Q4 of 65 Page 138

In Fig. 6.36, and ∠1 = ∠2. Show that Δ PQS ~ Δ TQR.

Given: In ΔPQR,

∠PQR = ∠PRQ


PQ = PR (i)


Given,



Using (i),


(ii)


In triangle PQS and TQR, we get


[Using (i)]


∠Q = ∠Q


Therefore,


Triangle PQS TQR (SAS similarity rule)


More from this chapter

All 65 →
2

In Fig. 6.35, ΔODC ~ ΔOBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB


 3

Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that

5

S and T are points on sides PR and QR of
Δ PQR such that ∠P = ∠RTS. Show that
Δ RPQ ~ Δ RTS

 6

In Fig. 6.37, if Δ ABE ≅Δ ACD, show that
Δ ADE ~ Δ ABC.