Question

A ladder reaches a window which is 12m above the ground on one side of the street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window 9m high. If the length of the ladder is 15m, find the width of the street.​

Answer 1

✬ Street = 21 m ✬

Step-by-step explanation:

Given:

• Height of windows are 12 m and 9 m above the ground respectively.
• Length of ladder is 15 m.

To Find:

• Width of street ?

Solution: Let AB be the first window and ED be the second. The distance between the foots of these two windows is DB. Foot of ladder on both the sides is common i.e C.

Here we have

• AB = 12 m
• ED = 9 m
• AC = EC = Ladder = 15 m

Now in right angled triangle ∆ABC

• AB {perpendicular}
• BC {base}
• AC {hypotenuse}

By using Pythagoras theorem

★ H² = Perpendicular² + Base² ★

$\implies{\rm }$ 15² = 12² + BC²

$\implies{\rm }$ 225 = 144 + BC²

$\implies{\rm }$ √225 – 144 = BC

$\implies{\rm }$ √81 = BC

$\implies{\rm }$ 9 = BC

Again similarly in right angled ∆EDC

• EC {hypotenuse}
• DC {base}
• ED {perpendicular}

$\implies{\rm }$ 15² = 9² + DC²

$\implies{\rm }$ 225 = 81 = DC²

$\implies{\rm }$ √225 – 81 = DC²

$\implies{\rm }$ √144 = DC

$\implies{\rm }$ 12 = DC

• Street is DB = DC + BC
• DB = 12 + 9 = 21 m

Hence, width of street is 21 m.

Answer 2

Answer:

Solution:–

$\rm \: →In \: ∆ADC, \\ \rm→AD²+AC²= CD² \\ \rm→9²+AC²=15² \\ \rm→AC²=225-81 \\ \rm→AC=12m \\ \rm→In \: ∆BEC, \\ \rm→EC²=BC²+BE² \\ \rm→15²=12²+BC² \\ \rm→225-144=BC² \\ \rm→BC= \: 9m \\ \rm \: Width \: of \: the \: road = \: AC+BC=12+9=21m$