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Yakvenalex [24]
4 years ago
15

Try and solve this math problem

Mathematics
1 answer:
weqwewe [10]4 years ago
8 0

Side CE is already known to be congruent to itself. Angle CBE is marked as congruent to angle CDE, so the missing part of the proof involves one of the other angles in the triangles. It is either

... ∠CEB ≅∠CED

or

... ∠DCE ≅∠BCE

Of these choices, only the first one appears in your answer list. The appropriate choice is ...

... ∠CEB ≅∠CED

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The diagram below models the layout at a carnival where G, R, P, C, B, and E are various locations on the grounds. GRPC is a par
mylen [45]

Answer:

The similar triangles are \triangle BPE and \triangle BCG

BE = 321 and PE =286

Step-by-step explanation:

Given

See attachment for the required figure

Solving (a): The similar triangles

The similar triangles are \triangle BPE and \triangle BCG

Solving (b): Why they are similar

Both triangles are similar because \triangle BCG is dilated (i.e. enlarged) and then reflected to give \triangle BPE.

Solving (c): Calculate BE and PE

The following are equivalent ratios

BP: BC = BE : BG

and

BP: BC = PE : CG

Solving for BE, we have:

BP: BC = BE : BG

Substitute the known values

250:350 = BE:450

Express as fraction

\frac{250}{350} = \frac{BE}{450}

Multiply both sides by 450

450 * \frac{250}{350} = BE

321 = BE

BE = 321 -- approximated

Solving for PE, we have:

BP: BC = PE : CG

Substitute known values

250:350 = PE:400

Express as fraction

\frac{250}{350} = \frac{PE}{400}

Multiply both sides by 400

400 * \frac{250}{350} = PE

286 = PE

PE =286

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