# Ex.6.5 Q11 Triangles Solution - NCERT Maths Class 10

## Question

An aeroplane leaves an airport and flies due north at a speed of \(1000\; \rm{}km\) per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of \(1200\; \rm{}km\) per hour. How far apart will be the two planes after \(\,1\frac{1}{2}\) hours?

**Diagram**

## Text Solution

**Reasoning:**

We have to find the distance travelled by aeroplanes, we need to use

\(\text{distance}\,\,\text{=}\,\,\text{speed}\,\,\text{ }\!\!\times\!\!\text{ }\,\,\text{time}\)

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 distance travelled by aeroplanes travelling towards north

\[\begin{align} A B &=1000\, \mathrm{km} / \mathrm{hr} \times 1 \frac{1}{2} \mathrm{hr} \\ &=1000 \times \frac{3}{2} \mathrm{km} \\ {AB} &=1500\, \mathrm{km} \end{align}\]

\(BC\) is the distance travelled by another aeroplane travelling towards south

\[\begin{align} BC&=1200\,\,\text{km/hr}\times 1\frac{1}{2}\,\,\text{hr} \\ \,\,\,\,\,\,\,\,\,\,&=1200\times \frac{3}{2}\text{hr} \\ BC&=1800\,\,\text{km} \\ \end{align}\]

Now, In \(\Delta ABC\,\,,\,\,\angle ABC={{90}^{0}}\)

\[\begin{align} AC{}^{2}&=A{{B}^{2}}+B{{C}^{2}}\,\,\left( \text{Pythagoras theorem} \right) \\ & ={{(1500)}^{2}}+{{(1800)}^{2}} \\ & =2250000+3240000 \\ A{{C}^{2}}&=5490000 \\ AC&=\sqrt{549000} \\ \,\,\,\,\,\,\,\,\,\,\,&=300\sqrt{61}\,\,\text{km} \end{align}\]

The distance between two planes after \(1\frac{1}{2}\text{hr}=300\sqrt{61}\,\,\text{km}\)