 105- 5\overline{BC}$

$\dpi{120} \Rightarrow 7\overline{BC} 105$

$\dpi{120} \Rightarrow \overline{BC} \frac{105}{7}$

$\dpi{120} \Rightarrow \overline{BC} 15$

Soon $\dpi{120} \overline{AB} 21 - \overline{BC} 21 - 15 6$ .

Resolution of question 4

Making a representative drawing, we can see that $\dpi{120} \overline{AB} + \overline{BC} + \overline{AC} 52$ .

Since the sides of the triangles are proportional, we have:

$\dpi{120} \frac{\overline{AB}}{2} \frac{\overline{BC}}{6} \frac{\overline{AC}}{5} r$

Being $\dpi{120} r$ the ratio of proportionality.

Furthermore, if the sides are proportional, their sum, that is, the perimeters, are also:

$\dpi{120} \frac{\overline{AB} + \overline{BC} +\overline{AC} }{2 + 6 + 5} r$

$\dpi{120} \Rightarrow \frac{52 }{13} r$

$\dpi{120} \Rightarrow r 4$

From the ratio of proportionality and the known sides, we obtain the measures of the sides of the other triangle:

$\dpi{120} \overline{AB} r\cdot \overline{A'B'} 4\cdot 2 8$

$\dpi{120} \overline{BC} r\cdot \overline{B'C'} 4\cdot 6 24$

$\dpi{120} \overline{AC} r\cdot \overline{A'C'} 4\cdot 5 20$

To download this list of exercises on proportional segments in PDF, click here!

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