Q5 of 70 Page 120

In ΔABC and ΔPQR, ABC ↔ QPR is a similarity. The perimeter of AABC is 15 and perimeter of ΔPQR is 27. If BC = 7 and QR = 9, find PR and AC.

Given, In ABC and PQR, ABCQPR is a similarity

The perimeter of ABC is 15 and perimeter of PQR is 27


And BC = 7 and QR = 9


AB + BC + CA = 15


AB + 7 + CA = 15


AB + CA = 8 ...... (i)


PQ + QR + RP = 27


PQ + 9 + RP = 27


PQ + RP = 18 ...... (ii)


Due to similarity



Hence,




PR =


PR = 14.4


Hence,



AC =


AC = 5


More from this chapter

All 70 →
3

The correspondence ABC ↔ PQR between Δ ABC and Δ PQR is a similarity. AB : PQ = 4 : 5. If AC = 6, then find PR. If QR = 15, then find BC.

 4

Δ PQR ~ ΔDEF for the correspondence PQR ↔ EDF. If PQ + QR = 15, DE + DF = 10 and PR = 6, find EF.

 6

In ΔXYZ ~ ΔDEF consider the correspondence XYZ ↔ EDF.

If find and .

 7

Using the definition of similarity prove that all the isosceles right angled triangles are similar.