(D) Let $\overline{AC}$ represent the ladder and $\overline{AB}$ represent the height of the wall.In $\Delta ABC$,$\angle B = 90^{\circ}$.By the Pytha

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(D) Let $\overline{AC}$ represent the ladder and $\overline{AB}$ represent the height of the wall.In $\Delta ABC$,$\angle B = 90^{\circ}$.By the Pythagorean theorem,$AC^2 = AB^2 + BC^2$.Given $AC = 6.5 \, m$ and $AB = 6 \, m$.$6.5^2 = 6^2 + BC^2$$42.25 = 36 + BC^2$$BC^2 = 42.25 - 36 = 6.25$$BC = \sqrt{6.25} = 2.5 \, m$.Thus,the lower end of the ladder is $2.5 \, m$ away from the wall.

Class 10 Mathematics Triangles Mix Examples - Triangles

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$A$ $6.5 \, m$ long ladder leans on a wall to reach a height of $6 \, m$ on the wall. Then,the lower end of the ladder remains $\ldots \ldots \ldots \ldots \, m$ away from the wall.

A
$5$
B
$4$
C
$3.5$
D
$2.5$

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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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In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. Then,$\angle BAM \cong \ldots$

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In $\Delta PQR$,$M$ and $N$ are the midpoints of $\overline{PQ}$ and $\overline{PR}$ respectively. If the area of $\Delta PMN$ is $24$,find the area of $\Delta PQR$.

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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 $m$)

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In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $D$ is a point on $\overline{BC}$. Prove that $AD^{2} + BC^{2} = AC^{2} + BD^{2}$.

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$A$ $6.5 \, m$ long ladder leans on a wall to reach a height of $6 \, m$ on the wall. Then,the lower end of the ladder remains $\ldots \ldots \ldots \ldots \, m$ away from the wall.

Similar Questions

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

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. Then,$\angle BAM \cong \ldots$

In $\Delta PQR$,$M$ and $N$ are the midpoints of $\overline{PQ}$ and $\overline{PR}$ respectively. If the area of $\Delta PMN$ is $24$,find the area of $\Delta PQR$.

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 $m$)

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $D$ is a point on $\overline{BC}$. Prove that $AD^{2} + BC^{2} = AC^{2} + BD^{2}$.

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