**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.

(i)

**Solution : **

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

(ii)

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.

**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 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.

**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 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.

**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.

After having gone through the stuff given above, we hope that the students would have understood, "Practice Questions on Similar Triangles".

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