Two trees are standing parallel each other, the bigger tree 8m high casts a shadow of an
6m
1)if AB and CD are the two trees and AE is the shadow of the longer tree than
a) AEB CED b) ABE CED AEB DEC AEB d) BEA DEC
2) Since AB : CD so by BPT we have
a) AE/CE =BD/DE b) ACIAE-DE/BE C) AE/CE =AB/CD d) AE/CE =BE/DE
3) If the ratio of height of two trees is 3:1 then the shadow of the smaller tree is
a) 2m b) 6m c) 8/3m d)8m
4)The distance of point B from Eis
a)10mb) 8m c)18m d) 10/3 m
Answers
1)if AB and CD are the two trees and AE is the shadow of the longer tree than
a) AEB CED b) ABE CED AEB DEC AEB d) BEA DEC
2) Since AB : CD so by BPT we have
a) AE/CE =BD/DE b) ACIAE-DE/BE C) AE/CE =AB/CD d) AE/CE =BE/DE
3) If the ratio of height of two trees is 3:1 then the shadow of the smaller tree is
a) 2m b) 6m c) 8/3m d)8m
4)The distance of point B from Eis
a)10mb) 8m c)18m d) 10/3 m
1) If AB and CD are the two trees and AE is the shadow of the longer tree then, ∆AEB is similar to ∆CED and, ∆BEA is similar to ∆DEC .
2) When AB ll CD, by BPT we have AE/CE = BE/DE .
3) If the ratio of height of two trees is 3:1, then the shadow of the smaller tree is 2 m .
4) The distance of point B from E is equal to 10 m .
Given :- Two trees are standing parallel each other, the bigger tree 8m high casts a shadow of 6m .
To Find :-
1) if AB and CD are the two trees and AE is the shadow of the longer tree than which two triangles are similar ?
2) Since AB ll CD so by BPT we have ?
3) If the ratio of height of two trees is 3:1, then the shadow of the smaller tree is ?
4) The distance of point B from E is ?
Solution :-
(1)
Now, given that both trees are standing parallel to each other .
So, in ∆AEB and ∆CED we have,
→ ∠BAE = ∠DCE { Each 90° }
→ ∠AEB = ∠CED { common }
then,
→ ∆AEB ~ ∆CED { By AA similarity } (Ans.)
also, we can write them as,
→ ∆BEA ~ ∆DEC (Ans.)
(2)
since AB || CD. So by basic proportionality theorem (BPT) we have,
→ EC/CA = ED/DB
or,
→ CA/EC = DB/ED
adding 1 both sides,
→ (CA/EC) + 1 = (DB/ED) + 1
→ (CA + EC)/EC = (DB + ED)/ED
→ AE/EC = BE/ED
→ AE/CE = BE/DE (Ans.)
(3)
Now, we have given that,
→ Bigger tree height : Smaller tree height = 3 : 1
→ Bigger tree shadow = 6 m
So,
→ Bigger tree height/Smaller tree height = Bigger tree shadow/Smaller tree shadow
→ 3/1 = 6/Smaller tree shadow
→ Smaller tree shadow = 2 m (Ans.)
(4)
In right angled ∆BAE, given that,
→ BA = 8 m
→ AE = 6 m
So,
→ BE² = BA² + AE² { By pythagoras theorem }
→ BE² = 8² + 6²
→ BE² = 64 + 36
→ BE² = 100
→ BE² = 10²
square root both sides,
→ BE = 10 m (Ans.)
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