Question

1. $\triangle pqr \sim \triangle cba$

1. $\triangle srq \sim \triangle bcd$

1. $\triangle cba \sim \triangle ced$

1. $\triangle jkl \sim \triangle jef$

Explanation:

Problem 7: $\triangle PQR \sim \triangle CBA$

Step1: Identify corresponding sides

Since $\triangle PQR \sim \triangle CBA$, the corresponding sides are proportional. So, $\frac{PQ}{CA}=\frac{QR}{AB}$.
$PQ = 60$, $CA = 12$, $QR = 55$, $AB = x + 5$.

Step2: Set up the proportion

$\frac{60}{12}=\frac{55}{x + 5}$
Simplify $\frac{60}{12}=5$, so $5=\frac{55}{x + 5}$

Step3: Solve for $x$

Cross - multiply: $5(x + 5)=55$
$5x+25 = 55$
Subtract 25 from both sides: $5x=55 - 25=30$
Divide by 5: $x = \frac{30}{5}=6$

Problem 8: $\triangle SRQ \sim \triangle BCD$

Step1: Identify corresponding sides

Since $\triangle SRQ \sim \triangle BCD$, $\frac{SQ}{BD}=\frac{SR}{BC}$.
$SQ = 60$, $BD = 12$, $SR = 3x + 1$, $BC = 8$.

Step2: Set up the proportion

$\frac{60}{12}=\frac{3x + 1}{8}$
Simplify $\frac{60}{12}=5$, so $5=\frac{3x + 1}{8}$

Step3: Solve for $x$

Cross - multiply: $5\times8=3x + 1$
$40=3x + 1$
Subtract 1: $3x=40 - 1 = 39$
Divide by 3: $x=\frac{39}{3}=13$

Problem 9: $\triangle CBA \sim \triangle CED$

Step1: Identify corresponding sides

Since $\triangle CBA \sim \triangle CED$, $\frac{CA}{CD}=\frac{CB}{CE}$.
$CA = 63$, $CD = 27$, $CB = 70$, $CE = 4x-6$.

Step2: Set up the proportion

$\frac{63}{27}=\frac{70}{4x - 6}$
Simplify $\frac{63}{27}=\frac{7}{3}$, so $\frac{7}{3}=\frac{70}{4x - 6}$

Step3: Solve for $x$

Cross - multiply: $7(4x - 6)=3\times70$
$28x-42 = 210$
Add 42 to both sides: $28x=210 + 42=252$
Divide by 28: $x=\frac{252}{28}=9$

Problem 10: $\triangle JKL \sim \triangle JEF$

Answer:

s:

• Problem 7: $x = 6$
• Problem 8: $x = 13$
• Problem 9: $x = 9$
• Problem 10: $x = 7$