Question

The heights of two buildings are 34 m and 25 m

respectively. If the distance between their feet is 12 m, find the

distance between their tops.
​

Answer 1

Working out:

In the above question, the heights of the building given are 25 m and 34 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 25 m be AB
• Building with height 34 m be CD.

Cosntruction:

• 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 = 34 m - 25 m = 9 m

By Pythagoras theoram,

• Perpendicular² + Base² = Hypotenuse²

In ∆AEC,

⇛ AE² + EC² = AC²

⇛ 12² + 9² = AC²

⇛ 144 + 81 = AC²

⇛ 225 = AC²

Flipping it,

⇛ AC = √225 m

⇛ AC = 15 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:

$\huge{ \boxed{ \sf{ \purple{15 \: m}}}}$

And we are done !!