Question 1 :
Check if the triangle given below is similar.
Solution :
Let us consider the triangles, AED and ACB
If two triangles are similar, then the ratio of its corresponding sides will be equal.
Condition :
AE/AC = AD/AB
2/(7/2) ≠ 3/5
4/7 ≠ 3/5
So, the triangles AED and ACB are not similar.
Question 2 :
Find the value of x in the picture given below.
In triangle PQC,
<PQC = 180 - 110
<PQC = 70
Now let us consider the triangles ABC and PQC.
<ABC = <PQC
<ACB = <PCQ
By using AA criterion, the above triangles are similar. Hence the ratio of their corresponding sides will be equal.
AB/PQ = BC/QC
5/x = (3+3)/3
5/x = 6/3
5/x = 2
x = 5/2 = 2.5
Question 3 :
A girl looks the reflection of the top of the lamp post on the mirror which is 6.6 m away from the foot of the lamppost. The girl whose height is 1.25 m is standing 2.5 m away from the mirror. Assuming the mirror is placed on the ground facing the sky and the girl, mirror and the lamppost are in a same line, find the height of the lamp post.
Solution :
In triangles ABD and CED
<ABD = <ECD
<ADB = <EDC
By AA the above triangles are similar,
AB/ EC = BD/CD
x/1.25 = 6.6/2.5
x = 6.6(1.25)/2.5
x = 3.3 m
Hence the height of the lamp post is 3.3 m.
Question 4 :
A vertical stick of length 6 m casts a shadow 400 cm long on the ground and at the same time a tower casts a shadow 28 m long. Using similarity, find the height of the tower.
Solution :
Let us draw a rough diagram based on the given information.
AB/ED = BC/DC
400 cm = 4 m
AB/6 = 28/4
AB = (28/4)(6)
AB = 42 m
Hence the height of the tower is 42 m.
Question 5 :
Two triangles QPR and QSR, right angled at P and S respectively are drawn on the same base QR and on the same side of QR. If PR and SQ intersect at T, prove that PT × TR = ST × TQ.
Solution :
In triangles PTQ, and STR
<QPT = <RST (A)
<PTQ = <STR (Vertically opposite angles) (A)
So, the triangles PTQ and STR are similar.
PQ/SR = PT/TS = QT/RT
PT/TS = QT/RT
PT x RT = QT x TS
Hence proved.
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