A 25m long ladder is placed against a vertical wall such that the foot of the ladder is 7m from the feet of the wall. If the top of the ladder slides down by 4m by how much distance will the foot of the ladder slide ?
Answer
610.8k+ views
Hint:Here we will find the height of the wall at which the ladder placed initially using the given length and horizontal distance. With the height we will find the height of the wall after sliding and with the help of the length of the ladder we will find the slided foot distance. Throughout the problem we will use the Pythagoras theorem.
Formula used: Pythagoras theorem
\[Hypotenus{e^2}\; = {\rm{ }}Perpendicula{r^2}\; + {\rm{ }}Bas{e^2}\;\]
Complete step-by-step answer:
Here, it is given that length of the ladder = 25 m
That is from the diagram we have \[AC = ED = 25\]
Also given that the distance between the wall and the foot of the ladder is 7m.
That is \[CB = {\rm{ }}7\]
Here it is also given that the ladder slides 4m in the wall.
That is \[AE = 4\]
Before sliding, the situation is represented by a right triangle with hypotenuse 25, and base 7,
Using Pythagoras theorem we have,
\[Hypotenus{e^2}\; = {\rm{ }}Perpendicula{r^2}\; + {\rm{ }}Bas{e^2}\;\]
We have to find the length of the wall, which is nothing but the perpendicular in the triangle.
\[{\rm{ }}Perpendicula{r^2}\; = Hypotenus{e^2} - {\rm{ }}Bas{e^2}\;\]
On comparing with the triangle we get,
\[A{B^2} = A{C^2} - C{B^2}\]
By substituting the values known we get,
\[A{B^2} = {25^2} - {7^2}\]
Let us square the terms in the right hand side we get,
\[A{B^2} = 625 - 49\]
By solving and taking square root on both the sides, we get
\[AB = \sqrt {576} \]
\[AB = 24{\rm{ }}m\]
Hence, the vertical arm that is the height of the wall is 24 m.
After sliding, the situation is represented by a right triangle with hypotenuse 25, and vertical arm has been slide 4 m then the vertical arm is of length \[24 - 4 = 20\]m,
Using Pythagoras theorem in this triangle, we get
\[Hypotenus{e^2}\; = {\rm{ }}Perpendicular{r^2}\; + {\rm{ }}Bas{e^2}\;\]
We have to find the distance between the wall and foot of the ladder after sliding, that is we have to find the base of the triangle.
\[{\rm{ }}Bas{e^2}\;\; = Hypotenuse{e^2} - {\rm{ }}Perpendicula{r^2}\]
On comparing with the triangle we get,
\[D{B^2} = D{E^2} - E{B^2}\]
By substituting the known values we get
\[D{B^2} = {25^2} - {20^2}\]
Let us square and solve right hand side of the equation, we get,
\[D{B^2} = 625 - 400 = 225\]
By taking square root on both the sides, we get
\[DB = \sqrt {225} \]
\[DB = 15{\rm{ }}m\]
So that the distance between foot of ladder and wall is \[15{\rm{ }}m\]
To find the distance changed in the ladder foot is \[15 - 7 = 8{\rm{ }}m\]
Hence, the distance from the foot of the ladder slide is 8 m
Additional Information:Pythagoras theorem states that “In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides“. The sides of this triangle have been named as Perpendicular, Base and Hypotenuse. Here, the hypotenuse is the longest side, as it is opposite to the angle \[{90^ \circ }\].
Note:We can come to a conclusion that when a ladder is sliding in the height then the distance on the foot of the ladder increases. While sliding in the height we should subtract the initial height and the slided height.
Formula used: Pythagoras theorem
\[Hypotenus{e^2}\; = {\rm{ }}Perpendicula{r^2}\; + {\rm{ }}Bas{e^2}\;\]
Complete step-by-step answer:
Here, it is given that length of the ladder = 25 m
That is from the diagram we have \[AC = ED = 25\]
Also given that the distance between the wall and the foot of the ladder is 7m.
That is \[CB = {\rm{ }}7\]
Here it is also given that the ladder slides 4m in the wall.
That is \[AE = 4\]
Before sliding, the situation is represented by a right triangle with hypotenuse 25, and base 7,
Using Pythagoras theorem we have,
\[Hypotenus{e^2}\; = {\rm{ }}Perpendicula{r^2}\; + {\rm{ }}Bas{e^2}\;\]
We have to find the length of the wall, which is nothing but the perpendicular in the triangle.
\[{\rm{ }}Perpendicula{r^2}\; = Hypotenus{e^2} - {\rm{ }}Bas{e^2}\;\]
On comparing with the triangle we get,
\[A{B^2} = A{C^2} - C{B^2}\]
By substituting the values known we get,
\[A{B^2} = {25^2} - {7^2}\]
Let us square the terms in the right hand side we get,
\[A{B^2} = 625 - 49\]
By solving and taking square root on both the sides, we get
\[AB = \sqrt {576} \]
\[AB = 24{\rm{ }}m\]
Hence, the vertical arm that is the height of the wall is 24 m.
After sliding, the situation is represented by a right triangle with hypotenuse 25, and vertical arm has been slide 4 m then the vertical arm is of length \[24 - 4 = 20\]m,
Using Pythagoras theorem in this triangle, we get
\[Hypotenus{e^2}\; = {\rm{ }}Perpendicular{r^2}\; + {\rm{ }}Bas{e^2}\;\]
We have to find the distance between the wall and foot of the ladder after sliding, that is we have to find the base of the triangle.
\[{\rm{ }}Bas{e^2}\;\; = Hypotenuse{e^2} - {\rm{ }}Perpendicula{r^2}\]
On comparing with the triangle we get,
\[D{B^2} = D{E^2} - E{B^2}\]
By substituting the known values we get
\[D{B^2} = {25^2} - {20^2}\]
Let us square and solve right hand side of the equation, we get,
\[D{B^2} = 625 - 400 = 225\]
By taking square root on both the sides, we get
\[DB = \sqrt {225} \]
\[DB = 15{\rm{ }}m\]
So that the distance between foot of ladder and wall is \[15{\rm{ }}m\]
To find the distance changed in the ladder foot is \[15 - 7 = 8{\rm{ }}m\]
Hence, the distance from the foot of the ladder slide is 8 m
Additional Information:Pythagoras theorem states that “In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides“. The sides of this triangle have been named as Perpendicular, Base and Hypotenuse. Here, the hypotenuse is the longest side, as it is opposite to the angle \[{90^ \circ }\].
Note:We can come to a conclusion that when a ladder is sliding in the height then the distance on the foot of the ladder increases. While sliding in the height we should subtract the initial height and the slided height.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 10 Social Science: Engaging Questions & Answers for Success

Master Class 10 Science: Engaging Questions & Answers for Success

Master Class 10 Maths: Engaging Questions & Answers for Success

Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Master Class 10 Computer Science: Engaging Questions & Answers for Success

Trending doubts
Explain the Treaty of Vienna of 1815 class 10 social science CBSE

In cricket, how many legal balls are there in a standard over?

Why is there a time difference of about 5 hours between class 10 social science CBSE

Who Won 36 Oscar Awards? Record Holder Revealed

What is the median of the first 10 natural numbers class 10 maths CBSE

What is deficiency disease class 10 biology CBSE

