Question

Two poles of 8m and 14m stand upright on a playground. If the distance between two tops is 10m, find the distance between their feet.​

Answer 1

Refer to the above diagram

Question

Two poles of 8 m and 14 m stand upright on a playground. If the distance between two tops is 10 m, find the distance between their feet .

Explanation

Consider (1 ) and (2) be two poles of height 14 m and 8 m respectively standing upright on the playground. Secondly, the distance between their tops is 10 m ( can be obtained by joining the tops of shown tower through dotted lines ). Now we have to calculate the distance between their feet i.e AB = ?

In below shown diagram we need to calculate the highlighted line AB which is the supposed distance between their feet.

Illustrated Diagram

$\boxed{(1) \sf14 m\begin{array}{c c} \sf A\rule{2pts}{70pts} \sf \pink{\rule{70pts}{2pts}}\rule{2pts}{50pts} \sf \:B\end{array}(2)8m}$

Solution

Let AD and BC be the two poles standing upright

Then , AD = 14 m & BC = 8m

Given that ,

the distance between two tops is 10 m

So, CD = 10 m

Let E be a point on pole AD such that

AE = BC

since , it forms a rectangle and thus

AE = 8 m

From Above Figure ,

DE = AD - AE = 14 - 8 = 6 m

We are supposed to find the distance between their feet i.e AB ( highlighted)

Now, DEC is a Right angled triangle .Thus , we will here apply the concept of pythagoras Theorem which states that In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides.

Therefore ,

$\sf(Hypotenuse)² = ( Perpendicular)² + (Base)²$

$: \implies \sf \: CD² = DE² + CE²$

$:\implies \sf{{(10)}^{2} = {(6)}^{2} + {CE}^{2} }$

$:\implies \sf{ {CE}^{2} = {(10)}^{2} - {(6)}^{2} }$

$\implies \sf{ {CE}^{2} = 100 - 36} \\\implies \sf{ {CE}^{2} = 64} \\ \implies \sf{CE = 8m}$

Now , from the diagram

AB = CE = 8 m

ANSWER

The distance between their feet = 8 m

• AB = 8 m

Thankyou