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frika

Answer:

8 units

Step-by-step explanation:

Consider triangles BPC and APD. These two triangles are similar, because

  • ∠DAP≅∠CBP (as corresponding angles);
  • ∠BCP≅∠ADP (as corresponding angles);
  • ∠APD≅∠BPC (by reflexive property of congruence).

Then, by AA theorem, [tex]\triangle BCP\sim \triangle ADP.[/tex]

Similar triangles have proportional sides lengths:

[tex]\dfrac{BP}{AP}=\dfrac{CP}{DP}=\dfrac{BC}{AD},\\ \\\dfrac{BP}{AP}=\dfrac{CP}{CP+3}=\dfrac{5}{8}.[/tex]

The last proportion gives you

[tex]8CP=5(CP+3),\\ \\8CP=5CP+15,\\ \\3CP=15,\\ \\CP=5\ un.[/tex]

Then [tex]DP=CP+CD=5+3=8\ un.[/tex]

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akposevictor

Using the theorem of similar triangles and applying the knowledge of proportion, the length of segment DP in the diagram given is: 8 units.

See attachment for the diagram of the trapezium that is given.

Applying proportion and the theorem for similar triangles, we will solve for the length of segment DP.

Given:

BC = 5

AD = 8

CD = 3

Note: [tex]\triangle ADP $ and $ \triangle BCP[/tex] are similar triangles, therefore, their side lengths will be proportional to each other.

Thus:

[tex]\triangle ADP $ and $ \triangle BCP\\\\\frac{AD}{BC} = \frac{DP}{CP} = \frac{AP}{BP}[/tex]

To find DP, use [tex]\frac{AD}{BC} = \frac{DP}{CP}[/tex]

  • Substitute

[tex]\frac{8}{5} = \frac{CP + 3}{CP}[/tex]

  • Cross multiply

[tex]CP \times 8 = 5(CP + 3)\\\\8CP = 5CP + 15[/tex]

  • Subtract 5CP from both sides

[tex]8CP - 5CP = 15\\\\3CP = 15[/tex]

  • Divide both sides by 3

CP = 5

Find DP:

DP = CD + CP

DP = 3 + 5

DP = 8 units

Therefore, using the theorem of similar triangles and applying the knowledge of proportion, the length of segment DP in the diagram given is: 8 units.

Learn more here:

https://brainly.com/question/10269775

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