Question

The height of two buildings is 50 cm and 40 cm respectively. If the distance between their base is 24 cm, then find the distance between their top.

Answer 1

Let the two buildings be AB and CD

height of the first building AB =40 cm

Height of the second building CD =50 cm

Distance between their foot = BD =24 cm

Draw AE || BD touching CD at E

Now CE = CD - DE

CE =50 - 40 =10 (DE = AB)

In $\Delta$ ACE, applying pythagoras theorem we get.

$A{C}^2=A{E}^2+C{E}^2$

$A{C}^2=(24{)}^2+(10{)}^2$

$A{C}^2=576+100=676$

AC =26 cm

Hence, the distance between the tops of the buildings i.e AC =26 cm