23.07.2020

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 parallelogram. Parallelogram GRPC with point B between C and P forming triangle GCB where GC equals 375 ft, CB equals 325 ft, and GB equals 425 ft, point E is outside parallelogram and segments BE and PE form triangle BPE where BP equals 225 ft. Part A: Identify a pair of similar triangles. (2 points) Part B: Explain how you know the triangles from Part A are similar. (4 points) Part C: Find the distance from B to E and from P to E. Show your work. (4 points)

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Step-by-step answer

17.02.2022, solved by verified expert
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A: CGB and BPE are similar triangles.

B: Their side lengths are proportionate

C: PE is The diagram below models the layout at a carnival, №15214265, 23.07.2020 03:31 ft long, and BE is The diagram below models the layout at a carnival, №15214265, 23.07.2020 03:31  ft long.

Step-by-step explanation:

We can tell that CGB and BPE are similar triangles because they are the only two triangles in this (lol) and we can see that they look similar. We can divide 325 / 255 to get the constant of proportionality.

The diagram below models the layout at a carnival, №15214265, 23.07.2020 03:31.

To find the side lengths of BE and PE, we divide CG and BG by The diagram below models the layout at a carnival, №15214265, 23.07.2020 03:31 to get their values.

So, we get that PE = The diagram below models the layout at a carnival, №15214265, 23.07.2020 03:31  and BE = The diagram below models the layout at a carnival, №15214265, 23.07.2020 03:31.

I'm not very sure about Part C however.

Hope this helped!

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Mathematics
Step-by-step answer
P Answered by Master

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


The diagram below models the layout at a carnival where G, R, P, C, B, and E are various locations o
Mathematics
Step-by-step answer
P Answered by Specialist

 A) ΔGCB ≅ ΔERG

              B) Angle-Angle (AA) Similarity Theorem

               C)  BE = 266 2/3         PE = 233 1/3

Step-by-step explanation:

GC || ER  therefore, by Alternate Interior Angles Theorem: ∠G ≡ ∠E

GR || CE  therefore, by Alternate Interior Angles Theorem: ∠C ≡ ∠R

Since we have two congruent angles, we can prove ΔGCB and ΔERG are congruent using the Angle-Angle (AA) Similarity Theorem.

Since it is a parallelogram, the sides are congruent (GC = RP and GR = CP)

GC = 350                  GB = 400                    CB = 300

ER = 350 + PE           EG = 400 + BE           RG = 300 + 200

Since the triangles are similar, their sides are proportional.

\dfrac{GC}{ER}=\dfrac{GB}{EG}=\dfrac{CB}{RG}

\dfrac{GC}{ER}=\dfrac{CB}{RG} \quad \implies \quad \dfrac{350}{350+PE}=\dfrac{300}{300+200}\\\\\\.\qquad \qquad \qquad \qquad \qquad 350(500)=300(350+PE)\\\\\\.\qquad \qquad \qquad \qquad \qquad 175,000=105,000+300PE\\\\\\.\qquad \qquad \qquad \qquad \qquad 70,000=300PE\\\\\\.\qquad \qquad \qquad \qquad \qquad \dfrac{70,000}{300}=PE\\\\\\.\qquad \qquad \qquad \qquad \qquad \dfrac{700}{3}=PE\\\\\\.\qquad \qquad \qquad \qquad \qquad \large\boxed{233\dfrac{1}{3} =PE}

\dfrac{GB}{EG}=\dfrac{CB}{RG} \quad \implies \quad \dfrac{400}{400+BE}=\dfrac{300}{300+200}\\\\\\.\qquad \qquad \qquad \qquad \qquad 400(500)=300(400+BE)\\\\\\.\qquad \qquad \qquad \qquad \qquad 200,000=120,000+300BE\\\\\\.\qquad \qquad \qquad \qquad \qquad 80,000=300BE\\\\\\.\qquad \qquad \qquad \qquad \qquad \dfrac{80,000}{300}=BE\\\\\\.\qquad \qquad \qquad \qquad \qquad \dfrac{800}{3}=BE\\\\\\.\qquad \qquad \qquad \qquad \qquad \large\boxed{266 \dfrac{2}{3}=BE}


The diagram below models the layout at a carnival where G, R, P, C, B, and E are various locations o
Mathematics
Step-by-step answer
P Answered by Specialist

\Delta CGB\sim \Delta  PEB By AA similarity criterion)

PE=259.6\,\,feet\,,\,EB=294.2\,\,feet

Step-by-step explanation:

Given: GRPC is a parallelogram.

GC = 375 feet, BC = 325 feet, BP = 225 feet, BG = 425 feet

To find: a pair of similar triangles. criterion of similarity and BE, PE

Solution:

Opposite sides of a parallelogram are parallel.

As GR||PC, \angle CGE=\angle PEB\\

(if lines are parallel then alternate interior angles are equal)

As GC||RE, \angle GCE=\angle EPB

(if lines are parallel then alternate interior angles are equal)

So, \Delta CGB\sim \Delta  PEB (by AA similarity criterion)

\frac{CG}{PE}=\frac{GB}{EB}=\frac{CB}{PB}\\\\\frac{375}{PE}=\frac{425}{EB}=\frac{325}{225}

(if two triangles are similar then their sides are proportional)

\frac{375}{PE}=\frac{325}{225}\,,\,\frac{425}{EB}=\frac{325}{225}\\PE=\frac{375\times 225}{325}\,,\,EB=\frac{425\times 225}{325}\\PE=259.6\,\,feet\,,\,EB=294.2\,\,feet

Mathematics
Step-by-step answer
P Answered by Specialist

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


The diagram below models the layout at a carnival where G, R, P, C, B, and E are various locations o
Mathematics
Step-by-step answer
P Answered by Master

Down below

Sorry if it's wrong...

Step-by-step explanation:

A and B. The similar triangles must be GBC and BPE because they are the only two.

You can tell that GBC = PBE because of vertical angles.

Then, you can see that GCP is an angle on the parallelogram.

GCP is also an angle in GBC

GCP + RPC = 180

RPC + BPE also = 180

GCP = BPE

GBC ≈ BPE by AA similarity.

C. So since those two triangles are congruent, you can find which sides are similar. 225 : 325 is the ratio.

Simplify: 9 : 13

Apply to the rest:

9 : 13 = x : 425

BE = \frac{3825}{13}

9 : 13 = x : 375

PE = \frac{3375}{13}

Mathematics
Step-by-step answer
P Answered by PhD
Answer: 440 grams for 1.54 is the better value
Explanation:
Take the price and divide by the number of grams
1.54 / 440 =0.0035 per gram
1.26 / 340 =0.003705882 per gram
0.0035 per gram < 0.003705882 per gram

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