# Ex.6.5 Q12 Triangles Solution - NCERT Maths Class 10

## Question

Two poles of heights \(6\;\rm{}m\) and \(11\;\rm{}m\) stand on plane ground. If the distance between the feet of the poles is \(12\;\rm{}m\), find the distance between their tops.

**Diagram**

## Text Solution

**Reasoning:**

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

**Steps:**

\(AB\) is the height of one pole \(= 6\;\rm{}m\)

\(CD\) is the height of another pole \(= 11\,\rm{}m\)

\(AC\) is the distance between two poles at bottom \(= 12\rm{}\,m\)

\(BD\) is the distance between the tops of the poles \(=\,?\)

Draw \(B E \| A C\)

Now consider,

In \(\Delta B E D\)

\[\begin{align}\angle B E D&=90^{\circ} \\ {BE}={AC}&=12 \mathrm{m}\\ {DE}&={CD}-\mathrm{CE} \\ {DE}&=11-6=5 \mathrm{cm}\\\end{align}\]

Now,

\[\begin{align} B D^{2} &=B E^{2}+D E^{2}\\ & [ \text {Pythagoras } \text{ theorem} ] \\\\ &=12^{2}+5^{2} \\ &=144+25 \\ B D^{2} &=169 \\ B D &=13 \mathrm{m} \end{align}\]

The distance between the tops of poles \(=13\rm{}\,m\)