A ladder 10, m long reaches a window 8, m abo | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $OA$ be the wall and $AB$ be the ladder,where $OA = 8\, m$ and $AB = 10\, m$.Since the wall is perpendicular to the ground,$	riangle AOB$ is

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) Let $OA$ be the wall and $AB$ be the ladder,where $OA = 8\, m$ and $AB = 10\, m$.Since the wall is perpendicular to the ground,$	riangle AOB$ is a right-angled triangle with $\angle AOB = 90^\circ$.By Pythagoras theorem,$AB^2 = OA^2 + OB^2$.Substituting the values: $(10)^2 = (8)^2 + OB^2$.$100 = 64 + OB^2$.$OB^2 = 100 - 64 = 36$.$OB = \sqrt{36} = 6\, m$.Therefore,the distance of the foot of the ladder from the base of the wall is $6\, m$.

$A$ ladder $10\, m$ long reaches a window $8\, m$ above the ground. Find the distance of the foot of the ladder from the base of the wall. (in $, m$)

Similar Questions

Observe the figure and find $\angle P$.

In the figure,altitudes $AD$ and $CE$ of $\Delta ABC$ intersect each other at the point $P$. Show that $\Delta AEP \sim \Delta CDP$.

In an equilateral triangle $ABC$,$D$ is a point on side $BC$ such that $BD = \frac{1}{3} BC$. Prove that $9 AD^{2} = 7 AB^{2}$.

In the figure,$ABD$ is a triangle right-angled at $A$ and $AC \perp BD$. Show that $AD^{2} = BD \cdot CD$.

In the figure,$\frac{PS}{SQ} = \frac{PT}{TR}$ and $\angle PST = \angle PRQ$. Prove that $\triangle PQR$ is an isosceles triangle.

Vedclass Test Series

Exam Paper Generator

Online Exam Module