Matemática Elementar

Problema 1

Triangle $ABC$ has vertices $A(0, 0)$ , $B(4, 0)$ , and $C(0, 3)$ . A dilation centered at the origin with scale factor $k = 2$ maps $\triangle ABC$ to $\triangle A'B'C'$ .

a) Find the coordinates of $A'$ , $B'$ , and $C'$ .

b) Verify that $\frac{A'B'}{AB} = \frac{A'C'}{AC} = \frac{B'C'}{BC} = 2$ .

c) Are the corresponding angles of $\triangle ABC$ and $\triangle A'B'C'$ congruent? Explain.

Problemas

1. Triangle $ABC$ has vertices $A(0, 0)$ , $B(4, 0)$ , and $C(0, 3)$ . A dilation centered at the origin with scale factor $k = 2$ maps $\triangle ABC$ to $\triangle A'B'C'$ .

a) Find the coordinates of $A'$ , $B'$ , and $C'$ .

b) Verify that $\frac{A'B'}{AB} = \frac{A'C'}{AC} = \frac{B'C'}{BC} = 2$ .

c) Are the corresponding angles of $\triangle ABC$ and $\triangle A'B'C'$ congruent? Explain.

2. In the figure, $\overline{DE} \parallel \overline{BC}$ in $\triangle ABC$ , where $D$ is on $\overline{AB}$ and $E$ is on $\overline{AC}$ . Given $AD = 6$ , $DB = 9$ , and $AE = 8$ .

a) Find $EC$ .

b) If $DE = 10$ , find $BC$ .

c) State the similarity criterion that guarantees $\triangle ADE \sim \triangle ABC$ and write the full similarity statement.

3. Two triangles are given: $\triangle PQR$ with $\angle P = 50°$ and $\angle Q = 70°$ , and $\triangle XYZ$ with $\angle Y = 70°$ and $\angle Z = 60°$ .

a) Find the missing angle in each triangle.

b) Are the triangles similar? If so, write the similarity statement and identify the criterion used.

c) If $PQ = 12$ and $XY = 18$ , find the ratio of their perimeters.

4. A flagpole casts a shadow $15$ feet long at the same time that a $6$ -foot person standing nearby casts a shadow $4$ feet long. (Assume the sun's rays are parallel.)

a) Draw a diagram showing the two similar triangles formed.

b) Write a proportion relating the heights and shadow lengths.

c) Find the height of the flagpole.

5. In $\triangle ABC$ , point $D$ lies on $\overline{BC}$ such that $\overline{AD} \perp \overline{BC}$ . Given $AB = 13$ , $AC = 15$ , and $BC = 14$ .

a) Find $BD$ and $DC$ using the Pythagorean theorem applied to $\triangle ABD$ and $\triangle ACD$ .

b) Find $AD$ .

c) Compute the area of $\triangle ABC$ two ways — using $AD$ as the altitude to base $BC$ , and using Heron's formula — and verify they agree.

Gabarito — Similarity

1. a) Under a dilation with center $(0,0)$ and scale factor $k = 2$ , each coordinate is multiplied by $2$ :

$A' = (0, 0), \quad B' = (8, 0), \quad C' = (0, 6)$

b) Compute the side lengths:

$AB = 4, \quad A'B' = 8, \quad \frac{A'B'}{AB} = \frac{8}{4} = 2$

$AC = 3, \quad A'C' = 6, \quad \frac{A'C'}{AC} = \frac{6}{3} = 2$

$BC = \sqrt{4^2 + 3^2} = \sqrt{25} = 5, \quad B'C' = \sqrt{8^2 + 6^2} = \sqrt{100} = 10$

$\frac{B'C'}{BC} = \frac{10}{5} = 2 \quad \checkmark$

All ratios equal the scale factor $k = 2$ .

c) Yes. A dilation preserves angle measures. Each angle in $\triangle A'B'C'$ is congruent to its corresponding angle in $\triangle ABC$ . (This can also be verified: both are right triangles with the same acute angle measures since the side ratios are identical.)

2. a) Since $\overline{DE} \parallel \overline{BC}$ , by the Triangle Proportionality Theorem (or AA similarity):

$\frac{AD}{AB} = \frac{AE}{AC}$

$AB = AD + DB = 6 + 9 = 15$

$\frac{6}{15} = \frac{8}{AC} \implies AC = \frac{8 \times 15}{6} = 20$

$EC = AC - AE = 20 - 8 = 12$

b) Since $\triangle ADE \sim \triangle ABC$ with ratio $\frac{AD}{AB} = \frac{6}{15} = \frac{2}{5}$ :

$\frac{DE}{BC} = \frac{2}{5} \implies BC = \frac{5 \times 10}{2} = 25$

c) Since $\overline{DE} \parallel \overline{BC}$ :

$\angle ADE \cong \angle ABC$ (corresponding angles)
$\angle AED \cong \angle ACB$ (corresponding angles)
$\angle A \cong \angle A$ (shared angle)

By AA Similarity, $\triangle ADE \sim \triangle ABC$ .

3. a) In $\triangle PQR$ :

$\angle R = 180° - 50° - 70° = 60°$

In $\triangle XYZ$ :

$\angle X = 180° - 70° - 60° = 50°$

b) Matching angles: $\angle P = \angle X = 50°$ , $\angle Q = \angle Y = 70°$ , $\angle R = \angle Z = 60°$ .

By the AA Similarity criterion, $\triangle PQR \sim \triangle XYZ$ .

c) The ratio of corresponding sides is:

$\frac{PQ}{XY} = \frac{12}{18} = \frac{2}{3}$

Since similar triangles have proportional sides, the ratio of their perimeters is also $\dfrac{2}{3}$ .

4. a) The flagpole and its shadow form a right triangle, as do the person and their shadow. The sun's parallel rays ensure the angles of elevation are equal, making the triangles similar.

b) Let $h$ be the height of the flagpole. The proportion (height to shadow) is:

$\frac{h}{15} = \frac{6}{4}$

c) Solve:

$h = \frac{6 \times 15}{4} = \frac{90}{4} = 22.5 \text{ feet}$

The flagpole is $22.5$ feet tall.

Verification: $\frac{22.5}{15} = 1.5$ and $\frac{6}{4} = 1.5$ $\checkmark$

5. a) Let $BD = x$ , so $DC = 14 - x$ . In $\triangle ABD$ :

$AD^2 + BD^2 = AB^2 \implies AD^2 + x^2 = 169 \quad \text{(i)}$

In $\triangle ACD$ :

$AD^2 + DC^2 = AC^2 \implies AD^2 + (14 - x)^2 = 225 \quad \text{(ii)}$

Subtract (i) from (ii):

$(14 - x)^2 - x^2 = 225 - 169$

$196 - 28x + x^2 - x^2 = 56$

$196 - 28x = 56$

$28x = 140 \implies x = 5$

So $BD = 5$ and $DC = 9$ .

b) From equation (i):

$AD^2 = 169 - 25 = 144 \implies AD = 12$

Verification: $AD^2 + DC^2 = 144 + 81 = 225 = 15^2 = AC^2$ $\checkmark$

c) Method 1 — using the altitude:

$\text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot BC \cdot AD = \frac{1}{2} \cdot 14 \cdot 12 = 84$

Method 2 — Heron's formula:

The semi-perimeter is:

$s = \frac{AB + AC + BC}{2} = \frac{13 + 15 + 14}{2} = 21$

$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21 \cdot (21-14) \cdot (21-15) \cdot (21-13)}$

$= \sqrt{21 \cdot 7 \cdot 6 \cdot 8} = \sqrt{7056} = 84$

Both methods give an area of $84$ square units. $\checkmark$

Note: $\triangle ABD$ is a $5$ - $12$ - $13$ right triangle and $\triangle ACD$ is a $3$ - $4$ - $5$ triple scaled by $3$ (sides $9$ - $12$ - $15$ ). Both are right triangles sharing the altitude $AD = 12$ .