Question

Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between
their feet is 12 m, find the distance between their tops.
5. Find the length of the diagonal​

Answer 1

Working out:

In the above question, the heights of the building given are 6 m and 11 m. And, the distance between their feet is 12 m. So, let's draw the figure of this system and do some constructions and naming.

Let,

• Building with height 6 m be AB
• Building with height 11 m be CD.

Construction:

• Join A to CD and name the point as E.
• Join A and C.

Now here we can see that, AE is parallel and equal to BD (Distance between their feet) and ∆AEC is a right angled triangle being formed.

So,

• AE = 12 m
• CE = CD - DE = 11 m - 6 m = 5 m

By Pythagoras theoram,

• Perpendicular² + Base² = Hypotenuse²

In ∆AEC,

⇛ AE² + EC² = AC²

⇛ 12² + 5² = AC²

⇛ 144 + 25 = AC²

⇛ 169 = AC²

Flipping it,

⇛ AC = √169 m

⇛ AC = 13 m

And we can see in the figure that the AC is the distance between the top of the building. So, the distance between the tops is:

$\huge{ \boxed{ \sf{ \purple{13\: m}}}}$

And we are done !!

Answer 2

Given:

Height of first pole= 6m

Height of second pole = 11m

Distance between their feet = 12m

━━━━━━━━━━━━━━━

Need to find :

Distance between their tops.

━━━━━━━━━━━━━━━

Solution:

Let:

AB= 6m

BD= 12 m

CD = 11m

Clearly from the figure,

BD= AX as they are parallel to each other.

• $\purple{\boxed{\boxed{BD=AX=12 m}}}$

━━━━━━━━━━━━━━━

Also from the figure,

AB= XD, As they are parallel to each other.

• $\boxed{\pink{AB= XD=6m---ii}}$

━━━━━━━━━━━━━━━

Now clearly,

CD= CX+ XD

⟹ 11m = CX + 6m (putting value of XD from ii)

⟹ $\purple{\boxed{\boxed{CX= 5m}}}$

━━━━━━━━━━━━━━━

In the right angled triangle, AXC,

Using Pythagoras theorem,

${AC}^{2} = {CX}^{2} + {AX}^{2}$

$\implies \: {AC}^{2} = {12}^{2} + {5}^{2}$

$\implies \ {AC}^{2} = 169$

$\implies \: {AC} \: = \sqrt{169}$

$\red{\bold{ \implies \: AC = 13m}}$

Therefore the distance between their tops is $\huge{\red{13m}}$