Practice Questions on Similar Triangles :
Here we are going to see some practice problems based on the concept similar triangles.
Question 1 :
Check whether the which triangles are similar and find the value of x.
Let us consider the triangles, AED and ACB
If two triangles are similar, then the ratio of its corresponding sides will be equal.
AE/AC = AD/AB
2/(7/2) = 3/5
4/7 = 3/5
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 2 :
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.
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 3 :
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.
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 4 :
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.
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
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