Question

1. rectangle with 40 (top), 30 (side); another rectangle with 36 (sides), y (top)
2. rectangle with 30 (top), 20 (side); another rectangle with 24 (sides), y (top)

solve for x. the triangles in each pair are similar.

1. δlmn ~ δlsr triangle lsb: 30, 12; triangle lmn: 18, 12x+3
2. triangle peg: 50, 55, 45; smaller triangle ced: 10, 11, 2x-17
3. triangle sqb: 7x+10, 39, 21, 28 (with segment ik parallel to sb)
4. δedc ~ δevu triangle evu: x+7, 12; triangle edc: 27, 27

solve for x. the polygons in each pair are similar.

1. triangle 1: 20, 16, 32; triangle 2: 2x+7, 20, 40
2. triangle 1: x+4, 16; triangle 2: 7, 3, 8
3. rectangle 1: 2x+6, 35; rectangle 2: 24, 42
4. parallelogram 1: 28, 20, 20; parallelogram 2: 3+4x, 25, 25

Answer:

Let's solve problem 9 first (since it's about similar triangles, we can use the property of similar triangles: corresponding sides are proportional).

Problem 9: $\triangle LMN \sim \triangle LSR$

Step 1: Identify corresponding sides

In similar triangles $\triangle LMN$ and $\triangle LSR$, the corresponding sides are proportional. So, $\frac{LM}{LS} = \frac{MN}{SR}$? Wait, no, let's check the labels. Wait, the first triangle has sides: $LB = 12$, $LN$? Wait, the first triangle is $\triangle LSB$? Wait, the first triangle: $L$ to $B$ is 12, $L$ to $S$? Wait, the first triangle: $L$ with sides 30 (base), 12 (side), and the second triangle: $N$ to $M$ is 18, $M$ to $L$ is $12x + 3$. Wait, maybe the correspondence is $\triangle LSB \sim \triangle LMN$? Wait, the problem says $\triangle LMN \sim \triangle LSR$. Let's assume the sides: in $\triangle LSB$ (first triangle), sides are $LS = 30$? Wait, no, the first triangle: $L$ to $B$ is 12, $L$ to $S$? Wait, the first triangle has a side of 30 (maybe $LS = 30$), $LB = 12$, and the second triangle: $LN$? Wait, the second triangle: $N$ to $M$ is 18, $M$ to $L$ is $12x + 3$, and $N$ to $L$? Wait, maybe the ratio of corresponding sides is $\frac{12}{12x + 3} = \frac{30}{18}$? Wait, no, let's re-express.

Wait, the first triangle (let's say $\triangle LSB$) has sides: $LS = 30$, $LB = 12$, and the second triangle $\triangle LMN$ has $LM = 12x + 3$, $LN = 18$? Wait, no, maybe the correct correspondence is $\frac{LB}{LM} = \frac{LS}{LN}$. So $LB = 12$, $LM = 12x + 3$, $LS = 30$, $LN = 18$. So the proportion is $\frac{12}{12x + 3} = \frac{30}{18}$.

Step 2: Solve the proportion

Cross-multiplying: $12 \times 18 = 30 \times (12x + 3)$

Calculate left side: $12 \times 18 = 216$

Right side: $30 \times (12x + 3) = 360x + 90$

So, $216 = 360x + 90$

Subtract 90 from both sides: $216 - 90 = 360x$

$126 = 360x$

Divide both sides by 360: $x = \frac{126}{360} = \frac{7}{20}$? Wait, that doesn't seem right. Maybe I mixed up the sides.

Wait, maybe the first triangle is $\triangle LSB$ with $LS = 30$, $SB = 12$, and the second triangle $\triangle LMN$ with $LM = 18$, $MN = 12x + 3$? Wait, no, the problem says $\triangle LMN \sim \triangle LSR$. Let's check the labels again. The first triangle: $L$, $S$, $B$ with $LS = 30$, $SB = 12$, $LB$? The second triangle: $L$, $M$, $N$ with $LM = 18$, $MN = 12x + 3$, $LN$? Maybe the ratio is $\frac{LS}{LM} = \frac{SB}{MN}$. So $LS = 30$, $LM = 18$, $SB = 12$, $MN = 12x + 3$. Then:

$\frac{30}{18} = \frac{12}{12x + 3}$

Simplify $\frac{30}{18} = \frac{5}{3}$

So $\frac{5}{3} = \frac{12}{12x + 3}$

Cross-multiplying: $5(12x + 3) = 3 \times 12$

$60x + 15 = 36$

Subtract 15: $60x = 21$

$x = \frac{21}{60} = \frac{7}{20} = 0.35$? That still seems off. Maybe I got the correspondence wrong.

Wait, maybe the first triangle is $\triangle LSB$ with $LS = 30$, $LB = 12$, and the second triangle $\triangle LMN$ with $LN = 30$? No, the second triangle has $MN = 18$, $LM = 12x + 3$. Wait, maybe the problem is $\triangle LMN \sim \triangle LSR$, so $LM$ corresponds to $LS$, $MN$ corresponds to $SR$, $LN$ corresponds to $LR$? Wait, maybe the first triangle is $\triangle LSB$ with $LS = 30$, $SB = 12$, and the second triangle $\triangle LMN$ with $LM = 12x + 3$, $MN = 18$. Then the ratio of similarity is $\frac{SB}{MN} = \frac{12}{18} = \frac{2}{3}$. So $LS$ (30) should correspond to $LM$ (12x + 3) with the same ratio. So $\frac{LS}{LM} = \frac{2}{3}$? Wait, no, if $\triangle LSB \sim \triangle LMN$, then $\frac{LS}{LM} = \frac{SB}{MN} = \frac{LB}{LN}$. Wait, maybe I need to look at the problem again.

Wait, the first triangle: vertices $L$, $S$, $B$ with $LS = 30$, $SB = 12$, $LB$? The second triangle: vertices $L$, $M$, $N$ with $LM = 12x + 3$, $MN = 18$, $LN$? Maybe the correct proportion is $\frac{12}{18} = \frac{30}{12x + 3}$. Let's try that.

$\frac{12}{18} = \frac{30}{12x + 3}$

Simplify $\frac{12}{18} = \frac{2}{3}$

So $\frac{2}{3} = \frac{30}{12x + 3}$

Cross-multiplying: $2(12x + 3) = 3 \times 30$

$24x + 6 = 90$

Subtract 6: $24x = 84$

Divide by 24: $x = \frac{84}{24} = \frac{7}{2} = 3.5$

Ah, that makes more sense. So the correct proportion is $\frac{SB}{MN} = \frac{LS}{LM}$, so $\frac{12}{18} = \frac{30}{12x + 3}$, leading to $x = \frac{7}{2}$ or 3.5.

Problem 10: Similar triangles with sides 50, 10, 11, 55, 45, $2x - 17$

Let's identify the similar triangles. The larger triangle has sides 50 (from E to P), 55 (from P to G), and 45 (from E to G). The smaller triangle has sides 10 (from E to C), 11 (from C to D), and $2x - 17$ (from E to D). So the ratio of similarity is $\frac{10}{50} = \frac{1}{5}$, or $\frac{11}{55} = \frac{1}{5}$. So the ratio is $\frac{1}{5}$. Therefore, the side $2x - 17$ (from E to D) should be $\frac{1}{5}$ of 45 (from E to G). Wait, 45 times $\frac{1}{5}$ is 9. So $2x - 17 = 9$.

Step 1: Set up the equation

$2x - 17 = 9$

Step 2: Solve for x

Add 17 to both sides: $2x = 9 + 17 = 26$

Divide by 2: $x = 13$

Problem 11: Similar triangles with sides 7x + 10, 39, 21, 28

The larger triangle has height 28, the smaller triangle (inside) has height 21. The ratio of heights is $\frac{21}{28} = \frac{3}{4}$. So the ratio of the sides should also be $\frac{3}{4}$. The side of the larger triangle is $7x + 10$, and the side of the smaller triangle is 39. So $\frac{39}{7x + 10} = \frac{3}{4}$.

Step 1: Cross-multiply

$3(7x + 10) = 4 \times 39$

Step 2: Simplify

$21x + 30 = 156$

Subtract 30: $21x = 126$

Divide by 21: $x = 6$

Problem 12: $\triangle EDC \sim \triangle EVU$

The triangles are similar, so corresponding sides are proportional. The sides: $ED = 27$, $EC = 27$? Wait, no, the diagram shows $VE = x + 7$, $UE = 12$, $CE = 27$, $DE = 27$. Wait, the triangles are vertical angles, so $\triangle EDC \sim \triangle EVU$ by AA similarity (vertical angles and maybe alternate interior angles). So the proportion is $\frac{CE}{VE} = \frac{DE}{UE}$? Wait, $CE = 27$, $VE = x + 7$, $DE = 27$, $UE = 12$. Wait, no, maybe $\frac{CE}{UE} = \frac{DE}{VE}$? Wait, $CE = 27$, $UE = 12$, $DE = 27$, $VE = x + 7$. So $\frac{27}{12} = \frac{27}{x + 7}$. Wait, that would imply $12 = x + 7$, so $x = 5$. Which matches the handwritten answer. Let's check:

$\frac{27}{12} = \frac{27}{x + 7}$

Cross-multiplying: $27(x + 7) = 27 \times 12$

Divide both sides by 27: $x + 7 = 12$

Subtract 7: $x = 5$

Problem 13: Similar polygons (triangles) with sides 20, 16, 32 and 2x + 7, 20, 40

The ratio of corresponding sides: $\frac{20}{2x + 7} = \frac{16}{20} = \frac{32}{40}$. Let's check $\frac{16}{20} = \frac{4}{5}$, $\frac{32}{40} = \frac{4}{5}$. So the ratio is $\frac{4}{5}$. So $\frac{20}{2x + 7} = \frac{4}{5}$.

Step 1: Cross-multiply

$4(2x + 7) = 20 \times 5$

Step 2: Simplify

$8x + 28 = 100$

Subtract 28: $8x = 72$

Divide by 8: $x = 9$

Problem 14: Similar triangles with sides x + 4, 16 and 7, 3, 8

Wait, the first triangle has sides $x + 4$ (one side), 16 (base). The second triangle has sides 7 (one side), 3 (another side), 8 (base). Wait, the ratio of the bases: $\frac{16}{8} = 2$. So the ratio of similarity is 2. Therefore, the side $x + 4$ should correspond to 7, so $x + 4 = 7 \times 2 = 14$? Wait, no, maybe the sides are $x + 4$ (corresponding to 7) and 16 (corresponding to 8). So $\frac{x + 4}{7} = \frac{16}{8}$.

$\frac{16}{8} = 2$, so $\frac{x + 4}{7} = 2$

Multiply both sides by 7: $x + 4 = 14$

Subtract 4: $x = 10$

Problem 15: Similar rectangles with sides 2x + 6, 35 and 24, 42

The ratio of the sides: $\frac{35}{42} = \frac{5}{6}$. So the other side $2x + 6$ should correspond to 24 with the same ratio. Wait, $\frac{2x + 6}{24} = \frac{35}{42}$

Simplify $\frac{35}{42} = \frac{5}{6}$

So $\frac{2x + 6}{24} = \frac{5}{6}$

Cross-multiplying: $6(2x + 6) = 24 \times 5$

$12x + 36 = 120$

Subtract 36: $12x = 84$

Divide by 12: $x = 7$

Problem 16: Similar parallelograms with sides 28, 20 and 3 + 4x, 25

The ratio of the sides: $\frac{20}{25} = \frac{4}{5}$. So the other side $28$ should correspond to $3 + 4x$ with the same ratio? Wait, no, $\frac{28}{3 + 4x} = \frac{20}{25}$

Simplify $\frac{20}{25} = \frac{4}{5}$

So $\frac{28}{3 + 4x} = \frac{4}{5}$

Cross-multiplying: $4(3 + 4x) = 28 \times 5$

$12 + 16x = 140$

Subtract 12: $16x = 128$

Divide by 16: $x = 8$

Problem 7: Similar rectangles with sides 40, 30 and y, 36

The ratio of the sides: $\frac{30}{36} = \frac{5}{6}$. So the other side $40$ should correspond to $y$ with the same ratio? Wait, $\frac{40}{y} = \frac{30}{36}$

Simplify $\frac{30}{36} = \frac{5}{6}$

So $\frac{40}{y} = \frac{5}{6}$

Cross-multiplying: $5y = 40 \times 6$

$5y = 240$

Divide by 5: $y = 48$

Problem 8: Similar rectangles with sides 30, 20 and y, 24

The ratio of the sides: $\frac{20}{24} = \frac{5}{6}$. So the other side $30$ should correspond to $y$ with the same ratio? Wait, $\frac{30}{y} = \frac{20}{24}$

Simplify $\frac{20}{24} = \frac{5}{6}$

So $\frac{30}{y} = \frac{5}{6}$

Cross-multiplying: $5y = 30 \times 6$

$5y = 180$

Divide by 5: $y = 36$

Let's summarize the solutions:

• Problem 7: $y = 48$
• Problem 8: $y = 36$
• Problem 9: $x = \frac{7}{2} = 3.5$ (or $\frac{7}{2}$)
• Problem 10: $x = 13$
• Problem 11: $x = 6$
• Problem 12: $x = 5$
• Problem 13: $x = 9$
• Problem 14: $x = 10$
• Problem 15: $x = 7$
• Problem 16: $x = 8$

If you need a specific problem solved, let me know which one, and I can provide a more detailed step-by-step for that.