Question

similar triangles and proportions congruent
the symbol ~ means \is similar to.\ if two triangles are similar, the measurements of their corresponding sides are proportional. in the box below, $\triangle abc \sim \triangle def$.
box with two triangles, proportions, and handwritten notes: \ab ≅ de\, \show work!!\, proportions $\frac{5}{10} = \frac{13}{e} = \frac{12}{f}$ and $\frac{5}{10} = \frac{13}{26} = \frac{12}{24}$, note \all of the corresponding sides have a ratio of 1:2.\
each pair of triangles below is similar. set up proportions to find the missing measurements. work on scratch paper if you need to.
1.
triangle with sides 4, 5, 6; similar triangle with sides 12, s, t
$s = \underline{\quad\quad\quad\quad}$ $t = \underline{\quad\quad\quad\quad}$
2.
triangle with sides 2, 8, 10; similar triangle with sides 3, f, g
$f = \underline{\quad\quad\quad\quad}$ $g = \underline{\quad\quad\quad\quad}$
3.
triangle with sides j, 6, k; similar triangle with sides 1, 1.5, 3
$j = \underline{\quad\quad\quad\quad}$ $k = \underline{\quad\quad\quad\quad}$
4.
triangle with sides 10, m, n; similar triangle with sides 8, 20, 20
$m = \underline{\quad\quad\quad\quad}$ $n = \underline{\quad\quad\quad\quad}$
5.
triangle with sides 100, 50, 150; similar triangle with sides y, 30, x
$x = \underline{\quad\quad\quad\quad}$ $y = \underline{\quad\quad\quad\quad}$
6.
triangle with sides 27, p, q; similar triangle with sides 9, 15, 5
$p = \underline{\quad\quad\quad\quad}$ $q = \underline{\quad\quad\quad\quad}$
69
fs-10218 introduction to geometry

Explanation:

Step1: Solve for \( s \) and \( t \) in Problem 1

The ratio of corresponding sides of similar triangles is equal. For the first pair, the ratio of the vertical sides is \( \frac{4}{12}=\frac{1}{3} \). So, \( \frac{5}{s}=\frac{1}{3} \) and \( \frac{6}{t}=\frac{1}{3} \). Solving \( \frac{5}{s}=\frac{1}{3} \), we get \( s = 15 \). Solving \( \frac{6}{t}=\frac{1}{3} \), we get \( t = 18 \).

Step2: Solve for \( f \) and \( g \) in Problem 2

The ratio of vertical sides is \( \frac{2}{3} \). So, \( \frac{8}{f}=\frac{2}{3} \) (cross - multiply: \( 2f = 24 \), \( f = 12 \)) and \( \frac{10}{g}=\frac{2}{3} \) (cross - multiply: \( 2g = 30 \), \( g = 15 \)).

Step3: Solve for \( j \) and \( k \) in Problem 3

The ratio of corresponding sides: \( \frac{6}{1.5}=4 \). So, \( \frac{j}{1}=4 \) ( \( j = 4 \)) and \( \frac{k}{3}=4 \) ( \( k = 12 \)).

Step4: Solve for \( m \) and \( n \) in Problem 4

The ratio of vertical sides: \( \frac{10}{8}=\frac{5}{4} \). So, \( \frac{m}{20}=\frac{5}{4} \) (cross - multiply: \( 4m = 100 \), \( m = 25 \)) and \( \frac{n}{20}=\frac{5}{4} \) (cross - multiply: \( 4n = 100 \), \( n = 25 \)).

Step5: Solve for \( x \) and \( y \) in Problem 5

The ratio of the top sides: \( \frac{50}{30}=\frac{5}{3} \). So, \( \frac{150}{x}=\frac{5}{3} \) (cross - multiply: \( 5x = 450 \), \( x = 90 \)) and \( \frac{100}{y}=\frac{5}{3} \) (cross - multiply: \( 5y = 300 \), \( y = 60 \)).

Step6: Solve for \( p \) and \( q \) in Problem 6

The ratio of the top sides: \( \frac{27}{9}=3 \). So, \( \frac{p}{15}=3 \) ( \( p = 45 \)) and \( \frac{q}{5}=3 \) ( \( q = 15 \)).

Answer:

1. \( s = 15 \), \( t = 18 \)
2. \( f = 12 \), \( g = 15 \)
3. \( j = 4 \), \( k = 12 \)
4. \( m = 25 \), \( n = 25 \)
5. \( x = 90 \), \( y = 60 \)
6. \( p = 45 \), \( q = 15 \)