How to prove `ratio of corresponding sides of similar triangles is equal to ratio of their perimeters` ?

Asked by Sanjay | 15th Aug, 2014, 10:39: AM

Expert Answer:

Consider two triangles ABC and PQR.
S i n c e space triangle A B C space a n d space triangle P Q R space a r e space s i m i l a r space t r i a n g l e s comma space w e space h a v e fraction numerator A B over denominator P Q end fraction equals fraction numerator B C over denominator Q R end fraction equals fraction numerator A C over denominator P R end fraction equals k T h u s comma space space w e space h a v e comma fraction numerator A B over denominator P Q end fraction equals fraction numerator B C over denominator Q R end fraction equals fraction numerator A C over denominator P R end fraction equals fraction numerator k P Q plus k Q R plus k P R over denominator P Q plus Q R plus P R end fraction equals fraction numerator k open parentheses P Q plus Q R plus P R close parentheses over denominator P Q plus Q R plus P R end fraction equals fraction numerator P e r i m e t e r space o f space triangle A B C over denominator P e r i m e t e r space o f space triangle P Q R end fraction

Answered by Vimala Ramamurthy | 16th Aug, 2014, 02:56: PM