A22m ladder and a 20m ladder were leaned against a building. the bottom of the longer ladder was 4m farther from the building than the bottom of the shorter ladder, but both ladders reached the same distance up the building. find this distance to the nearest tenth.

12.2m

15.3m

18.1m

19.2m

Both ladder reaches 18.1 m up the building ⇒ 3rd answer

Step-by-step explanation:

* Lets study the information to solve the problem

- There are two ladders

- The lengths of them are 22 m and 20 m

- The bottom of the longer was 4 m farther than the bottom of the

  shorter from the building

- Both of them reached the same distance up the building

* Lets solve the problem

- Let the distance between the bottom of the shorter ladder to the

 building is x

∵ The bottom of the longer ladder is farther by 4

∴ The distance between the bottom of the longer ladder and the

   building is x + 4

- Let the ladders reached the distance h up the building

* Now we have two right triangles

# Their hypotenuses are 22 and 20

# Their heights are h

# Their bases are x + 4 , x

- Lets find h in each triangle using the rule of Pythagoras

∵ (hypotenuse)² = (leg 1)² + (leg 2)²

# The longer ladder

∵ hypotenuse = 22

∵ leg 1 = x + 4

∵ leg 2 = h

∴ (22)² = (x + 4)² + h² ⇒ simplify

∴ 484 = (x + 4)² + h² ⇒ subtract (x + 4)² from both sides

∴ h² = 484 - (x + 4)² ⇒ (1)

# The shorter ladder

∵ hypotenuse = 20

∵ leg 1 = x

∵ leg 2 = h

∴ (20)² = (x )² + h² ⇒ simplify

∴ 400 = x² + h² ⇒ subtract x² from both sides

∴ h² = 400 - x² ⇒ (2)

- Equate (1) , (2) to find x

∴ 484 - (x + 4)² = 400 - x² ⇒ Add (x + 4)² and subtract 400 in both sides

∴ 84 = (x + 4)² - x² ⇒ open the bracket

∴ 84 = x² + 2(4)(x) + 4² - x² ⇒ simplify

∴ 84 = 8x + 14 ⇒ subtract 16 from both sides

∴ 68 = 8x ⇒ divide both sides by 8

∴ x = 8.5

- Substitute this value of x in (1) or (2) to find h

∵ h² = 400 - x²

∴ h² = 400 - (8.5)² = 327.75 ⇒ take √ for both sides

∴ h = 18.1

* Both ladder reaches 18.1 m up the building

6/20

step-by-step explanation:

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