(C) Let $AB$ be the vertical wall and $AC = 10\,m$ be the length of the ladder.The distance of the foot of the ladder from the base of the wall is $BC

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(C) Let $AB$ be the vertical wall and $AC = 10\,m$ be the length of the ladder.The distance of the foot of the ladder from the base of the wall is $BC = 6\,m$.In the right-angled $\triangle ABC$,by the Pythagoras theorem:$AC^2 = AB^2 + BC^2$$(10)^2 = AB^2 + (6)^2$$100 = AB^2 + 36$$AB^2 = 100 - 36 = 64$$AB = \sqrt{64} = 8\,m$.Thus,the height of the point on the wall where the top of the ladder reaches is $8\,m$.

Class 10 Mathematics Triangles Mix Examples - Triangles

Medium

Foot of a $10\,m$ long ladder leaning against a vertical wall is $6\,m$ away from the base of the wall. Find the height of the point on the wall where the top of the ladder reaches. (in $m$)

A
$2$
B
$6$
C
$8$
D
$10$

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Mix Examples - Triangles

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In $\Delta ABC$,$AB + BC = 23$,$BC + AC = 32$,and $AB + AC = 25$. Show that $\Delta ABC$ is a right-angled triangle.

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In $\Delta ABC$,$\overline{AC}$ is the hypotenuse and $\overline{BE}$ is a median. Prove that $AB^{2} + BC^{2} + AC^{2} = 8AE^{2}$.

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Foot of a $10\,m$ long ladder leaning against a vertical wall is $6\,m$ away from the base of the wall. Find the height of the point on the wall where the top of the ladder reaches. (in $m$)

Similar Questions

If the measures of the sides of a triangle are $\ldots \ldots \ldots,$ then it is not a right-angled triangle.

In $\Delta XYZ$,$X-S-Y$,$X-T-Z$ and $\overline{ST} \parallel \overline{YZ}$. If $XS = 4$,$XT = 8$,$SY = x-4$ and $TZ = 3x-19$,find the value of $x$.

In $\square ABCD$,$\overline{AB} \parallel \overline{CD}$ and $\overline{AC} \cap \overline{BD} = \{M\}$. If $MA = 3x - 19$,$MB = x - 3$,$MC = x - 5$,and $MD = 3$,find the value of $x$.

In $\Delta ABC$,$AB + BC = 23$,$BC + AC = 32$,and $AB + AC = 25$. Show that $\Delta ABC$ is a right-angled triangle.

In $\Delta ABC$,$\overline{AC}$ is the hypotenuse and $\overline{BE}$ is a median. Prove that $AB^{2} + BC^{2} + AC^{2} = 8AE^{2}$.

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