Question

Two buildings I and II are of heights 19m and 40 m respectively 20 m apart. the distance between their tops is:​

Answer 1

Answer:

From the diagram, PB=CD=12 m

AP=AB−PB=(22−12) m=10 m

Also CP=BD=24  m

Since △APC is a right angled triangle,

AC2=AP2+PC2

AC2=102+242

AC2=676

∴AC=26  m

Hence, distance between their tops is 26 m.

Answer 2

Answer:

the distance between their tops is $29$ m

Step-by-step explanation:

Given:

Two buildings I and II are of heights $19$m and $40$ m respectively $20$ m apart.

We need to find the distance between their tops

the remaining from the first building taken vertically will be

$40-19=21$

By assuming a right-angled triangle at the top we get the base as $20$

So the hypotenuse will be the distance between the tops

So we get as

$d=\sqrt{(20)^2+(21)^2} \\\\d=\sqrt{400+441} \\\\d=\sqrt{841} \\\\d=29$