Question

Two poles of height 12m and 22 m stand on a plane ground. If the distance between their tops is 26
m, the distance between their feet is
A) 12metres
B) 22metres
C) 24metres
D) 26metres

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

A) 12 m
B) 9 m
C) 18 m
D) 21 m​

Answer 1

Question 1:

Two poles of height 12m and 22 m stand on a plane ground. If the distance between their tops is 26 m, the distance between their feet is ?

Solution 1:

First of all, refer to attached diagram 1:

AB || CD , and AB = DE ,therefore we can say:

CE = CD - DE

=> CE = CD - AB

=> CE = 22 - 12

=> CE = 10 m.

Now, applying Pythagora's Theorem in ∆ACE:

AC² = AE² + CE²

=> 26² = AE² + 10²

=> AE² = 676 - 100

=> AE² = 576

=> AE = √576

=> AE = 24 m.

Now , AE || BD and AE = BD = 24m

So, distance between the base of 2 buildings is 24 metres.

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Question 2:

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

Solution 2:

In ∆ ABC:

BC² = AB² + AC²

=> 15² = 12² + AC²

=> AC² = 225 - 144

=> AC² = 81

=> AC = 9 m.

In ∆ CED

CE² = DE² + CD²

=> 15² = 9² + CD²

=> CD² = 225 - 81

=> CD² = 144

=> CD = 12 m.

So, width of street = AD

=> width = AC + CD

=> width = 9 + 12

=> width = 21 m.

Hence, width of street is 21 metres.

$\star$ Hope It Helps.

Answer 2

Solution 1) :-

from image we have,

• AD = 12m .
• EC = 22m .
• AE = 26m .

in Right angle ∆EBA , we have,

→ EB = EC - BC

→ EB = 22 - 12

→ EB = 10m .

So, By pythagoras theorem,

→ AE² = EB² + AB²

→ 26² = 10² + AB²

→ AB² = 676 - 100

→ AB² = 576

→ AB = 24 cm. (Option C) (Ans.)

Hence, distance between their feet is 24 cm.

Solution 2) :-

from image we have,

• AB = 12m .
• AC = CE = 15m.
• ED = 9m.

So, By Pythagoras theorem in right ∆ABC,

→ BC = √[(15² - 12²]

→ BC = √(225 - 144)

→ BC = √(81)

→ BC = 9m.

similarly, By Pythagoras theorem in right ∆EDC,

→ CD = √[(15² - 9²]

→ CD = √(225 - 81)

→ CD = √(144)

→ CD = 12m.

Therefore,

→ BD = BC + CD

→ BD = 9 + 12

→ BD = 21 m. (Ans.) (Option D)

Hence, width of the street is 21m.

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