Answer:
see explanation
Step-by-step explanation:
(5)
Since the triangles are similar then the ratios of corresponding sides are equal, that is
[tex]\frac{DE}{PQ}[/tex] = [tex]\frac{EF}{QR}[/tex], substitute values
[tex]\frac{x}{6}[/tex] = [tex]\frac{10}{4}[/tex] ( cross- multiply )
4x = 60 ( divide both sides by 4 )
x = 10
AND
[tex]\frac{PR}{DF}[/tex] = [tex]\frac{EF}{QR}[/tex], that is
[tex]\frac{y}{20}[/tex] = [tex]\frac{10}{4}[/tex] ( cross- multiply )
4y = 200 ( divide both sides by 4 )
y = 50
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(6)
Since the figures are similar then the ratios of corresponding sides are equal
[tex]\frac{AD}{PS}[/tex] = [tex]\frac{AB}{PQ}[/tex], substitute values
[tex]\frac{x}{5.2}[/tex] = [tex]\frac{4.5}{3}[/tex] ( cross- multiply )
3x = 23.4 ( divide both sides by 3 )
x = 7.8
AND
[tex]\frac{RS}{CD}[/tex] = [tex]\frac{AB}{PQ}[/tex], that is
[tex]\frac{y}{6}[/tex] = [tex]\frac{4.5}{3}[/tex] ( cross- multiply )
3y = 27 ( divide both sides by 3 )
y = 9
