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jolli1 [7]
3 years ago
5

Point F is on line segment \overline{EG}

Mathematics
1 answer:
Pachacha [2.7K]3 years ago
8 0

Answer:

4 Units

Step-by-step explanation:

EF=x

FG=3x-5

EG=5x−8

Since F is on line segment \overline{EG}

EG=EF+FG

5x−8=x+3x-5

5x-8=4x-5

Collect like terms

5x-4x=-5+8

x=3 Units

Therefore:

\overline{FG}=3x-5\\=3(3)-5\\=9-5\\$=4 Units

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Answer:

Δ QRS ≈ Δ QST ≈ Δ SRT ⇒ 3rd answer

Step-by-step explanation:

From the given figure

In Δ QRS

∵ m∠S = 90°

∵ m∠S = m∠QST + m∠RST

∴ m∠QST + m∠RST = 90° ⇒ (1)

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∴ m∠Q + m∠R = 90° ⇒ (2)

In Δ QST

∵ m∠QTS = 90°

- By using the fact above

∴ m∠Q + m∠QST = 90 ⇒ (3)

- From (1) and (3)

∴ m∠QST + m∠RST = m∠Q + m∠QST

- Subtract m∠QST from both sides

∴ m∠RST = m∠Q

In Δ SRT

∵ m∠STR = 90°

- By using the fact above

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- From (1) and (4)

∴ m∠QST + m∠RST = m∠R + m∠RST

- Subtract m∠RST from both sides

∴ m∠QST = m∠R

In Δs QRS and QST

∵ m∠S = m∠QTS ⇒ right angles

∵ m∠R = m∠QST ⇒ proved

∵ ∠Q is a common angle in the two Δs

∴ Δ QRS ≈ Δ QST ⇒ AAA postulate of similarity

In Δs QRS and SRT

∵ m∠S = m∠STR ⇒ right angles

∵ m∠Q = m∠RST ⇒ proved

∵ ∠R is a common angle in the two Δs

∴ Δ QRS ≈ Δ SRT ⇒ AAA postulate of similarity

If two triangles are similar to one triangle, then the 3 triangles are similar

∵ Δ QRS ≈ Δ QST

∵ Δ QRS ≈ Δ SRT

∴ Δ QRS ≈ Δ QST ≈ Δ SRT

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The area of the circle is 706.5 ft².

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