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Foot of a $ 10 \mathrm{~m} $ long ladder leaning against a vertical wall is $ 6 \mathrm{~m} $ away from the base of the wall. Find the height of the point on the wall where the top of the ladder reaches.
Given:
Foot of a \( 10 \mathrm{~m} \) long ladder leaning against a vertical wall is \( 6 \mathrm{~m} \) away from the base of the wall.
To do:
We have to find the height of the point on the wall where the top of the ladder reaches.
Solution:
Let $\mathrm{AB}$ be a vertical wall and $\mathrm{AC}$ be a ladder.
$AC=10\ m$
The top of the ladder reaches $\mathrm{A}$ and the distance of the ladder from the base of the wall $B C$ is $6 \mathrm{~m}$.
In right angled triangle $A B C$,
$A C^{2}=A B^{2}+B C^{2}$
$(10)^{2}=A B^{2}+(6)^{2}$
$100=A B^{2}+36$
$A B^{2}=100-36$
$AB^2=64$
$A B=\sqrt{64}$
$AB=8 \mathrm{~m}$
Hence, the height of the point on the wall where the top of the ladder reaches is $8 \mathrm{~m}$.
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