Question

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

• A. 25 cm
• B. 26 cm
• C. 24 cm
• D. 27 cm

Answer 1

The correct option is B

26 cm

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