Question

A ladder 20 m long reaches a window which is 16 m 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.​

Answer 1

SOLUTION:

Given,

One side height of window= AC=16m

Other side height of window=XY=12m

Length of ladder= 20m

To find:

The width of the street.

Solution:

In ∆XCY,

∠C= 90°

=) ∆XCY is a right angle ∆

So, Using Pythagoras theorem:

$(hypotenuse) {}^{2} = {(height)}^{2} + {(base)}^{2} \\ = > ( {XC)}^{2} = ( {XY)}^{2} + ( {CY)}^{2} \\ \ = > ( {CY)}^{2} = ( {XC)}^{2} - {(XY)}^{2} \\ = > (CY) {}^{2} = {20}^{2} - {12}^{2} \\ = > ( {CY)}^{2} = 400 - 144 \\ = > CY = \sqrt{256} = 16m$

Therefore, CY= 16m

Similarly,

In ∆ABC,

$(AB) {}^{2} = ( {AC)}^{2} + ( {BC)}^{2} \\ = > ( {20)}^{2} = ( {16}^{2} + ( {bc)}^{2} \\ = > ( {BC)}^{2} = 400 - 256 \\ = > ( {BC)}^{2} = 144 \\ = > BC = \sqrt{144} = 12$

So, BC= 12m

Therefore, the width of the street,

=) BY= BC+ CY

=) BY= 12+ 16

=) BY= 28m

Hope it helps ☺️

Answer 2

Answer:

Step-by-step explanation:

Step-by-step explanation:

ladder ht. same in both case = 15 m

in Δ APC by Pythagoras theorem

 CA² + AP² = CP² ⇒ CA² = 225 - 144 = 81 ⇒ CA = 9 m

similarly in ΔBPD

DB² + BP² = DP² ⇒  BP² = 225-81 = 144 ⇒ BP = 12m

width of street AB = AP+BP = 12+9= 21m

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