Walls of two buildings on either side of a street are parallel to each other.A ladder 5.8m long is placed on the street such that its top just reaches the window of a building at the height of 4m.On turning the ladder over to the other side of the street,its top touches the window of the other building at a height 4.2m.Find the width of the street.
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Answers
Answer:
The width of street is 8.2 m.
Step-by-step explanation:
We are given that the two buildings are parallel to each other.
Since we are given that a ladder 5.8m long is placed on the street such that its top just reaches the window of a building at the height of 4m
So. AC = EC = 5.8 m
And AB = 4 m.
On turning the ladder over to the other side of the street its top touches the window of the other building at a height of 4.2m.
So, ED = 4.2 m
So, let BC = x and CD = y
We are required to calculate the width of street i.e. x+y
So, in ΔABC , use Pythagorean Theorem
Hypotenuse^{2} =Perpendicular^{2}+ Base^{2}Hypotenuse
2
=Perpendicular
2
+Base
2
AC^{2} =AB^{2}+ BC^{2}AC
2
=AB
2
+BC
2
5.8^{2} =4^{2}+ x^{2}5.8
2
=4
2
+x
2
33.64 =16+x^{2}33.64=16+x
2
33.64 -16 =x^{2}33.64−16=x
2
17.64 =x^{2}17.64 =x
2
\sqrt{17.64}=x
17.64
=x
4.2=x4.2=x
So, in ΔEDC , use Pythagorean Theorem
Hypotenuse^{2} =Perpendicular^{2}+ Base^{2}Hypotenuse
2
=Perpendicular
2
+Base
2
EC^{2} =ED^{2}+ DC^{2}EC
2
=ED
2
+DC
2
5.8^{2} =4.2^{2}+ y^{2}5.8
2
=4.2
2
+y
2
33.64 =17.64+y^{2}33.64=17.64+y
2
33.64 -17.64 =y^{2}33.64−17.64=y
2
16 =y^{2}16=y
2
\sqrt{16}=y
16
=y
4=y
So, the width of street = x+y = 4.2+4 =8.2 m
Hence the width of street is 8.2 m.
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