Question

A 5m long ladder reaches a window of height 4m on one side of the road; the ladder is then
turned over to the opposite side of the road and is found to reach another window of height
3m. Find the width of the road.​

Answer 1

See the figure:

Here AB=AD= 5m

In ∆ABC, By pythagorus theorem:

AB² = BC² + AC²

BC² = AB² - AC²

BC² = 25 - 16

BC = √9

BC = 3m

In ∆ACD, By pythagorus theorem:

CD² = ED² - EC²

CD² = 25 - 9

CD = √16

CD = 4m

Therefore the width of the road= BC + CD = 3m +4m = 7m

Answer 2

Answer:

In triangle ABC

1. AC is ladder(hupotenuse)
2. AB is window's height from ground
3. BC is distance between the road and ladder

By applying pythagorus

AC²= AB²+BC²

5²=4²+BC²

5²–4²=BC²

25–16=BC²

9=BC

3=BC

• So the distance between the window wall and ladder is 3m.

Now in triangle PQR,

1. PQ is ladder(hypotenuse)
2. PQ is window's height from ground
3. QR is distance between the road and ladder

By applying pythagorus

PR²=PQ²+QR²

5²=3²+QR²

5²–3²=QR²

25–9=QR²

16=QR²

4=QR

• So the distance between the road and the window wall is 4m.

• Total width of road is distance between both the walls which is

BC and QR

BC=3m

QR=4m

3+4=7m

• So,the width of the road is 7m