$A$ guy wire attached to a vertical pole of height $18\, m$ is $24\, m$ long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) Let $OB$ be the pole and $AB$ be the wire. The pole is vertical,so $\triangle AOB$ is a right-angled triangle with $\angle AOB = 90^{\circ}$.
By Pythagoras theorem,
$AB^{2} = OB^{2} + OA^{2}$
$(24\, m)^{2} = (18\, m)^{2} + OA^{2}$
$576\, m^{2} = 324\, m^{2} + OA^{2}$
$OA^{2} = (576 - 324)\, m^{2} = 252\, m^{2}$
$OA = \sqrt{252}\, m = \sqrt{36 \times 7}\, m = 6\sqrt{7}\, m$
Therefore,the distance from the base is $6\sqrt{7}\, m$ (approximately $15.87\, m$).

Explore More

Similar Questions

$A$ 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 $, m$)

In the figure,$\frac{QR}{QS} = \frac{QT}{PR}$ and $\angle 1 = \angle 2$. Show that $\Delta PQS \sim \Delta TQR$.

$A$ vertical pole of length $6\, m$ casts a shadow $4\, m$ long on the ground and at the same time a tower casts a shadow $28\, m$ long. Find the height of the tower. (in $m$)

In the figure,two chords $AB$ and $CD$ intersect each other at the point $P$. Prove that $AP \cdot PB = CP \cdot DP$.

$ABC$ and $BDE$ are two equilateral triangles such that $D$ is the mid-point of $BC$. The ratio of the areas of triangles $ABC$ and $BDE$ is

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo