Question

Two trees 9cm and 4m high stands upright on a ground are 12m apart, then the distance between their top is ​

Answer 1

let AC be a height of first tree which is 9cm.

also ED be a height of second tree which is 4cm.

CD be a base of both tree which is parallel to EB

So

length of AB=AC- BC

=9cm - 4cm (BC//ED)

= 5cm

in ∆ABE

by the phythagoras theorem

(AB)^2 +(EB)^2=(EA)^2

(5)^2 + +(12)^2 = (EA)^2

25+144 = (EA)^2

169 = (EA) ^2

EA = (169)^1/2

EA = 13cm

Answer 2

Answer:

let AC be a height of first tree which is 9cm.

also ED be a height of second tree which is 4cm.

CD be a base of both tree which is parallel to EB

So

length of AB=AC- BC

=9cm - 4cm (BC//ED)

= 5cm

in ∆ABE

by the phythagoras theorem

(AB)^2 +(EB)^2=(EA)^2

(5)^2 + +(12)^2 = (EA)^2

25+144 = (EA)^2

169 = (EA) ^2

EA = (169)^1/2

EA = 13cm

Step-by-step explanation:

let AC be a height of first tree which is 9cm.

also ED be a height of second tree which is 4cm.

CD be a base of both tree which is parallel to EB

So

length of AB=AC- BC

=9cm - 4cm (BC//ED)

= 5cm

in ∆ABE

by the phythagoras theorem

(AB)^2 +(EB)^2=(EA)^2

(5)^2 + +(12)^2 = (EA)^2

25+144 = (EA)^2

169 = (EA) ^2

EA = (169)^1/2

EA = 13cm