Question

A road is 8m wide. Two poles measuring 10m and 16m are fixed on two sides across the road. Find distance between their upper ends $(AC)$.

Answer 1

Hint : In above question we need to find the length of AC.
For that we will assume this figure. As a combination of two figures. Below one is a rectangle with sides ABDE and above one is the right triangle AEC.
We have all the dimensions of the rectangle, we will find dimensions of triangles.


Formulas/Properties used
1. Pythagorean Theorem -
According to this theorem ${(Hypotenuse)^2} = $ sum of square of legs
${(AB)^2} = {(AC)^2} + {(BC)^2}$
2. Property of rectangle – Opposite sides of rectangle are equal and parallel to each other. And its vertices make an angle of $90^\circ $.


Complete step by step solution :
Let’s divide the given figure in two parts.

Two parts of given figure one is $\Delta ACE$ and other one is rectangle ABDE
Here in $\Delta ACE$

$EA = 8m$
$CE = CD = ED$
$ = (16 - 10)m$
$CE = 6m$
In rectangle ABDE

If $AB = 10m$ given
$ED = 10m$ (opposite sides of rectangle are equal)
$DB = 8m$ (given)
$EA = 8m$ (opposite sides of rectangle)
$\Delta ACE$ is a right angled triangle, so by using Pythagora's theorem.
${(Hypotenuse)^2} = {(Base)^2} + {(Perpendicular)^2}$
${(AC)^2} = {(CE)^2} + {(AE)^2}$
${(AC)^2} = {(6)^2} + {(8)^2}$
${(AC)^2} = 36 + 64$
${(AC)^2} = 100$
$AC = \sqrt {100} $
$AC = 10m$
Therefore distance between upper ends of given figure is $AC = 10m$

Note: While using Pythagoras theorem do not make the mistake of adding the values of both the legs first and then squaring the result. It is incorrect .