the base of the ladder is leaning against a vertical wall 1.6 M away from the Wall if the ladder is 6.5 M long find how high the top of the ladder is above the ground
Answers
Answer:
In fig ΔDLW & ΔERW is a wall DL and RE are two position of ladder of lengths 5m.
Refer image.
In right angled ΔLWD,
DW
2
+LW
2
=DL
2
(By Pythagoras)
DW
2
=DL
2
−LW
2
⇒DW
2
=5
2
−4
2
=25−16=9
⇒DW=3
Now, RW=DW−DR
=3−1.6=1.4m
In right angled triangle RWE,
EW
2
+RW
2
=RE
2
(BY Pythagoras)
EW
2
=RE
2
−RW
2
=5
2
−1.4
2
=25−1.96
=23.04
EW=
23.04
=4.8m
∴ The distance by which the ladder shifted upward=
EL=EW−LW=4.8m−4m=0.8m.
Step-by-step explanation:
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Answer:
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Step-by-step explanation:
In fig ΔDLW & ΔERW is a wall DL and RE are two position of ladder of lengths 5m.
Refer image.
In right angled ΔLWD,
DW
2
+LW
2
=DL
2
(By Pythagoras)
DW
2
=DL
2
−LW
2
⇒DW
2
=5
2
−4
2
=25−16=9
⇒DW=3
Now, RW=DW−DR
=3−1.6=1.4m
In right angled triangle RWE,
EW
2
+RW
2
=RE
2
(BY Pythagoras)
EW
2
=RE
2
−RW
2
=5
2
−1.4
2
=25−1.96
=23.04
EW=
23.04
=4.8m
∴ The distance by which the ladder shifted upward=
EL=EW−LW=4.8m−4m=0.8m.