Question

If the top of a 4.50 m ladder reaches twice as far up a vertical wall as the foot of the

ladder is out from the base of the wall, how far up the wall does the ladder reach?
Draw a diagram. Give your answer to two decimal places

Answer 1

The ladder reaches 4.02m up the wall.

Explanation:

Check the attachment for the diagram.

A right-angled triangle with is formed.

Hypotenuse is |AC|

Opposite sides are |AB| and |BC|.

Where x is the distance of the foot of the ladder from the wall.

We are required to find the distance |AB|

Applying Pythagora's Rule,

|AC|² = |AB|² + |BC|²

(4.5)² + (2x)² + x²

20.25 = 4x² + x²

20.25 = 5x²

Divide both sides by 5

x² = 20.25/5

x² = 4.05

Taking square roots of both sides

x = ±√4.05

We are only interested in the positive part, as we deal with distance.

= √4.05

x ≈ 2.01 to two decimal places.

The distance |AB| = 2x = 2×2.01 = 4.02m

Answer 2

4.02 meters.

Explanation:

In the diagram, the length of the ladder is |AC|.

If the foot of the ladder is x meters from the base of the ladder

Then the distance of the ladder up the wall, AB=2x meters.

Using Pythagoras Theorem

`|AC|^2=|AB|^2+|BC|^2\\4.5^2=(2x)^2+x^2\\4.5^2=4x^2+x^2\\20.25=5x^2\\$Divide both sides by 5\\x^2=20.25/ 5\\x^2=4.05\\x=√(4.05)=2.01 feet`

Therefore, the distance of the ladder up the wall,

AB=2 X 2.01 =4.02 meters.