Question

1. $\triangle cde \sim \triangle fge$; find $ce$

1. $\triangle kmz \sim \triangle kdh$; find $x$

1. $\triangle agn \sim \triangle fls$; find $x$

1. $\triangle cse \sim \triangle yje$; find $ey$

1. if $\triangle klm \sim \triangle pqr$ with a scale factor of $3:5$, find the perimeter of $\triangle pqr$.

1. if $\triangle tsr \sim \triangle tfe$, find the perimeter of $\triangle tfe$.

Explanation:

Problem 7

Step1: Set up proportion for similar triangles

Since $\triangle CDE \sim \triangle FGE$, corresponding sides are proportional:
$\frac{CE}{FE} = \frac{DE}{GE}$

Step2: Substitute given values

$\frac{8x-1}{21} = \frac{8x+3}{17}$

Step3: Cross-multiply to solve for $x$

$17(8x-1) = 21(8x+3)$
$136x - 17 = 168x + 63$
$136x - 168x = 63 + 17$
$-32x = 80$
$x = \frac{80}{-32} = -2.5$

Step4: Calculate $CE$

$CE = 8x - 1 = 8(-2.5) - 1$

Step1: Set up proportion for similar triangles

Since $\triangle KMZ \sim \triangle KDH$, corresponding sides are proportional:
$\frac{KZ}{KH} = \frac{MZ}{DH}$

Step2: Substitute given values

$\frac{32}{32+28} = \frac{x+8}{(x+8)+21}$
$\frac{32}{60} = \frac{x+8}{x+29}$

Step3: Cross-multiply to solve for $x$

$32(x+29) = 60(x+8)$
$32x + 928 = 60x + 480$
$928 - 480 = 60x - 32x$
$448 = 28x$
$x = \frac{448}{28}$

Step1: Set up proportion for similar triangles

Since $\triangle AGN \sim \triangle FLS$, corresponding sides are proportional:
$\frac{AG}{FL} = \frac{GN}{LS}$

Step2: Substitute given values

$\frac{24}{x} = \frac{x+7}{6}$

Step3: Cross-multiply to solve for $x$

$24 \times 6 = x(x+7)$
$144 = x^2 + 7x$
$x^2 + 7x - 144 = 0$

Step4: Solve quadratic equation

Use quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ where $a=1, b=7, c=-144$:
$x = \frac{-7 \pm \sqrt{49 + 576}}{2} = \frac{-7 \pm \sqrt{625}}{2} = \frac{-7 \pm 25}{2}$
Take positive root: $x = \frac{-7 + 25}{2}$

Answer:

$CE = -21$

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Problem 8