Question

A.How far is the library from the park?
B.How far is the park from the football field?

Answer 1

A. The distance between the library and the park is equal to
`2√(10)` miles.

B. The distance between the park and the football field is
`2√(35)` miles.

Therefore, the correct answer option is D.
`2√(10)` miles and
`2√(35)` miles.

According to the geometric mean (leg) theorem, the length of the leg of a right-angled triangle is the geometric mean between its hypotenuse and the segment of the hypotenuse which is adjacent to that leg;

Hypotenuse/leg = leg/part or altitude/left = right/altitude

where;

left = 10 miles

right = 4 miles

Part A.

By applying geometric mean (leg) theorem, the distance between the library and the park (x) can be calculated as follows;

`x^(2) =10 * 4\\\\x=√(40) \\\\x=2√(10) \;miles.`

By applying geometric mean (leg) theorem, the distance between the park and the football field (y) can be calculated as follows;

(10 + 4)/y = y/10

14/y = y/10

`y^2=√(140) \\\\y=2√(35)\;miles`

Answer 2

a. The distance from the library to the park is 2√5 miles

b. The distance from the park to the football field is 2√35 miles

The last answer ⇒ 2√5 miles , 2√35 miles

Explanation:

* Lets revise the rules in the right angle triangle when we draw the

perpendicular from the right angle to the hypotenuse

- In triangle ABC

# Angle B is a right angle

# The hypotenuse is AC

# BD ⊥ AC

∴ (AB)² = AD × AC

∴ (BC)² = CD × AC

∴ (BD)² = AD × CD

∴ BD × AC = AB × BC

* Lets use one of these rules to solve the problem

- Vertex B represents the position of the park

- Vertex A represents the position of the home

- Vertex C represents the position of the football field

- Vertex D represents the position of the library

- AB represents the distance from the home to the park

- CB represents the distance from the football field to the park

- DB represents the distance from the library to the park

- AC represents the distance from the home to the football field

- AD represents the distance from the home to the library

- CD represents the distance from the football field to the library

* Now lets solve the problem

a.

∵ The distance from the home to the library is 4 miles

∴ AD = 4

∵ The distance from the football field to the library is 10 miles

∴ CD = 10

∵ The distance from the library to the park represented by BD

- Lets use the rule (BD)² = AD × CD

∴ (BD)² = 4 × 10 = 40 ⇒ take √ for both sides

∴ BD = √40 = 2√5 miles

* The distance from the library to the park is 2√5 miles

b.

∵ The distance from the park to the football field represented by BC

* Lets use the rule (BC)² = CD × AC

∵ CD = 10

∵ AC = 4 + 10 = 14

∴ (BC)² = 10 × 14 = 140 ⇒ take √ for both sides

∴ BC = √140 = 2√35 miles

* The distance from the park to the football field is 2√35 miles